Física A P French Newtonian mechanics

Newtonian mechanics A. R French Introductory ll'KHHS ooooo 531.01 FRE senes OOOOOOO OOOOOOOOOOO OOOOOOOOOOO OOOOOOOOOOO OOOOOOOOOOO OOOOOOOOOOO OOOOOOOOO OOOOOOO I his tcxl prescnts the import m bot h cUlMlol and modern physics as relevani today, atartlng fiom ibout niotion and ending with an problcms in murinml dynamics. lu- prime empha the lopment and usc of New. i i i ii llret-yaar a cornpletc tnechanka couim and contains manj iiiggesdona further rcadh Aboul \ tbC i<>r Aulhor eived bot h his B. A. Ph. D. degrees from ( amhridge i nivenlty, where he cootinued asa Physics tepartment from 1948 undi 1955. b ic joincd the staff of the i member of the l m. i South : PhyalGS and became l»*)2 hc wenl to Profc mi I i was appointed t wo years later. has devoted ai half his timc during the past ala lopment of the \t i i . Introdw but his eapertaaca purel . h bj do dm He hai had carch assignments with extc the Cavendlah aboratoriea, the i Manhattan Project, and the U.K. Vtomk Eoergj Reseercb iblishmcnt THE LIBRARY THE HARRIS COLLEGE CORPORATION STREET, PRESTON AH liooks must be Returned to «he College Library or later iban the last date sbown below. Renewed not Re. NGV !974 -6. OCT. Re W 19J 975 n< FEB. -? ! 197L. M 197 19. . m. [991 -3. MAY 1976 2 -07. 21. I FED. 1997 j _ The M.I.T. Introductory Physics Series Special RelatiVity A.P.FRENCH Vibrations and Waves A.P.FRENCH Newtonian Mechanics A.P.FRENCH TheM.I.T. Introductory Physics Series Newtonian Mechanics A. E French _ PROFESSOR OF MATHEMATICS, THE MASSACHUSETTS INSTITUTE OF TECHNOLOGY Nelson » TIIOMAS NELSON AND SONS LTD 36 Park Street London wIy 4de po Box 18123 Nairobi Kenya THOMAS NELSON (AUSTRALIA) LTD 597 Little Collins Street Melbourne 3000 THOMAS NELSON AND SONS (CANADA) LTD 8 Curlew Drive Don Mills Ontario THOMAS NELSON (NIGERIA) LTD po Box 336 Apapa Lagos THOMAS NELSON AND SONS (SOUTH AFR1CA) (PROPRIETARY) LTD 51 Commissioner Street Johanncsburg 1 © Copyright 1971, 1965 by the Massachusetts Institute of Technology First published in Great Britain 1971 sbn 17 771074 8 (paper) L^bn 17 761075 1 (boards) PROTON per ic' 47809 —— 551 we Made and printed by William Clowes & Sons Ltd, Beccles, Colchester and London Contents Preface xi Prologue PART I 1 A 3 THE APPROACH TO NEWTONIAN DYNAMICS universe of particles 21 The particulate view Eleclrons and nucleons Atomic nuclei 25 Atoms 26 21 24 Molecules; licing cells Sand and dust Other 32 lerrestrial objecls Planels and satellifes Stars Galaxies 33 35 36 PROBLEMS 2 28 31 38 Space, time, and motion What is motion? 43 Frames ofreference 46 Coordinale systems 48 43 Combination ofvector displacements The 53 56 resolution ofvectors 59 Vector addilion and the properties ofspace Time 61 Uni is and standards oflength and time 66 Space-time graphs 63 67 Velocily 68 lnslantaneous velocily Relatlve velocily and relative motion 72 Planetary motions: Ptolemy versus Copernicus PROBLEMS 3 74 78 85 Accelerated motions 85 87 The analysis of straight-line motion 93 A comment on exlraneous roots 95 Trujectory problems in two dimensions 98 Free fail of individual aloms 102 Other features of motion in free fail Acceleralion 105 Uniform circular motion and acceleralion 108 PROBLEMS Velocily 4 in polar coordinales 106 115 Forces and equilibrium 116 Forces in static equilibrium 118 Units offorce Ekuilibrium conditions; forces as vectors Action and reaction in the contact of objects 124 Rotational ekuilibrium; loraue Forces without contact; weight 119 123 128 130 and slrings problems 132 Pulleys 5 The various 139 forces of nature 139 The hasic lypes offorces 140 Gravitatkmal forces Electric and magnetic forces Nuclear forces 145 147 Forces between neutral atoms Con laci forces 1 50 Frictional contact forces problems VI 152 154 Concluding remarks 154 148 6 Force, inertia, and motion The 161 161 of inertia mass: Newlon's law 164 Some comments on Newlon's law 167 principle Force and Scales inertial ofmass and force 170 The effect of a continuing force The incariance of Newton's law; 173 Incar iance with specific force laws Newlon's law and time reuersal Concluding remarks 180 problems PART 7 II 173 relatioily 176 178 181 CLASSICAL MECHANICS AT WORK Using Newton's law Some I 188 inlroduclory examples Motion Mo 187 in ion in two dimensions a 194 198 circle Curoilinear motion with changing speed 200 Circular pai lis of charged particles in uniform magnelic fields Charged particle in a magnelic fielcl 202 205 Mass spectrographs 206 The fracture ofrapidly rotating objects Motion against resistive forces 210 Deiailed analysis ofresisted motion Motion gocerned by viscosily 218 Growlh and decay ofresisted motion 208 213 221 Air resistance and "independence of molions" Simple harmonic motion 226 More about simple harmonic motion problems 234 8 225 231 Universal gravitation 245 The discovery of universal gravitation The orbit s of the planets 246 Planetary periods 249 Kepler's third law 252 The moon and the apple 256 Finding the dislance to the The gracitational Other satellites moon attraciion of the eartli 259 of a large sphere 265 The ualue of G, and the mass of the earth Local uariations of g 270 The mass of the sun 274 Vll 245 261 268 275 Finding the dislance to the sun Mass and weight Weightlessness 279 285 286 Learning about other planets 288 The moons ofJupiter 291 The discovery ofNeptune 295 Gravilation outside the solar system Einstein's theory 299 of gravilation PROBLEMS 301 \ 9 307 Collisions and conservation laws 308 The laws of impact The conservation of linear momentum 310 Momentum as a vector guantity 313 Action, reaction, and impulse 309 Extending the principie of momentum conservation 321 The force exerted hy a stream ofparticles 324 Reaction from a fluid jet 327 Rocket propulsion Collisions and frames ofreference 331 333 335 Kinetic energy in collisions The zero-momentum frame Collision processes in two dimensions 339 342 Elastic nuclear collisions Inelastic and explosive processes What a collisionl is 346 351 lnteracting particles subject t o external forces The pressure of a gas 356 The neutrino 357 PROBLtMS 10 Energy conservation vibrational motions in 318 352 354 dynamics; 367 367 368 of motion Work, energy, and power Introduction lntegrals 373 Gravitational potential energy More about 376 one-dimensional situations 379 The energy methodfor one-dimensional motions 384 Some examples of the energy method The harmonic oscillator by the energy method 395 Small oscillations in general The linear oscillator as a two-body problem Collision processes inuolving energy storage The diatomic molecule 411 PROBLEMS Vlll 405 381 393 397 400 1 1 Conservative forces and motion in space Extending the concepl of conservative forces 425 Acceleration of two connected masses 426 Object moving in a verlical circle An experimenl by Galileo Mass on a puraboiic track 423 423 429 431 434 The simple pendulum 437 The pendulum cis a harmonic oscillator 440 The pendulum with larger amplitudo Universal gravitation: a conservative central force 446 A gravitating spherical shell 450 A gravitating sphere 453 Escape velocities More Fields aboul the cri teria for conservative forces and the gradien! of potential energy Motion in conservative fields The effect of dissipative forces 473 Gauss's law Applications of Gauss's theorem PROBLEMS 12 III 457 461 Equipotential surfaces PART 442 463 466 470 476 478 SOME SPECIAL TOPICS Inertial forces 493 and non-inertial frames Motion observed from unaccelerated frames Motion observed front an accelerated frame 497 Accelerated frames and inertial forces Accelerometers 501 Accelerating frames and gravity Centrifugal force 494 495 504 507 511 514 518 Dynamics on a merry-go-round 519 General ecjuation of motion in a rotating frame 524 The earth as a rotating reference frame The tides 531 535 Tidal heights; effect of the sun 538 The search for a fundamental inertial frame 542 Speculations on the origin of inertia Centrifuges Coriolis forces PROBLEMS 13 Motion under 546 555 central forces Basic features of the problem The law ofequal areas 557 555 The conservation of angular momentum IX 560 Energy conseruation Vse of the 563 molions in ceniral force 565 effectiue potential-energy citrues 568 Bounded orbtts Unbcnmded orbits 569 572 Circular orbtts in an inuerse-square force field 574 Small perturbation of a circular orbit 577 The elliptic orbits ofthe planels 583 Deducing the inuerse-scpiare law from the eltipse 585 Elliptic orbits: analylical treatment 589 Energy in an elliptic orbit 591 Molion near the earth's surface 592 Interplcmetciry transfer orbits 595 Calculaling an orbit from inilial condilions 596 A family of relai ed orbits 598 Central force motion as a two-body problem Deducing the orbit from the force law 604 Rutherford scattering 609 Cross sections for scattering Alpha-particle scattering (Geiger Magazine excerpts) 615 An historical note problems 617 14 600 and Marsden, Philosophical 612 Extended systems and rotational dynamics 627 Momentum and kinetic energy of a many-particle system 632 Angtilar momentum 636 Angular momentum as a fundamental quanlity 639 Conseruation of angular momentum 643 Moments of inertia of extended objects 647 Two theorems concerning moments of inertia 651 Kinetic energy ofrotaling objects Angular momentum conseruation and kinetic energy and rigid pendulums Motion under combined forces and tonjues Impu/sioe forces and torgues 668 Bachground to gyroscopic motion 671 Torsional oscillations Gyroscope More in steady precession about precessional motion Gyroscopes as gyroscopes Gyroscopic motion in teriris The precession of the equinoxes problems 700 Appendix 709 Bibliography lndex 686 of F = »ia 691 Niitation Answers 677 680 683 in nauigation Atoms and nuclei to 713 problems 733 659 664 723 694 688 654 628 Preface the work of the Education Research Center at M.I.T. (formerly the Science Teaching Center) is concerned with curriculum im- and aids provement, with the process of instruction with the learning process established by M.I.T. Friedman as its undergraduate in 1960, Director. and primarily with respect to students itself, at the college or university thereto, level. The Center was with the late Professor Francis L. Since 1961 the Center has been supported mainly by the National Science Foundation; generous support has also been received from the Kettering Foundation, Companies Foundation, the Victoria Foundation, the Grant Foundation, and the Bing Foundation. T. The M.I.T. Introductory Physics Series, a direct outgrowth the Shell W. of the Center's work, is designed to be a set of short books that, taken collectively, span the main areas of basic physics. The series seeks to emphasize the interaction of experiment and intuition in generating physical theories. The books in the series are intended to provide a variety of possible bases for introductory courses, ranging from those which physics to those which embody chiefly emphasize a considerable classical amount of atomic and quantum physics. The various volumes are intended to be compatible in level and style of treatment but are not conceived as a tightly knit package; on the contrary, each to book is designed be reasonably self-contained and usable as an individual component in many different course structures. XI The text material in the present volume is designed to be a more or less self-contained introduction to Newtonian mechanics, such that a student with little or no grounding in the subject can, A considerable proficicncy. the book Approach suggested by is its division into three parts. Newtonian Dynamics, to poses. First, it is Part The I, intended to serve two pur- does discuss the basic concepts of kinematics and dynamics, more or less from study of mechanics squarely phenomena and of is brought gradually to a level of rough guide to the possible use of at the beginning, be by beginning scratch. Second, in the it seeks to place the context of the world of physical This neccssarily imperfect physical theories. a conscious reaction, on the author's part, against the preserta- tion of mechanics as "applied mathematics," with the divorce- ment from and the misleading impression of rigor that reality has engendered in generations of brought up in the British this students (especially, alas, those educational system). The student who Newton's laws will find little of an analytical or quantitative sort to learn from Part I, but he may still derive some value and interest from reading through it some already has for its expertise in using broader implications. Part Classical II, heart of the book. relegate Part emphasis is I Mechanics at Work, Some is undoubtedly the instructors will wish to begin here, to the status of background reading. The on Newton's second law applied to individual and initial objccts. Later, the emphasis shifts to systems of two or more particlcs, and to the conservation laws for momentum and energy. A fairly lengthy chapter is place in the whole gravitation and its devoted to the subject that deserves pride of Newtonian scheme— the theory of universal successes, which can still be appreciated as a pinnacle in man's attempts to discover order in the vast universe in which he finds himself. Part III, Some Special Topics, concerns itself with the problems of noninertial frames, central-force motions, and rotational dynamics. Most of this material, except perhaps the fundamental motion and angular momentum, could be regarded as optional if this book is used as the basis of a genuinely introduetory presentation of mechanics. Undoubtedly the book as a whole contains more material than could in its entirety be features of rotational covered in a one-term course; one could, however, consider using Parts and XII I III and II as a manageablc paekage for beginners, and Parts as a text for students having some prior preparation. II One of the great satisfactions of classical mechanics the vast range and variety of physical systems to which ples can be applied. make The attempt has been made its in this lies in princi- book to such applications and, as in other books "document" the presentation with appropriate from original sources. Enriched in this way by its own explicit reference to in this series, to citations history, classical mechanics has an excitement that author's view, surpassed is not, in this by any of the more recent fields of physical thory. This book, like the others in the series, owes much to the and suggestions of many people, both students and instructors. A special acknowledgment in connection with the present volume is due to Prof. A. M. Hudson, of Occidenthoughts, criticisms, Los Angeles, who worked with the present author in the preparation of the preliminary text from which, five years Grateful thanks are also due to later, this final version evolved. tal College, Eva M. Hakala and William H. Ingham in preparing the manuscript for their invaluable help for publication. A. P. Cambridge, Massachusetts July 1970 XIII FRENCH Newtonian mechanics In the Beginning was Mechanics. max von laue, / offer this work as the History ofPhysics (1950) mathematical principles of philosophy, for the whole burden of philosophy seems from the nature, phenomena ofmotions and thenfrom — to consist in this to investigate theforces of these forces to demonstrate the other phenomena. newton, Preface to the Principia (1686) Prologue one of the most prominent features of the universe is motion. Galaxies have motions with respect to other galaxies, all stars have motions, the planets have distinctive motions against the background of the stars, the events that capture our attention most quickly in everyday life are those involving motion, and even the apparently inert book that you are now reading is made up of atoms in rapid motion about their equilibrium positions. "Give me matter and motion," said the seventeenth-century French philosopher Rene Descartes, "and I will construct the universe." There can be we must no doubt that motion learn to deal with at all levels we if the world around us. Isaac is a phenomenon are to understand . Newton developed a precise and powerful theory j regarding motion, according to which the changes of_motion of any object are the result of Jorces acting created the subject with "wmcrTthis is called classical or book on itfln so doing he coBeerned and which is Newtonian mechanics. the history of science, because it I It was a landmark in replacea a merely descriptive account of phenomena with a rational and marvelously successful scheme of cause and effect. Indeed, the strict causal nature of Newtonian mechanics had an impressive influence in the development of Western thought and civilization generally, provoking fundamental questions about the interrelationships of science, philosophy, and religion, with repercussions in social ideas and other areas of human endeavor. Classical mechanics character. For it starts is a subject with a fascinating dual out from the kinds of everyday experiences a : that are as old as mankind, yet it some brings us face to face with of the most profound questions about the universe in which we find ourselves. Is it not remarkable that the fiight of a thrown pebble, or the fail of an apple, should contain the clue to the mechanics of the heavens and should ultimately involve some of the most basic questions that we are able to formulate about the nature of space and time? though it Sometimes mechanics is presented as consisted merely of the routine application of evident or revealed truths. self- Nothing could be further from the a superb example of a physical theory, slowly evolved and refined through the continuing interplay between observation case; it is and hypothesis. The is richness of our first-hand acquaintance with mechanics impressive, hand we and through the partnership of mind and eye and solve, direct action, innumerable dynamical problems by without benefit of mathematical analysis. Like Moliere's famous character, M. Jourdain, who learned that he had been speaking prose all his life without realizing it, every human being an is expert in the consequences of the laws of mechanics, whether or not he has ever seen these laws written down. The skilled sportsman or athlete has an almost incredible degree of judgment and control of the amount and achieve a desired result. It direction of muscular effort needed to has been estimated, for example, that championship would have changed hands in 1962 if one crucial swing at the ball had been a mere But experiencing and controlling the motions millimeter lower. of objects in this very personal sense is a far cry from analyzing World the Series baseball ' terms of physical laws and equations. It is the task of classical mechanics to discover and formulate the essential principles, so that they can be applied to any situation, par- them in ticularly to inanimate objects interacting with one another. intimate familiarity with our consequences, although it own muscular actions and Our their represents a kind of understanding (and an important kind, too), does not help us much here. The greatest triumph of classical mechanics was Newton's own success in analyzing the workings of the solar system — feat immortalized in the famous couplet of his contemporary and admirer, the poet Alexander Pope P. 4 Kirkpatrick, Prologue Am. J. Phys., 31, 606 (1963). . Nature and Nature's Laws Iay hid God Men said "Let Newton and be," night in was all light. 1 had observed the motions of the heavenly bodies since time They had noticed various regularities and had immemorial. learned to predict such things as conjunctions of the planets and eclipses of the sun and moon. Then, in the sixteenth century, the Danish astronomer Tycho Brahe amassed meticulous records, of unprecedented accuracy, of the planetary motions. His assistant, Johannes Kepler, after wrestling with formation for years, found that all this enormous body of in- the observations could be summarized as follows: 1 2. The The planets move in ellipses having the sun at one focus. joining the sun to a given planet sweeps out line equal areas in equal times. 3. mean The square of distance a planet's year, divided by the cube of from the sun, the is same its for all planets. This represented a magnificent advance in man's knowledge of the mechanics of the heavens, but Why? was rather than a theory. it was still the question that an answer. Then came Newton, with his a description still looked for concept of force as the cause of changes of motion, and with his postulate of a particular — the inverse-square law of gravitation. law of force how he demonstrated Using these Kepler's laws were just one consequence of a scheme of things that also included the falling apple and other terrestrial tails of I f motions. this (Later in this universal gravitation had planetary periods and distances, did theory. from which tions of a theory phenomena, or familiar shall go into the de- it done no more than to would still is, it it it was deduced. Investigating the predic- fit unsuspected an already new framework. In either searching tests, by which it must into the case the theory is 'To which there the almost equally famous, although facetious, riposte: is It subjected to did not last; the Devil, howling "Ho, Let Einstein be!" restored the status quo. (Sir 5 had could be applied to situations besides may involve looking for hitherto it may involve recognizing that phenomenon must relate have been a splen- But, like any other good theory in physics, predictive value; that the ones book we great achievement of Newton's.) Prologue John Squire) stand or With Newton's theory of fail. almost entirely tests resided but what a list! gravitation, the initial in the analysis of known effects— Here are some of the phenomena for which Newton proceeded to give quantitative explanations: 1. the earth and Jupiter because of their The bulging of rotation. 2. The variation of the acceleration of gravity with latitude over the earth's surface. 3. The generation of the tides by the combined action of sun and moon. 4. 5. The paths of the comets through the solar system. The slow steady change in direction of the earth's axis gravitational torques from the sun and of rotation produced by moon. years, (A complete cycle of this variation takes about 25,000 and the so-called "precession of the equinoxes" is a manifestation of it.) This marvelous illumination represented the last it in of the workings of nature part of Newton's program, as he described our opening quotation ". . . and then from these forces to demonstrate the other phenomena." ceals not only the This modest phrase conimmensity of the achievement but also the magnitude of the role played by mathematics ment. Newton had, in in this the theory of universal develop- gravitation, created what would be called today a mathematical model of the solar system. And having once made the model, he followed out a host of its The working out was purely step— the test of the conclusions— other implications. mathematical, but the final involved a return to the world of physical experience, in the detailed checking of his predictions against the quantitative data of astronomy. Although Newton's mechanical picture of the universe was amply confirmed of its in his own greatest triumphs. was the use of his laws to time, he did not live to see some Perhaps the most impressive of these identify previously unrecognized mem- By a painstaking and lengthy analysis known planets, it was inferred that disof the of the motions other planets must be at work. Thus it turbing influences due to bers of the solar system. was that Neptune was discovered in 1846, and Pluto in 1930. In each case it was a matter of deducing where a telescope should be pointed to reveal a new planet, identifiable through its changing position with respect to the general background of the stars. 6 Prologuc - What more the theory striking . and convincing evidence could Probably everyone who reads this book has acquaintance with classical mechanics and with And mathematically precise statements. to realize that, as with was not there be that works? this some may make any other physical theory, prior expression in its its it hard development a matter of mathematical logic applied undisWas Newton inexorably driven to the inverse-square Iaw? By no means. It was the result of guesswork, intuition, and imagination. In Newton's own words: just criminatingly to a mass of data. "I began to think of gravity extending to the orbit of the Moon, fromKepler's Ruleof the periodictimesofthePlanets and I deduced that the forces which keep the Planets in their orbits . . . . . must be reciprocally as the squares of their distances from the centers about which they revolve; and thereby compared the force requisite to keep the Moon in her orbit with the force of and found them to answer leap of this sort although seldom gravity at the surface of the Earth, pretty nearly." An intellectual as great as Newton's or model. It is — is — involved in the creation of any theory a process of induction, and facts immediately at hand. Some facts it goes beyond the may even be temporarily brushed aside or ignored in the interests of pursuing the idea, for a partially correct theory is often better than at all. And at all stages there is main no theory a constant interplay between experiment and theory, in the process of which fresh observations are continually suggesting themselves and modification of the theory is an ever-present possibility. The following diagram, the relevance of which goes beyond the realm of classical mechanics, suggests this pattern of man's investigation of matter and motion. Laws of Motion — INDUCTION 1 Laws of Force Observations and Experiments •DEDUCTION- J Mathematical Models Predictions The enormous success of classical mechanics I made it seem, more was needed to account for the whole world of physical phenomena. This belief reached a pinnacle toward the end of the nineteenth century, when some at one stage, that nothing 7 Prologue optimistic physicists felt that physics was, in principle, complete. They could hardly have chosen a more unfortunate time at which to form such a conclusion, for within the next few decades physics underwent its greatest upheaval since Newton. The discovery of radioactivity, of the electron and the nucleus, and the subtleties ideas. of electromagnetism, called for fundamentally new that Newtonian mechanics, like Thus we know today every physical theory, has its The fundamental limitations. analysis of motions at extremely high speeds requires the use of modified descriptions of space and time, as spelled out by Albert In the analysis of phe- Einstein's special theory of relativity. nomena on the atomic or subatomic scale, the still modifications described by quantum theory are more drastic required. And Newton's particular version of gravitational theory, success, has had to admit modifications embodied in Einstein's for all its general theory of relativity. in But an enormous range and this does not alter the fact that, variety of situations, Newtonian means to analyze and predict the from electrons to galaxies. Its range mechanics provides us with the motions of physical objects, of validity, and its limits, are indicated very qualitatively in the figure below. In developing the subject of classical mechanics in this book, we shall try to indicate how the horizons of its physical world, and the horizons of one's Mechanics, as we gradually broadened. not at all application to the own view, can be shall try to present a cut-and-dried subject that would justify its Cosmological Physics 10-'° 10 !0 m Galaxy m Atom Size 8 Proloeuc it, is description game as "applied mathematics," in which the rules of the given at the outset and in which one's only concern We ing the rules to a variety of situations. which ferent approach, in sumptions that cannot be rigorously Newton essence of doing physics. beginning of Book III with apply- wish to offer a dif- one can be conscious of at every stage partial or limited data working with is are and of making use of But justifled. this is himself said as much. as- the At the of the Principia he propounds four "Rules of Reasoning in Philosophy," of which the last runs as follows: "In experimental philosophy we are to look upon propositions inferred by general induction from phenomena as accurately or very nearly true, notwithstanding any contrary hypotheses that may be imagined, by which they may till such time as other phenomena occur, made more either be The person who exceptions." waits for complete information on the way to dooming himself never to to construct a useful theory. accurate, or liable to is an experiment or finish Lest this should be taken, however, as an encouragement to slipshod or superficial thinking, we shall end this introduction with a little fable due to George Polya. 1 He writes as a mathematician, but the moral for physicists (and others) is clear. The Logician, the Mathematician, and the Physicist, "Look at this that the what he "A by than 100 and less calls induction, that all physicist divisible mathematician," said the logician. 99 numbers are first the Engineer numbers are less believes," said the 1, 2, 3, 4, 5, and 6. "He observes infers, hence, by than a hundred." mathematician, "that 60 He examines a few more is cases, taken at random (as he says). Since 60 is also divisible by these, he considers the experimental evidence such as 10, 20, sufficient." and 30, "Yes, but look at the engineers," said the physicist. all odd numbers are prime numbers. At any rate, 1 can be considered as a prime number, he argued. Then there come 3, 5, and 7, all indubitably primes. Then there comes 9; an awkward case; it does not seem to be a prime num- "An engineer suspected that 'This cautionary tale is to be found in a book entitled Induction and Analogy Mathematics, Princeton University Press, Princeton, N.J., 1954. This volume and its companion, Patterns of Plausible Inference, make delightful in reading for any scientist. 9 Prologuc Yet ber. said, 'I and 1 1 'Corning back to 13 are certainly primes. conclude that 9 must be an experimental error.' 9,' he " But having done his teasing, Polya adds these remarks. only too obvious that induction can lead to error. Yet It is it is remarkable that induction sometimes leads to truth, since the chances of error seem so overwhelming. Should we begin with the study of the obvious cases in which induction fails, or with the study of those remarkable cases in which induction succeeds? The study of precious stones is understandably more attractive than that of ordinary pebbles and, moreover, was much more it the precious stones than the pebbles that led the mineralogists to the wonderful science of crystallography. With that encouragement, we shall, in Chapter approach to the study of classical mechanics, which most end perfect and polished gems this Prologue, l, is begin our one of the in the physicist's treasury. We however, with some preparatory exercises. EXERCISES-HORS D'OEUVRES meaning of the phrase "hors d'oeuvre" is "outside The exercises below correspond exactly to that the work." definition, although it is hoped that they will also whet the The literal appetite as hors d'oeuvres should. They deal mostly with order- power of 10) an important role in a physicist's approach to problems but seldom get emphasized or systematically presented in textbooks. For example, of-magnitude estimates (i.e., estimates to the nearest and judicious approximations— things that play everybody learns the binomial theorem, but how many students think of it as a useful tool for obtaining a quite good value for the hypotenuse of a right triangle, by the approximation ^ + b^'â{\+^ where we assume b < b = wrong by only about 6 percent 1.5 instead it takes practice and some conscious a, the result is of 1.414 . . . .) a? (Even in the worst possible case, with — Moreover, develop the habit of assessing, quite crudely, the magnitudes of quantities and the relative importance of various possible effects in a physical system. For example, in dealing with effort to objects 10 moving through Prologue liquids, can one quickly decide whether 9 viscosity or turbulence is going to be the chief source of resistance for an object of given speed of the effects of changes and the properties of systems. well-known essay by Size," which J. linear An dimensions? awareness of scale can give valuable insights into [A beautiful example of this is the "On Being the Right B. S. Haldane, reprinted in The World of Mathematics, Vol. is Newman, New II Simon and Schuster, methods and ways of thought one can deepen one's appreciation of physical phenomena and can improve one's feeling for what the world is like and how it behaves. (J. R. By the use of such It ed.), how much one surprising is relatively small stock York, 1956.] can do with the help of a of primary information — which might include such items as the following: Physical Magnitudes Gravitational acceleration (g) Densities of solids and liquids 2 10 m/sec' kg/m 3 3 4 10 -10 kg/m 3 Density of air at sea level 1 Length of day 10 Length of year 3.16 X Earth's radius 6400 km 5 (approx.) sec (approx.) 10 7 sec « I0 75 Angle subtended by finger thickness 1° (approx.) at arm's length mm (approx.) Thickness of paper 0.1 Mass of 0.5 g (approx.) a paperclip Highest mountains, deepest oceans 10 km Earth-moon separation 3.8 Earth-sun separation 1.5 X X Atmospheric pressure Equivalent to weight of 1 (approx.) 10 10 5 8 km km kg/cm 2 or a 10-m column of water Avogadro's number 6.0 Atomic masses 1.6 Linear dimensions of atoms 10 Molecules/cm Atoms/cm 3 3 i n gas at in solids Elementary charge (e) 4 X _,0 X 2.7 23 10 1.6 Electron mass Speed of 3 light light EKcrcises- hors d'oeuvres 6 X X 10 23 10- 27 kgto _25 kg 10 m (approx.) 10 13 (approx.) X 10~ 30 Wavelength of 11 STP X X lO -1 C kg (approx.) 8 10 m/sec -7 10 m (approx.) sec Malhematical Magnitudes w2 e log, o 3 1 = (radians) « rad 0.16 X m arc 1 ~ sr 0.08 X log,,, 4 log, o e full log, « logc 10 0.48 length/radius. full circle Solid angle (steradians) m 0.60 ~ 0.43 T«0.50 10 2.7 2«0.30 log 10 Angle m « = « 2.3 = circle Full 2jrrad. 57°. area/(radius) 2 . = Full sphere 4*- sr. sphere. Approximations Binomial theorem: Forx« (1 1, e.g., (1 (1 For b « (a a, + m « 2 -x)" a + + b)" x) x) - n 1 + «* 1 + 3* 3 fl*(l 1 - **« ~ + jjjY (1 an +x)- 1/2 (l+n?\ Othcr expansions: For « 6 1 sin rad, fl e w 3 > 6 cos d e « 2 1 1 2 For a: « log« (1 1, log, No + (l ~ x +Jc)« 0.43x x) answers are given to the problems that follow. For most of them, you yourself will be the best judge. You may want to turn to an encyclopedia or other reference book some of your assumptions or conclusions. If you are not prepared at this point to tackle return to / What them is all, don't worry; you can always the order of magnitude of the its number of times that was formed? the axis since the solar system During the average lifetime of a human being, how many heart- beats are there ? 3 them later. earth has rotated on 2 to check How many Make reasoned breaths ? estimates of (a) the total number of ancestors ytou would have (ignoring inbreeding) since the beginning of the human race, and (b) the number of hairs on your head. 4 The (a) 12 present world population How many Prologue (human) is about 3 X 10°. squarc kilometers of land are there per person? — ; How many (b) If feet long is the side of a square of that area? one assumes that the population has been doubling every 50 years throughout the existence of the human race, when did Adam start it all ? If the doubling every 50 years were to continue, and Eve how long would over it be before people were standing shoulder to shoulder land area of the world? all the 5 Estimate the order of magnitude of the mass of (a) a speck of dust (b) a grain of sak (or sugar, or sand); (c) a water corresponding to (e) the (0 a small hill, 500 ft 1 in. high; and (g) mouse; Mount an elephant; (d) of rainfall over square mile; 1 Everest. Estimate the order of magnilude of the number of atoms in (a) a 6 pin's head, (b) a human and being, (c) the earth's atmosphere, (d) the whole earth. now 7 Estimate the fraction of the total mass of the earth that the form of 8 Estimate (a) the total volume of ocean water on the earth, and mass of sah (b) the total 9 in all the universe. in oceans. estimated that there are about 10 80 protons It is is living things. If all these (known) the in were lumped into a sphere so that they were what would the radius of the sphere be? Ignore the spherical objects are packed and takc the radius of a proton to be about 10~ 15 m. just touching, spaces left when 10 The sun is losing mass (in the form of radiant energy) at the rate of about 4 million tons per second. What fraction of its mass has it lost during the lifetime of the solar system 11 Estimate the time in minutcs that of about 1000 people to use up ing were sealed. 10% it ? would take for a theatre of the available oxygen The average adult absorbs about one if audience the build- sixth of the oxygen that he or she inhales at each breath. 2 falls on the earth at the rate of about 2 cal/cm /min. repremegawatts or horsepower, amount of power, in Estimate the sented by the solar energy falling on an area of 100 square miles 12 Solar energy about the area of a good-sized city. How would power requirementsofsuch a city? hp = 746 W.) total 1 (1 cal = this comparc with the = U/sec; 1 4.2 J; W 13 Starting from an estimate of the total mileage that an automobile tire will give before wearing out, estimate what thickness of rubber is one revolution of the wheel. Consider the possible physical significance of the result. (With acknowledgment to E. M. Rogers, Physics for the lnquiring Mind, Princeton University Press, worn off during Princeton, N. J., 1%0.) 14 13 An ine.vpensive wristwatch (a) What Exercises is its is found to fractional dcviation — hors d'oei v re s lose 2 min/day. from the correct rate? By how much could the length of a ruler (nominally 1 ft long) in. and still be fractionally as accurate as the (b) differ from exactly 12 watch ? 15 The astronomer Tycho Brahe made observations on the angular posilions of stars at its center and planets by using a quadrant, with one peephole of curvature and another peephole mounted on the arc. One such quadrant had a radius of about 2 ments could usually be trusted to m, and Tycho's measure- minute of arc 1 What diameter (^g°). of peepholes would havc been needed for him to attain this accuracy ? 16 Jupiter has a mass about 300 times that of the earth, but density (a) is only about one What fifth its mean that of the earth. radius would radius would a planet of a planet of Jupiter's mass and earth's density have? (b) What earth's mass and Jupiter's density have ? 17 Identical spheres of material are tightly packed in a given volume of space. (a) Consider why one does not need to know the radius of the spheres, but only the density of the material, in order to calculate the total mass contained in the volume, provided that the linear dimensions of the volume are large compared to the radius of the individual spheres. (b) Consider the possibility of packing may be chosen and used. Show that the total surface area more material if two sizes of spheres (c) of the spheres of part (a) does depend on the radius of the spheres (an important consideration in the design of such things as filters, which absorb in proportion to the total exposed surface area within a given volume). 18 Calculate the ratio of surface area to volume for radius r, (b) a cube of edge a, and (c) (a) a sphere of a right circular cylinder of d. For a given value of the volume, which of these shapes has the greatest surface area ? The least surface diameter and height both equal to area? 19 How many at the sun? seconds of arc does the diameter of the earth subtend a football be At what distance from an observer should placed to subtend an equal angle? 20 From the time the lower limb of the sun touches the horizon it sun to disappear beneath the horizon. (a) Approximately what angle (exprcssed both in degrees and in radians) does the diameter of the sun subtend at the earth ? (b) At what distance from your eye does a coin of about ^-in. takes approximately 2 diameter (c) What 14 Prologue for the a dime or a nickel) just block out the disk of the sun? solid angle (in steradians) does the sun subtend at the (e.g., earth? min 21 How many inches per mile does a terrestrial great circle (e.g., a meridian of longitude) deviate from a straight line ? 22 A crude measure of the roughness of a nearly spherical surface could be defined by Ar/r, where Ar the height or depth of local is irregularities. Estimate this ratio for an orange, a ping-pong ball, and the earth. 23 What is the probability (expressed as sized meteorite falling to earth would 1 chance in 10") that a good- strike a man-made structure? A human ? 24 Two want to measure the speed of sound by the following positioned some distance away from the students One of them, procedure. The second student starts a stopwatch and stops it when he hears the bang. The speed roughly 300 m/scc, and the students must admit the other, sets off a firecracker. when he sees the flash of sound in air possibility is of an error (of undetermined sign) of perhaps 0.3 sec in the elapsed time recorded. If they wish to keep the error in the measured the minimum distance over which sides of length 5 m and m adjoining the right speed of sound to within 5%, what is they can perform the experiment? 25 A right triangle has 1 Calculate the length of the hypotenuse from the binomial ex- angle. pansion to two terms only, and estimate the fractional error approximate 26 The radius of a sphere What is in this result. is measured with an uncertainty of 1%. volume? the percentage uncertainty in the 27 Construct a piece of semilogarithmic graph paper by using the graduations on your slide rulc to mark off the X function y = 2 ruler to the abscissa. mark On off the ordinates this piece and a normal of paper draw a graph of . 28 The subjective sensations of loudness or brightness have bcen judged to be approximately proportional to the logarithm of the intensity, so that equal mulliples of intensity are associated with equal (For example, arithmetic increases in sensation. tional to 2, 4, 8, intensities and 16 would correspond to equal increases tion.) In acoustics, this has led to the measurement of sound proporin sensa- intensities Taking as a reference value the intensity /o of the faintest audible sound, the decibel level of a sound of intensity / is defined by in decibels. the equation dB = lOlogio (a) what An © intolerable noise level is intensity /o? 15 represented by about 120 dB. factor does the intensity of such a Exercises —hors d'oeuvres By sound exceed the threshold (b) A similar logarithmic scale stars (as seen brightness of is used to describe the relative from the earth) in terms of magnitudes. "one magnitude" have a ratio of apparent brightness Stars differing by Thus equal to about 2.5. a "first-magnitude" (very bright) star times brighter than a second-magnitude star, (2.5) than a third-magnitude star, and so on. largely to differences of distance.) 200-in. The 2 is 2.5 times brighter (These differences are due faintest stars detectable with the Palomar telescope are of about the twenty-fourth magnitude. us from such a star less By what factor is the amount of light reaching than we receive from a first-magnitude star ? 29 The universe appears to be undergoing a general expansion in which the galaxies are receding from us at speeds proportional to their disThis tances. is described by Hubble's law, v = «r, where the con- becoming equal to the speed of light, c (= 3 X 10 8 m/sec), at r « 10 26 m. This would imply that the mean mass per unit volume in the universe is decreasing with time. (a) Suppose that the universe is represented by a sphere of volume stant a corresponds V at any time is instant. to Show v that the fractional increase of volume per unit given by 1 dV V dt = 3a (b) Calculate the fractional decrease of mean density per second and per century. 30 The table lists the mean orbit radii of successivc planets expressed in terms of the earth's orbit radius. The planets are numbered Planet r/rg 2 Mercury Venus 0.72 1.00 1 (a) n Make abscissa. is 3 Earth Mars 1.52 5 Jupiter 5.20 6 Saturn 7 Uranus 9.54 19.2 is ordinate and the (Or, alternatively, plot values of logarithmic paper.) On this samc graph, r /re against 7, 8). The points representing the seven ably well fitted by a straight 16 (i.e., at n = 6, planets can thcn be reason- line. is taken to represent the asteroid between the orbits of Mars and Jupiter, what value of r/rE would (b) If belt = number n on semi- replot the points for Jupiter, Saturn, and Uranus at values of n increased by unity and order («): 0.39 4 a graph in which \og(r/rE) in n Prologue 5 in the revised plot your graph imply for this ? Compare with the actual mean radius of the asteroid belt. (c) If n = 9 is taken to suggest an orbit radius for the next planet (Neptune) beyond Uranus, what value of r/re would your graph imply ? (d) Compare with the observed value. Consider whether, in the light of (b) and (c), your graph can be regarded as the expression of a physical law with predictive value. (As a matter of history, it was so used. See the account of the discovery of Neptune near the end of Chapter 8.) PHILOSOPHLE NATURALIS PRINCIPI A MATHEMATICA Autore J S. NEWTON, S. & Socictatis Regalis E P Y S, Reg. 5. Juiii Soaetatis Rcgi* ac title was officially page oflhe firsl edilion of Newion's It may be seen thal Ihe work accepted by Ihe Royal Sociely of London when ils president was thefamous diarist Samuel Pepys (who was also Secretary to Ihe Admirally at Ihe time). E I N S. I, Strealcr. Proftat Amo MDCLXXXVII. Principia (published 1687). in July, 1686, JE S 1686. Typis Jofepbi plures Bibliopolas. Facsimile oflhe P Soc. ND L Juflii Sodalt. IMPRIMATUR R P Mathefeos Irin. Coli. Cantab. Soc. Profeflbre Lucafiano, apud PartI The approach Newtonian dynamics to // seems probable Matter in solid, to me, that God in the Beginningfortrid massy, hard, impenetrable, moveable Particles .... newton, Opticks (1730) — 1 A universe of particles THE PARTICULATE VI EW the essence of the Newtonian approach to mechanics the motion of a given object which it outset we A is by subjected its is is that analyzed in terms of the forces to Thus from the very environment. are concerned with discrete objects of various kinds. special interest attaches to objects that can be treated as they are point masses; such objects are called particles. strictest sense there is Nevertheless, nothing you have in nature that lived for years in a fits 1 if In the this definition. world of particles — and electrons, atoms, baseballs, earth satellites, stars, galaxies is. If you have read George Orwell's famous political satire Animal Farm, you may remember the cynical proclamation "Ali animals are equal, have an excellent idea of what a particle : but some animals are more equal than others." the same way, you may protons, for example) are feel that more some particles (electrons or particulate than others. any case the judgement as to whether something only be made in terms of In somewhat is a — specific specific questions But particle in can kinds of experiments and observations. And the answer to the question "Is such and such an object a particle?" is not a clear-cut yes or no, but "It depends." For example, atoms and atomic nuclei will look 'Actually, might (i.e., behave) like Newton himself now call reserved the word "particle" to denote what we "fundamental particles"—atoms and other such natural — but the building blocks usage has since changed. 21 — Fig. 1-1 Photograph of a portion of the night sky. (Photograph from the Hale Obsercatories.) particles if you don't hit Planets and stars will them too hard. look like particles (both visually and in behavior) enough away from them (see Fig. 1-1). objects has spatial extension will and an if you get far But every one of these internal structure, and there always be circumstances in which these features must be taken into account. Very often this will be done by picturing a given object not as a single point particle but as an assemblage of such ideal particles, more or another. (If the possible to make less firmly connected to one connections are sufficiently strong, use of another fiction it may be — the ideal "rigid body" that further simplifies the analysis of rotational motions, in particular.) 22 A For the moment, however, we universe of particles shall restrict ourselves to a consideration of objects that exist as recognizable, individual entities and behave, in appropriate circumstances, as particles in the idealized dynamical sense. What sort of information description of a particle? we write down without any (or, for that matter, 1. Mass 2. Size 3. Shape do we need to build up a good Here are a few obvious items, which suggestion that the list is exhaustive sharply categorized): 4. Internal structure Electric charge 5. 6. Magnetic properties 7. Interaction with other particles of the 8. Interaction with though that Partial same kind different sorts of particles list may be, it is already formidable, and would not be realistic to tackle it all at once. So we ask a more modest question What is the smalkst number of properties that it : suffkes to characterize a particle? we If are concerned with the so-called "elementary" particles (electrons, mesons, etc), the state of charge (positive, negative, or neutral) is datum, along with the mass, and these two many an important may be sufficient to Most other composed of large numbers of atoms, are normally electrically neutral, and in any event the mass alone is for many identify such a particle in circumstances. objects, purposes the only property that counts in considering a particle's dynamic behavior — provided we take being independently specified. ' the forces acting at least approximately, the size also. Not only this is most informative pieces of data concerning any magnitude may to be filled in later, if we want laws of interaction The many one of our object, but its of the finer details will have shall begin with a not exhaustive or detailed. interactions of the 23 we particles are objects possessing 'Of course, as reasonably be treated as a point mass. Recognizing, then, that is it will help to tell us whether, in given circumstances, the particle which on however, useful to know, It is, On minimal description mass and size. the contrary, particulate view (e.g., is we have sought from characteristic by gravitatton), then the the subject of Chapter 5. to treat the forces as being derived body with its surroundings must also be known. That in Our survey to reduce minimum, to a it consistent with illustrating the gen- scheme of things, by considering only the masses and the eral linear dimensions of some typical particles. We the smallest and least massive particles and go shall begin with up the scale until You to be a fundamental limit. we reach what appears appreciate that this account, brief though it is, will draws upon the of a tremendous amount of painstaking observation and results research in diverse fields. A note on units In this book we second (MKS) with most frequently employ the meter-kilogram- at least for the basic it, If not, shall metric system. you should learn it You are probably already familiar measures of mass, length, and time. at this time. occasional use of other measures. We shall, however, make In mechanics the conversion from one system of measurement to another presents no problem, because (This a matter of applying simple numerical factors. contrast to electromagnetism, where the particular it is is in just choice of primary quantities affects the detailed formulation of the theory.) A tabulation of MKS and other units is given in the Appendix. ELECTRONS AND NUCLEONS The principal building blocks of matter of physics and from the standpoint chemistry are electrons, protons, and neutrons. Protons and neutrons are virtually equivalent as constituents of atomic nuclei and are lumped together under the generic nucleons. mentary The amount of research on and on the structure of particles, title the so-called ele- vast nucleons, has not brought forth any evidence for particles notably smaller (or notably less massive) than those that were known to science 50 years ago. Thus, although the study of subatomic particles field of very great richness and complexity, filled is a with bizarre and previously unsuspected phenomena, the microscopic limits of the physical world are still well represented by such familiar particles as electrons and protons. Theelectron,withamassofaboutl(r to be 24 A more precise), is by far the universc of particles 30 kg(9.1 lightest (by X l(T more than 31 kg three — (The elusive 10) of the familiar constituents of matter. powers of no neutrino, emitted in radioactive beta decay, appears to have mass at This puts all. of the electron it -15 m. size not sharply or uniquely defined for is however, we regard the electron as a sphere of If, electric charge, its radius 10 The not something that can be unequivocally stated. is Indeed, the concept of size any object. a rather special category!) in can be estimated to be of the order of In our present state of knowledge, the electron can properly be regarded as a fundamental particle, in the sense that there is no evidence that it can be analyzed The nucleon, with a mass of 1.67 basic ingredient of atoms. proton — is it (like the In — it kg, is the other — the charged form electron) completely stable; that cannot survive its electrically and a neutrino. The 13 neutral is, it form but decays radio- isolation in about actively (with a half-life of electron, into other constituents. 10~ 27 electrically its survives indefinitely in isolation. In the neutron X min) into a proton, an fact that neutrons spontaneously hydrogen atoms has led some give birth to the constituents of cosmologists to suggest that neutrons represent the true primeval particles of the universe — but that have a diameter of about X 3 is 10 just a speculation. -15 m — by Nucleons which we mean that the nuclear matter appears to be confined within a moderately well defined region of this size. Unlike electrons, nucleons seem to have a quite structure, in complex internal of mesons are incorporated. which various types But from the standpoint of atomic physics they can be regarded as primary particles. ATOMIC NUCLEI The combination of protons and neutrons to form nuclei provides the basis for the various forms of stable, ordinary matter as we know it. The individual proton. (that of 10 -25 238 kg. smallest The U)— contains and lightest nucleus is of course the heaviest naturally occurring nucleus 238 nucleons and has a mass of 4.0 All nuclei have about the X same mass per unit volume, so that their diameters are roughly proportional to the cube roots of the numbers of the nucleons. cover a range from about 3 A X Thus nuclear diameters 10~ 15 to 2 X 10~ 14 m. unit of distance has been defined that when dealing with nuclear dimensions. 25 Alomic nuclei is It is very convenient named after the 1 Enrico Fermi Italian physicist lfermi(F) = H)" 18 m= : 10- 13 cm Thus the range of nuciear diameters The density of nuciear matter is from about 10 17 kg/m 3 . This is so vast (it is larger, than the density of water) that although we now have we Given that the enormous. is uranium nucleus has a mass of about 4 X 10 of about 10 F, you can deduce (do it!) that really evidence that -25 its 3 to 20 F. kg and a radius density is about 14 by a factor of 10 , cannot apprehend some astronomical it, objects (neutron stars) are composed of this nuciear matter in bulk. ATOMS A great deal was learned about atomic masses long before From possible to count individual atoms. it was the concepts of valence and chemical combinations, chemists established a relative mass The mole was scale based on assigning to hydrogen a mass of 1 introduced as that amount of any element or compound whose mass in grams was equal numerically to its relative mass on this . Furthermore, from the relative proportions of elements combined to form compounds, it was known that a mole of any substance must contain the same unique number of atoms the number known as (or molecules in the case of compounds) number was itself unknown. But this Avogadro's constant. Obviously, if the number could be determined, the mass of an scale. that — individual The atom could be found. mass existence of characteristic transfers in electrolysis gave corroborative evidence on relative atomic masses but also pointed the way clear that the electrolytic teristic mass determinations, for it seemed phenomena stemmed from a charac- to absolute atomic charge was necessary was unit. Ali that to establish the size of this unit (e)—a feat finally achieved in Millikan's precision measurements mass values are listed in in 1909. Some representative atomic Table 1-1. and 'E. Fermi (1901-1954) was the greatest Italian physicist since Galileo one of the most distinguished scientists of the twentieth century, gifted in both theoretical and experimental work. He achieved popular fame as the man who produced the first self-sustained nuciear chain reaction, at the University of Chicago in 1942. 26 A universe of particles TABLE ATOMIC MASSES 1-1: Atomic Electrotytic mass kg/C Element H C 1.04 X lO" 8 10- 8 10- 8 10~ 7 8.29 Na Al K Zn Ag mass. kg 1 1.67 2e 12 2.00 2e 16 2.66 e 23 3.81 3e 27 4.48 e 39 6.49 65 107 1.09 e X X 2.38 X 9.32 X lO" 8 4.05 X 10" 7 3.39 X 10- 7 6 1.118 X 106.22 O Approximate relative mass Charge per ion transfer, 2e e 1.79 X X X X X X X X lO" 27 10" 26 lO" 26 K)- 26 -26 10 -26 lO 10- 2S 10~ 2S Modern precision measurements of atomic masses are based on mass spectroscopy (see p. 206 for an account of the principles) and are quoted in terms of an atomic mass unit (amu). This is now defined as tV of the mass of the isotope carbon 12. lamu = X 1.66043 Since almost all 10" 27 kg the mass of any atom is concentrated in its nucleus (99.95% for hydrogen, rising to 99.98% for uranium), we can say that to a mass of just the as we have approximation the mass of an atom nucleus. its represents a leap of eters, first many But, in terms of orders of magnitude. just seen, are of the order of 10 the size, is atom Nuclear diam- -14 m. Atomic — 4 diameters are typically about 10 times larger than this i.e., of -1 ° the order of 10 m. One way of getting a feeling for what this factor this means page is is to consider that if the dot you letter i on taken to represent a medium-weight nucleus, the outer boundary of the atom fine on a printed is about 10 ft away. Think of a grain of sand suspended in the niiddle of your bedroom or study, and will get a feeling for what that means in three dimensions. (Nuclei are really very small.) It is very convenient to take 10 -10 m as a unit of distance in describing atomic sizes or interatomic distances in solids and other condensed states in which the atoms are closely packed. The A. unit named after the nineteenth-century Swedish physicist, Angstrom: J. 1 It is is angstrom (A) = 10 -10 m= 10~ 8 cm = 10 5 F noteworthy that the heaviest atoms are not markedly bigger than the lightest ones, although there are systematic variations, 27 Atoms H ' Fig. 1-2 ' Relative atotnic radii (iiiferred from atomic volumes) versus atomic mass number, A. with pronounced peaks at the alkali atoms, as one progresses through the periodic table of the elements (see Fig. 1-2). Atoms are so small that it is hard to develop any real ap- enormous numbers of atoms present in even objects. For example, the smallest object that can be preciation of the the tiniest seen with a good microscope has a diameter of perhaps a few tenths of a micron anda mass of the order of 10~ 7 to 10~ fi kg. This minuscule object nevertheless contains something like 1 billion atoms. Or (to take another example) a very good labora- vacuum may contain residual gas at a pressure of a few times of atmospheric. One cubic centimeter of such a vacuum tory 10 -1 ' would likewise contain about 1 billion atoms. The atoms or molecules of a gas at normal atmospheric pressure are separated from one another, on the average, by about 10 times their diameter. This justifies (although only barely) the picture of a gas as a collection of particles that move independently of one another most of the time. MOLECULES; LIVING CELLS Our first introduction to molecules is likely to be in an elementary chemistry course, which very reasonably limits its attention to — simple molecules made up of small numbers of atoms C0 2) Na 2 S0 4 C 6 H 6 , , the order of 10 or 100 28 A 2 0, with molecular weights of and the like, and with diameters of a few angstroms. universe of particles Edge of Bacterium coli Foot-and- mouth virus Bushy stunt virus 10,000,000 Yellow fever virus' |Tobacco mosaic virus Fig 1-3 Sizes ,000 half-le ( Hemocyanin molecule 16,000,000 of microscopic and sub- mo Hemoglobin molecule (63,000) microscopic objects, m A bumin m0 |ecule (40,000) | from , J. bacteria down to . „ r, i t -a A. V. Butler, Inside Amino acjd chain _ 10 units (1|30 0) * Su 6ar molecule (350) _ ' the Living Cell, George Allen _ ,,„„ % Smgle am,n0 aC,d moleCule (130) & 0.1 Vnwin, London, 1959.) u.= 10,000 These then, do not represent much of an advance, either in size or in mass, on the individual atoms we have just been discussing. But through the development of biochemistry and biophysics we have come to know of molecules of remarkable size and com- We plexity. can feel justified in regarding them as particles on the strength of such features as a unique molecular weight for all molecules of a given type. The biggest objects that are de- scribable as single molecules have molecular weights of the order 20 7 kg and lengths hence masses of the order of 10~ of 10 amu— of the order of 10~ 7 m. Such objects are, however, far more important for their structure, and for their involvement logical processes, particles. The in bio- than for their rather precarious status as particle dynamics of a protein molecule is a pretty — limited perhaps to the behavior of the molecule structure a study a centrifuge — whereas the elucidation of in slim subject its that requires (and merits) the chemists and crystallographers. is most intensive efforts of brilliant It would be both presumptuous and inappropriate to attempt to discuss such matters here, but it is perhaps worth indicating the range of magnitudes of such particles with the help of Fig. 1-3. 29 Molecules; living cells A convenient unit of length for describing biological systems 1 The micron (m) = lO" 9 m = largest object across and would be limit of resolution of shown is 10 4 in Fig. visible in a about is 0.2/n the micron: A 1-3 (a bacterium) good microscope — rather less is (for about 1m which the than one wavelength light). Figure 1-3 includes some viruses, which have a peculiar status between living and nonliving definite size and mass, — possessed of a rather isolatable (perhaps as a crystalline substance), yet able to multiply in a suitable environment. Figure an electron-microscope photograph of some virus particles. These are almost the smallest particles of matter of which we can form a clear image in the ordinary photographic sense. (You 1-4 is have perhaps seen "photographs" of atomic arrangements as observed with the device called a field ion microscope. These are not direct images of individual atoms, although the pattern Fig. 1-4 particles Sphericat of polio virus. [C. E. Schwerdt et al., Proc. Soc. Exptl. Biol. Med., 86, 310 (1954). Photograph courtesy of Robley C. Williams.] 30 A. universe of particles ' 1 does reveal their spatial relationships.) we go one step further along this biological road, then of course we come to the living cell, which has the kind of significance for a biologist that the atom has for a physical scientist. If Certainly appropriate to regard biological cells as particles, it is most of them albeit of such a special kind that the study of outside physics. venient reference points on our scale of physical magnitudes, that is lies They do, however, provide us with some con- — except, our only reason for mentioning them here and perhaps, for the matter of reminding ourselves that biological systems also belong within framework defined by the fundamental a atomic interactions. Although some single cells may be than less 1/* (certain more than 1 cm (e.g., the yolk of a hen's egg), the cells of most living organisms have diameters of the order of I0~ 5 m (1<V) on the average. Thus a human being, with a volume of about 0. 1 m 3 contains about 10 4 cells, each of which bacteria) or , (on the average) contains about 10 l4 atoms. SAND AND DUST Vast areas of our earth arc covered with particles that have come from the breaking down of massive rock formations. and are ically very inert word These predominantly of quartz (crystalline Si0 2 ), are chem- particles, just the kind of objects to "particle" applies in ordinary speech The and inanimate. earth's surface, and — small the that surface, are loaded with such particles. of (say, of the order ground. 1 The Others, orders of magnitude smaller, combined The fail. effect visible, atmosphere above biggest ones mm across) rest more or less firmly on the motes of dust dancing tendency to which the but in fate of wind may be seen as the sunlight, apparently showing no of a given particle depends on the (or air resistance) and gravity. Ac- cumulations of windblown sand are found to be made up of particles maximum of about mm. Below that size from a of about 0.01 airborne, and 1 mm diameter to a minimum the material tends to remain the smallest dust particles are of the order of 'Improvemcnts in electron- and proton-microscope technique reach further and further down into the world of small particles. In 1969 the helical structure of a protein molecule was photographed in an electron microscope. 31 Sand and dust Fig. 1-5 Size distribulion (After R. A. Bagnold, The of particles found on or above the earlh's surface. Blown Sand and Desert Dunes, Physics of Meihuen, London, 1941.) 10 -4 mm Figure 1-5 shows an approximate (=0.1/*) diameter. classification of particle sizes for dust, sand, and other small particles. OTHER TERRESTRIAL OBJECTS man-made In a range from the smallest sand grains to the largest objects we are in the realm of our most immediate experience: things that are large enough to be apprehended by the unaided sight or touch, yet not so big that ready a large number, we cannot human and a factor of 1000 achieve a rather terms, 100 is al- up or down in linear In ordinary direct awareness of them. from human dimensions brings us close to the scale limits of anything that can properly be regarded as a fuli, first-hand contact with the physical world. chiefly upon we depend Outside that domain indirect evidence, imagination and analogy. Since the densities of most solid materials (as we find them in the earth's crust) are of the order of magnitude of a few thousand kg/m 3 , the range of diameters from a grain of sand (1 mm) to a cliff or a dam (1 km) implies a range of masses from about 1 mg 2 to 10' kg (i.e., 10 9 (or the weight associated with tons). it Actually, if mass itself under the gravitational conditions at the earth's surface) were to be our criterion, then the range we have just stated goes far beyond the span of perceptions. In terms of weights, direct experience gives us about 10 mg some or as heavy as 1000 it would be fair to human say that our feeling for objects as light as kg— i.e., up to a factor of 10 ±4 with respect to a central value of the order of 0. 1 kg (or a few ounces). 32 A universc of particles PLANETS AND SATELLITES For an earthdweller going about his ordinary daily affairs, it is almost impossible— as well as unrealistic—to regard the earth as a particle. We cannot help but be aware, primarily, of the vastness of the earth and of the fact that man's greatest edifices are, at least in terms tions of its of physical into space for different point size, totally insignificant if And all placed on the same footing as the other planets. can, on be regarded as particles moving in this scale, about a much more massive of them their orbits In Kepler's mathe- particle, the sun. and subsequently matical description of the orbits of the planets, in modifica- we can imagine ourselves backing off 100 million miles or so, we can arrive at a very of view. The earth loses its special status and is Yet surface. Newton's dynamical theory of these motions, the planets could be regarded as point particles— their extent and legitimately internal structure were in no way relevant. The reason, of course, was the simple one that on the scale of the solar system the more than mere points, as is obvious whenever we look up into the night sky. The earth's diameter is only about -4 of the distance between earth and sun and the sun's own 10 planets are little diameter only 10 — that a is good -2 of the same distance. No wonder, then, approximation to the dynamics of the solar first system can be obtained by taking such ratios to be zero. But we should not indeed, we cannot. rest content with When Newton such an approximation; turned his attention from the planetary orbits to the tides, the physical extent of the earth became a key feature, because it was only this that made possible the existence of tide-producing forces, through a significant change of the moon's gravitational effect over a distance equal to the earth's diameter. In any case, once we free ourselves of the restriction of dealing with terrestrial objects and terrestrial phenomena, the particles that comprise the solar system are among the first to claim our attention. The planets can be described as roughly spherical particles, somewhat flattened as a result of their own rotation. Their range of diameters Mercury and is considerable Jupiter. — a factor of nearly 30 between Since the really drastically (the densest is mean densities do not differ our earth, at 5.5 times the density of water), the masses cover a very great range indeed. Again the extremes are Mercury and Jupiter, and there 6000 between their masses. 33 Planets and satellites is a factor of about Figure 1-6 depicts these principal Fig. 1-6 Relative sizes of llre planets and the planets in correct relative scale. The sun. which has size of the sun, about the same density as Jupiter but 10 times the diameter and about 1000 times the mass, is indicated for comparison. The same data are presented TABLE 1-2: in Table 1-2. DATA ON THE PLANETS Mean radius, X km M/ME R/Re 10 3 Mean density, 6.37 1.00 1.000 5400 5100 5520 (Moon) Mars (1.74) (0.27) (0.012) (3360) 3.37 0.53 0.108 3970 Jupiter 69.9 10.98 317.8 Saturn 58.5 9.20 95.2 680 Uranus Neptune 23.3 3.66 14.5 1600 17.2 2250 Mercury Venus Earth 2.42 6.10 0.38 0.054 0.96 0.815 22.1 3.48 Pluto 3.0 0.47 (Sun) (6.96 X 10 6 ) 1330 0.8? (3.33 (1093) kg/m 3 ? X 10 5) (1400) The nine major planets represent almost all the mass of matter around the sun (and Jupiter alone accounts for almost two thirds of it), but the number of other captive objects we is ignore There are, first, the natural man's contributions, there are about 30 known satellites of the planets, most of them extremely tiny in comparison with the enormous. satellites. planets to which they are tied. the largest, but even the earth.) terest, These it (Our own moon has only a satellites little A because from their motions universe of particles 1% is relatively of the mass of have a very special dynamical it the masses of the planets themselves. 34 over If in- has been possible to infer The planetary by the minor satellites are, however, vastly outnumbered planets, also called planetoids or asteroids. Tens ' of thousands of them are orbiting the sun in the region between Mars and Jupiter. In size they range from 500 miles diameter down to 1 mile or less, and they most probably come the orbits of in from the breakup of a larger body. Baade's description of them all — "the If the astronomer Walter vermin of the skies" representative, they are not greatly beloved — is at by professional skywatchers. STARS A star is a magnificently complex structure, almost inconceivably by human standards and with a fascinating gigantic tromagnetism. Yet when we gaze out into space we interior and dynamics that involves nuclear reactions, gravitation, see nothing — we look with the eye and the instruments astrophysicist. Instead, we see the stars as luminous of unless this, which (in contrast to the planets) sources — of an points, continue to appear as point when examined through even most powerful the In relation to their diameters, in fact, stars are scopes. farther elec- away from each other than tele- much are the planets of our solar system. A convenient unit for specifying astronomical distances vacuum the Hght-year, the distance light travels in a 1 light-year The - 9.46 X is year: 1 m 10 15 nearest star to the sun about 25,000,000,000,000 in is about 4 light-years away, or A miles! number of other stars that are near neighbors of the sun have mutual separations of the order of 10 light-years or 10 17 diameter to separation about 10 of stars is m, which makes the ratio of -7 -8 or 10 . A cluster or galaxy thus an excellent example of a system of massive point of the particles. The particles, despite the fantastically large size best vacua attainable in the laboratory scarcely approach a corresponding emptiness as given by the ratio of interparticle spacing to particle diameter. One must look to the extreme vacua of interstellar space, where there are regions containing one hydrogen atom per cubic centimeter, to find a less still than emptier kind of system. •The word "asteroid" derives from the objects as seen through a telescope. 35 Stars star-like appearance of these tiny Most of the objects that are recognizable as stars are within 2 the range of 10* solar masses. regarded as an average or typical In this sense our sun can be In terms of size or lumi- star. nosity (total radiation) the range of variation is much very greater, but a worthwhile account of these features would call really for some discussion of the evolution and interior mechanism of stars; and this is certainly not the place to attempt shall content ourselves, therefore, with 1 it. remarking that the We stars, regarded simply as aggregations of matter, with masses between 32 about 10 28 and 10 kg (thus containing something of the order of 10 54 to 10 58 atoms) can still be regarded as particles when we discuss their motions through space, because of the distances that separate immense them from one another. GALAXIES In 1900 the words "galaxy" and "universe" were regarded as being synonymous. Our universe appeared to consist primarily of a huge number of stars— many billions of them— scattered through space (see Fig. 1-1). Here and there, however, could be seen cloudy objects — nebulae — near enough or big enough to have an observable extent and even structure, as contrasted with the pointlike appearance of the stars. By 1900 many thousands of nebulae were known and catalogued. But what were they? To quote the astronomer Allan Sandage: "No one knew before 1900. Very few people knew in 1920. AH astronomers knew 2 For it was in 1924 that the great astronomer after 1924." Edwin Hubble produced the conclusive proof that the nebulae were, in the picturesque phrase, "island universes" far outside the region of space occupied by the Milky Way, and that our G) was only one of innumerable systems of the same general kind. The first suggestion for such a picture of the universe was in fact put forward by the own Galaxy (distinguished with a capital Kant as long ago as 1755, but of course more than a pure hypothesis. was no philosopher Immanuel at that time it For extensive discussion, see (for example) F. Hoyle, Fronliers of Aslronomy, Harper & Row, New York, 1955. 2The story of this development is fascinatingly told by Sandage in his iniropublished duction to a beautifui book entitled The Hubble Alias of Galaxies, Hubble's See also Washington, D. C. of Institution the Carnegie in 1961 by own classic work, The Realm of the Nebulae, Dover Publications, New York, ; 1958. 36 A universe of particles Fig. 1-7 Cluster of galaxies in the constellation Borealis. Corona Dislance about 600 million light-years. graphfrom (Photothe Hale Observatories.) To quote Sandage again: "Galaxies are the largest single aggregates of stars in the universe. They are to astronomy what atoms are to physics." And so far as we can tell at present, they represent the largest particles in the observable scheme of things. A 10 ' ' single galaxy stars. may contain anywhere from about 10 6 to Our own Galaxy appears with a diameter of about 10 21 m to be one of the larger ones, (10 5 light-years). As we have already seen, the stars within an individual galaxy are very widely spaced indeed, so that the average density of matter in a galaxy 20 is very, very low only about 10~ kg/m 3 But even so, galaxies — . represent notable concentrations of matter. spacing between galaxies although there is particles, by On the average the about 100 times their diameter, a tendency for them to separations perhaps galaxies is exist in clusters with 10 times less than this. may, on an appropriately large scale, Thus even the be regarded as and the interactions between them may be approximated treating them as points (see Fig. 1-7). Astronomical surveys indicate that space contains a roughly uniform concentration of tinues to be true galaxies. If up to the theoretical we assume that this con- limits of observation, we can make some estimate of the content of the universe as a whole. For it would appear that the universe can be represented by a sphere with a radius of about 10 37 Galaxies 10 light-years (10 26 m). The galaxies in general appear to be receding from us with speeds proportional to their distances, and at a distance of 10 l0 light- years the recessional speed would reach the speed of light. At this speed, because of the "galactic red shift" (a form of the wavelength and frequency —called source in for radiation shift general the Doppler of energy back to us would effect), from a moving the transmission to zero, thus setting a natural fail limit to the extent of the knowable universe. If we take the average density of matter throughout space to be about 10 -a(i kg/m 3 (10 -6 of the density within an individ- volume of the universe to be we arrive at a total mass ) 79 hydrogen atoms). 52 (equivalent to about 10 kg of about 10 il galaxies, This would then correspond to a total of about 10 ual galaxy) and assume the of the order of (10 26 3 each containing about , i.e., 10" total 10 78 stars. m3 , Such numbers are of course so stupendous that they defy any attempt to form a mental image of the universe as a whole. But it scems that we can be assured of one thing, at least, which is that the basis of our description of the — what we described at the outset as "the particulate view"— flnds some justiflcation over the entire range of our experience, from the nucleus to the cosmos. And the fact that this approach to the description of nature makes sense, while physical world embracing a span of about 10 40 not merely aesthetically pleasing; in distance it and 10 80 in mass, is also suggests that something, we at least, of the physical description give. to the behavior of atoms will be found applicable to the behavior of galaxies, too. If we want to find a unifying theme for the study of nature, especially for the applications of classical mechanics, this particulate view is a strong candidate. PROBLEMS 1-1 Make a tabulation of the orders of magnitude of diameter, volume, mass, and density for a wide selection of objects that you regard as being of physical importance e.g., nucleus, cell, and star. For the diameters and masses, rcprcsent the data by labclcd points on — a straight line marked off logarithmically in successive powers of 10. This should give you a useful overview of the scale of the universe. 1-2 Estimate the number of atoms in: (a) (b) 38 A The The smallest speck of matter earth. universe of partieles you can see with the naked eye. Calculate the approximate mass, in tons, of a teaspoonful of 1-3 nuclear matter 1-4 — closely packed nuclei or neutrons. James Jeans once suggested that each time any one of us is a good chance that this lungful of air contains onc molecule from the last breath of Julius Caesar. Make your Sir draws a breath, there at least own calculation to test this hypothesis. mm of 6 evacuates a bottle to a pressure of 10~ A vacuum pump 1-5 -9 of atmospheric pressure). mercury (about 10 Estimate the magnitudes of the following quantities: The number of molecules per cubic centimeter. The average distance between molecules. (a) (b) In a classic experiment, E. Rutherford and T. Royds 1-6 [Phil. Mag., showed that the alpha particles emitted in radioactive decay are the nuclei of normal helium atoms. They did this by collecting the gas resulting from the decay and measuring its spectrum. They 17, 281 (1909)] started with a source of the radioactive gas radon (itself a decay product The half-life of radon is 3.8 days; i.e., out of any given number of atoms present at any instant, half are left, and half have of radium). decayed, 3.8 days later. When the experiment started, the rate of alpha- by the radon was about particle emission 5 X 10 9 per second. Six enough helium gas had been collected to display a complete helium spectrum when an electric discharge was passed through the tube. What volume of helium gas, as measured at STP, was collected days later, in this experiment? function of time is [The number of surviving radioactive atoms as a given by the exponential decay law, N(t) The number of disintegrations per unit time at any instant \Noe- u Given the value of the half-life, you can deduce = is . Noe-*'. \dN/dt\ = the value of the constant, X.] 1-7 law The general expansion of [o{r) volume = the universe as described by Hubble's amount of mass per ar] implies that the average in the universe should be decreasing by about 1 unit 17 part in 10 per second (see Hors d'oeuvres no. 29, p. 16). According to one theory (no longer so strongly held) this Ioss is being made up by the continuous creation of matter in space. (a) Calculate the approximate cubic meter per year that would, if number of hydrogen atoms per produced throughout the volume of the universe, bring in the necessary amount of new mass. (b) It has of new matter, been hypothesized (by if it J. G. King) that the creation occurs, might well take place, not uniformly throughout space, but at a rate proportional to the amount of old matter present within a given volume. the On number of hydrogen atoms equivalent this hypothesis, calculate to the per day in a vessel containing 5 kg of mercury. 39 Problcms (J. new mass created G. King concluded on this basis that a test sure of 1-8 of the hypothesis was feasible.) Torr 1 (1 mm of mercury). Theodore Rosebury, in his book Life on 1969), remarks that the total bulk of all Man (Viking, New York, the microorganisms that live on the surface of a human being (excluding a is bottom of a thimble. taken to be 5 y. (1 n = If the 10 -6 mean on the accommodated far larger quantity inner surfaces of the intestinal system) could easily be in the Convert your an equivalent volume of hydrogen gas as measured at a pres- result to radius of these microorganisms m), what the order of magnitude of is the population of such organisms that each of us carries around the time ? Compare all human population of the the result with the total globe. 1-9 It seems probable that the planets were formed by condensation from a nebula, surrounding the sun, whose outer diameter corresponded roughly to the orbit of Pluto. (a) If this nebula were assumed to be thickness equal to about one tenth of been its mean density in kg/m 3 ? (Do in the its form of a disk with a radius, what would have not include the mass of the sun in the calculation.) (b) A better picture appears to be of a gas cloud whose thickness Suppose that Jupiter increases roughly in proportion to the radius. were formed from a ring-shaped portion of the cloud, extending radially for half the distance from the asteroid belt to the orbit of Saturn and with a thickness equal to this radial extension. the mean What would have been density of this portion of the nebula ? (In terms of the radius of the earth's orbit, the orbit of Pluto has a radius of about 40 units, the asteroid belt is at about 3 units, and the orbit of Saturn has a radius of about 9.5 units.) 1-10 The core of a large globular cluster of stars about 30,000 stars within a radius of about (a) stellar ratio Estimate the ratio of the A typically contain separation of stars to the diameter in such a core. (b) At approximately what degree of vacuum would the same of separation to particle diameter be obtained for the molecules of a gas? 40 mean may 5 light-years. universe of particles / do not define time, space, place and motion, since they are well known to all. newton, Principia (1686) Space, time, and motion WHAT IS MOTION? you are undoubtedly familiar with motion in all kinds of mani- what would you say if you were asked to define it? The chances are that you would find yourself formulating a statement in which the phrase "a change of position with time," festations, but or something equivalent to that, expressed the central thought. For it seems that our depends in ability to give any precise account of motion an essential way on the use of the separate concepts We say that an object is moving if it occupies of space and time. different positions at different instants, and any stroboscopic photograph, such as that shown in Fig. 2-1, gives vivid expression to this mental picture. good Newtonians in the sense that about space and time are closely in harmony Ali of us grow our intuitive ideas up to be The following paragraphs are a deliberate attempt to express these ideas in simple terms. The with those of and himself. may appear description many Newton natural and plausible, but it embodies notions which, on closer scrutiny, will turn out to be naive, difficult or impossible to defend. So the account below apart with square brackets to emphasize should not be accepted at its its (set provisional status) face value but should be read with a healthy touch of skepticism. Newton's view, is absolute, in the sense that it exists permanently and independently of whether there is any matter in the space or moving through it. To quote Newton's [Space, in own words in the Principia: "Absolute space, in its own nature, 43 Fig. 2-1 Slrobo- scopic photograph of a motion. {Front PSSC Physics, D. C. Healh, Lexington, Massachusetis, 1965.) without relation to anything external, remains always similar and immovable." [Space is thus a sort of stationary three-dimensional matrix into which one can place objects or through which objects can move without producing any the space. in space Each and change of its interaction between the object and object n the universe exists at a particular point time. i An object in motion undergoes a continuous position with time. And although it would not be one can imagine the charting of positions with the help of a vast network of meter sticks, laid out end to end in a One can three-dimensional, cubical array throughout space. practicable, conceive of extending such measurements to any point in the universe. In other words, the space is there, and we simply have a practical task of attaching markers to it. Moreover, our physical measurements agree with the theorems of Euclidean geometry, and space is thus assumed to be Euclidean. [Time, in Newton's view, 44 Space, time, and motion is also absolute and flows on with- Again quoting from out regard to any physical object or event. and mathematical the Principia: "AbsoLute, true, and from its own name thing external, and by another language is said in the time, of itself, nature, flows equably without relation to anyis The As Newton called duration." elegant but delightfully uninformative. remark quoted at the beginning of this chapter, he did not attempt to define either space or time. [One can neither speed up time nor slow down its rate, and throughout the universe. If we this flow of time exists uniformly imagine the instant "'now" as planet and occurs simultaneously on every it and an hour star in the universe, of this 60-minute interval, we assume mark later the end that such a time interval has been identical for every object in the universe, as could (in by observations of physical, chemical, or biological processes at various locations. As an aid to measuring principle) be verified time intervals, it would be possible, in principle, to place identical clocks at each intersection of a meter-stick framework and to synchronize these clocks so that they indicate the same time at a common, simultaneous would Being identical clocks, they instant. thereafter correctly mark off the flow of absolute time and remain synchronized with each other. [Space and time, although completely independent of each other, are in a sense interrelated insofar as we find it impossible to conceive of objects existing in space for no time at existing for a finite time interval but "nowhere" all, in space. or Both — to have no space and time are assumed to be infinitely divisible ultimate structure.] The preceding five paragraphs describe in everyday language some commonsense notions about the nature of space and time. Embedded in these notions are many assumptions that we adopt, either knowingly or unconsciously, the universe. many correct they seem, we mentioned in the at very high speeds, This first became apparent Prologue) in connection with motions approaching or equaling the speed of and with the phenomena of electromagnetism and ; in his development of special some of the most important cluding Newton's how own relativity theory, it light, was Einstein, who exposed limitations of classical ideas, in- ideas about relativity, and then showed they needed to be modified, especially with regard to the concept of time. 45 our picture of of these ideas have consequences that are inconsistent with experience. (as in developing fascinating therefore that, however intuitively It is What is motion '? The crux of the matter is that it is one thing to have abstract concepts of absolute space and time, and it is another thing to have a way of describing the actual motion of an object in terms of measured changes of position during measured intervals of time. Newton himself understood this very well, at least as far Thus in the Principia "But because the parts of space cannot we find him remarking: be seen, or distinguished from one another by our senses, therefore in their stead we use sensible [i.e., observable] measures measurements were concerned. as spatial of them . . . And use relative ones rest, and motions, we so, instead of absolute places . . . For it may be that there is no body to which the places and measures of others ferred." space, ments. it If there is really at may be re- any knowledge to be gained about absolute can only be by inference from these relative measureThus our attention turns to the only basis we have for describing motion — observation of what a given object does in relation to other objects. FRAMES OF REFERENCE If you should hear somebody say "That car certain that would be quite what is is moving," you being described is a change of position of the car with respect to the earth's surface and any buildings and the like that may be nearby. Anybody who announced "There is motion between that car and the regarded as a tiresome pedant. But this relative earth" would be rightly does not alter the fact that it takes the pedantic statement to We express the true content of the colloquial one. local surroundings and therefore at accept the — a collection of objects attached to the earth rest relative to one another— as defining a frame ofreference with respect to which the changes of position of other objects can be observed It is clear and measured. that the choice of a particular frame of reference to which to refer the motion of an object of taste and convenience, but it is is entirely a matter often advantageous to use a reference frame in which the description of the motion is simplest. ship, for example, is for many purposes a self-contained world A within which the position or path of any person is most efficiently described in terms of three perpendicular axes based on the up and down the ship itself uniform motion of The (according to deck number). directions fore 46 and aft, port and starboard, and Space, time, and motion with respect to a frame of reference attached to the earth may be unnoticed or even ignored by the passengers, as long as they can rely on the navigation officers. Is the earth itself at rest? become accustomed We would not say so. We have to the fact that the earth, like the other planets, is continually changing its position with respect to a greater frame of reference represented by the stars. And since the stars constitute an almost completely unchanging array of 'i» • • ' ' JK f' ] ' Fig. 2-2 Apparent circular molions of the stars. (Carillon ' ff - T , 'f f i ir Tower of Wellesley College, Wellesley, Mass., from J. C. Duncan, Astronomy, 19.) %ti<$ 47 Frames of reference 5th edilion, Harper &Row,J954,p. ntjaKKm reference points in the sky, we regard the totality of them as representing a fixed frame of reference, within which the earth both rotates and moves bodily. It remains true, however, that our primary data are only of relatioe positions and displacements; the belief that it really rotating on makes more sense its to assume that the earth is axis once every 24 hours, rather than that is going around us, is one that cannot be by primary observations alone (see Fig. 2-2). With our present knowledge of the great masses and enormous numbers of the stars, it does, to be sure, simply seem more reasonable to attribute the motion to our puny earth. But it would be hard to the system of stars justified elevate that subjective Later we judgment into a physical law. shall see that there are powerful theoretical reasons for preferring some reference frames to others. The "best" choice of reference frame becomes ultimately a question of dynamics i.e., dependent on the actual laws of motion and — But the choice of a particular reference frame is often made without regard to the dynamics, and for the present we shall just concern ourselves with the purely kinematic problems force. of analyzing positions and motions with respect to any given frame. COORDINATE SYSTEMS frame of reference, as we have said, is defined by some array of physical objects that remain at rest relative to one another. Within any such frame, we make measurements of position and A displacement by setting up a coordinate system of some kind. doing this we have In a free choice of origin and of the kind of coordinate system that is best suited to the purpose at hand. Since the space of our experience has three dimensions, we must in general specify three separate quantities in order to fix uniquely the position of a point. However, most of the problems that we shall consider will be of motion confined to a single plane, so us first let consider the specification of positions and displacements two dimensions only. As you are doubtless well aware, the position of a point in a plane is most often designated with respect to two mutually perpendicular straight lines, which we call the x and y axes of a The position coordinate system, intersecting at an origin O. of the point P [Fig. 2-3(a)] relative to O is then described by a position vector r, as shown, characterized by a specific length and in 48 Space, time, and motion Fig. 2-3 (a) Square grid; the basis of Carlesian coordinates in a plane. (b) Plane polar coordinate grid. Using our perpendicular a specific direction. uniquely define r by ing r is 2-3(b). r makes in terms of polar coordinates Here with the positive x designating the position of axis, as P r two dimcnsions) r as shown in Fig. (r, 0), P from O and r is the distance of measured wise (conventionally positive) direction. (In we can onto the x and y However, another important way of specify- ordinates (x, y), which are projections of axes, respectively. axes, the pair of rectangular (Cartesian) co- d is in the angle that a counterclock- The two schemes of are related as follows: 2 = x2 + y 2 y = - tan (2-1) X x = r cos d y = = r sin Figure 2-4 shows examples of the use of these coordinate systems. We shall on various occasions be making use of unit vectors that represent displacements of unit length along the basic coordinate directions. shall we In the rectangular (Cartesian) system in the x and y directions by i and j, The position veetor r can be written as the sum of its denote the unit vectors respectively. two veetor components: (In two dimensions) = r xi In the polar coordinate system, denote a unit veetor 6 in and the symbol ee we in the direction (2-2) shall use the symbol e r to of increasing r at constant to denote a unit veetor at right angles to r the direction of increasing purpose comes from 49 + y\ Coordinate systems the 9. (The use of the symbol e for German word for unit, this which is /*>" '""'V * y>«iui.i«ns< (b) (a) — Example of Cariesian coordinales in use a portion of midtown Manhattan, New York City. (b) Example of plane polar coordinales in fig. 2-4 use—a North (a) radar scopeface with is shown afew incipient thunderslorms (June 3, 1970). as O" azimuth. The heavy circles ofr = const are at 100 km, 200 km; the lighter circles are spaced by 25 km. (Photograph courtesy of Department of Meteorology, M.l.T.) "Einheit.") In this polar-coordinate system, the vector equal to re r , and one might wonder introduced at important as As we soon as we all. shall see, why r is simply the unit vector e« however, it is becomes very consider motions rather than static displacements, for motions will often have a component perpendicular to r. Although the above coordinate schemes are the most familiar ones — and are the only ones we shall be using in this book for — two-dimensional problems it is worth noting that any mapping of the surface that uniquely fixes the position of a point is a possible system. — one Figure 2-5 shows two examples a nonorthogonal system based on straight axes, and the other an orthogonal system based on two sets of intersecting curves. Such systems are introduced to capitalize on the kinds of symmetry that particular physical systems 50 Spacc, time, and motion may possess. Fig. 2-5 Ihal is freguently (a) Oblique coordinate system, time diagrams) in special relativity. curvilinear coordinate system, confocal ellipses If (b) Orthogonal made o/ intersecting sets of and hyperbolas. necessary to specify it is ofthe kind used in Minkowski diagrams (space- all three of the spatial coordinates of a point, the most generally useful coordinate systems are the three-dimensional rectangular (Cartesian) coordinates (x, y, and spherical polar coordinates (r, 6, <p). These are both z), illus- The Cartesian system is almost always be right-handed, by which we mean that the positive z trated in Fig. 2-6(a). chosen to direction is chosen so that, looking upward along of rotating from the positive x it, the process direction toward the positive direction corresponds to that of a right-handed screw. It y then follows that the cyclic permutations of this operation are also — from -\-y to +z, looking along +x, and from +x, looking along +y. You may note that the two- right-handed +z to dimensional coordinate system, as shown in Fig. 2-3(a), would on this convention bc associated with a positive z axis sticking up toward you out of the plane of the paper. Introducing a unit vector k in the +z direction, analogous to the i and j of Eq. (2-2), we can write the vector r as the sum of its three Cartesian vector components: r 51 = xi + yi + zk Coordinate systems (2-3) 2-6 Fig. (a) Coordinates ofa poinl in three dimensions, showing both spherical polar and right-handed Carlesian coordinates. (lalitude ofa (6) Point located and longitude) on a by angular coordinates sphere, and the unit vectors local Carlesian coordinate system at the point in question. The description of the position or displacement in spherical polar coordinates makes use of one distance and two angles. (Notice that three dimensions requires three independent co- form they may ordinates, whatever particular is, take.) as with plane polar coordinates, the distance r The distance from the chosen shown as in Fig. 2-6(a)] is simply the angle between the vector r and the positive z axis; The other angle represents the it is known as the polar angle. angle between the zx plane and the plane defined by the z axis and r. It can be found by drawing a perpendicular PN from the One origin. end point P of the angles [the one of r onto the xy plane and measuring the angle between the positive x axis and the projection ON. This angle (p) The geometry of the figure shows that the is called the azimuth. rectangular and spherical polar coordinates are related as follows: x = y = z = If 52 we set r sin cos r sin sin tp (2-4) <p r cos = tt/2, we make z Space, time, and motion = and so get back to the two- dimensional world of the xy plane. The first two equations of (2-4) then give us = r cos y = r sin x ¥> v? very unfortunate that a long-established tradition uses the It is symbol d, the vector as r we ourselves did earlier, to denote the angle between and the x axis two-dimensional case. in this special This need not become a cause for confusion, but one does need to be on the alert for the inconsistency of these conventions. have all grown up with one important use of spherical We mapping of the earth's surface. This is indicated in Fig. 2-6(b). The longitude of a given point is just the angle ip, and the latitude is an angle, X, equal to jr/2 — 6. (This entails calling north latitudes positive and south latitudes polar coordinates, the negative.) At any given point on the earth's surface a set of three mutually orthogonal unit vectors defines for us a local coordinate system; the unit vector e r points vertically upward, the vector ee points due south, parallel to the surface, and the third unit vector, e^,, points due east, also parallel to the surface. the plane polar coordinates, the vector r is As with given simply by re r . COMBINATION OF VECTOR DISPLACEMENTS Suppose we were at a point P\ on a flat horizontal plane [Fig. 2-7 (a)] and wished to go to another point chose to by Fig. 2-7 (a) make the +x and +y in P g. Imagine Suc- cessiue displacements on a plane; the final position is independen! of the order in which the displacements are made. {b) Addition of secerat displacement vectors in a plane. 53 that we trip by moving only east and north (represented the figure). We know there are two particularly Combination of vector displacements Fig. 2-8 X Seatar 2r multiples of a given veetor r, including negative multiples. IHnBBHBBHHBBHBB^HEHBHiBI straightforward ways of doing this: (1) travel a certain distance and then a certain distance sz due su due north, followed by cast su due north, or (2) travel sx due east. The order in which we component displacements does not matter; we same point P 2 in either casc. Our representation of the take these two rcach the veetor yj is r in sum Eq. (2-2) as the of the individual veetors xi and an example of such a combination. This simple and familiar property of linear displacements exemplifies an essential feature of all those quantities combinations we illustrate head to tail, we how veetors call and is not confined to Thus, for example, in Fig. 2-7(b) at right anglcs. three veetor displacements, A, B, and C, placed can be combined into a single veetor displacement S drawn from the original starting point to the final end point. This is what we mean by adding the veetors A, B, and C. The order in which veetors are added is of no consequence; thus successive displacements of an object can be combined according to the veetor addition law, without regard to the sequence in which the displacements are made. quantity, in general, is that What we mean by a veetor a direeted quantity obeying the it is same laws of combination as positional displacements. We shall often be concerned with forming a numerical A multiple of a given veetor. we changc the ing is its means that length of the veetor by the faetor n without chang- The negative of a veetor (multiplication by — 1) mean a veetor of equal magnitude but in the opposite direetion. defined to added to the direetion, so that negative multiplier, tion positive multiplier, n, and changed —n, then in length original veetor it gives zero. A defines a veetor reversed in diree- by the faetor n. These operations are illustrated in Fig. 2-8. Subtracting one veetor from another is accomplished by noting that subtraetion basically involves the addition of a negative quantity. we form Thus the veetor A - B = A 54 I if veetor —B and B is add (-B) Space, time, and motion to be subtraeted it to A: from veetor A, Fig. 2-9 Addilion and subtraction of Iwo given vectors. Note that the magnitude ofthe vector difference may be {as here) larger than the In Fig. 2-9 we show both We veetors. the sum and the difference of two given have deliberately chosen the direetions of to be such that the vector this will help to A — B emphasize the is A longer than the vector faet that vector B and A + B; combination is something rather different from simple arithmetical combination. The evaluation of the vector distance from a point P^ to a point P 2 when originally the positions of these points are given , separately with respect to an origin O direct application of vector subtraction. relative to P\ is = r2 — ri2 given by the vector r 12 [see Fig. The (2-10)], such that is to be read as "one-two" and is notation in the deseription of two-particle systems.) r, Fig. 2-10 tion of the (i> 2 ) r2 . P! Clearlyr 2 P2 relative to i = — r 12 is Construc- ofone with respect to another point OPi). 55 a common Similarly, given by the vector . relative position vector point — a P2 ri (The subseript "12" the position of is position of Combination of vector displacements r2 i = Fig. 2-11 (a) Com- ponents of a given vector in two different rectangular coordinate systems related by an angular dis- placemenl 9 in the xy plane. (b) Com- ponents of a given vector in a rectangular coordinate system and in an obligue coordinate system. THE RESOLUTION OF VECTORS 1 In discussing the description of a given vector in terms of components, we have indicated that There operated in either direction. of the vector into or there is its components this is is its a process that can be the analysis (resolution) in a given coordinate system, the synthesis (addition) of the vector components to reconstitute the original vector. There is, however, an important The vector sum of the components is unique— it is the particular vector that we are considering but the process of resolving the vector into components can be done in an infinity of ways, depending on the difference between thcse two operations. — choice of coordinate system. we are using a coordinate system based on orthogonal axes (whether Cartcsian or polar or anything else), the comIf ponents of the vector are easily found by multiplying the length of the vector by the cosine of the angle that the vector makes with cach of the coordinate axes in turn. Thus, for example, if we had a vector A confined to the xy plane [see Fig. 2-1 l(a)] its components in the xy coordinate system are given by A„ = A x = Acosa We know that separate angles 'U is hoped that /3 = /4cos<3 (tt/2) - «> but by introducing the two that lends itself to being we have a formalism this seetion may be helpful in a general way, but the only is the scalar produet of two arbi- fcaturc that will be specifically needed later trary veetors. 56 Space, (ime, and motion extended to the case of three dimensions, making use of the three and T that the vector makes with the three more general case there is no simple connection between the angles themselves, but we have the relationship separate angles a, 0, In this axes. cos 2 a + + cos 2 /3 cos 27 = 1 Our two-dimensional case corresponds to putting 7 = w/2. The total vector A in Fig. 2- 11 (a) can of course be written A = AJ + A v\ = + {A cos d)\ (A cos ;8)j A very convenient way of expressing such results is made possible by introducing what This is A tween any two vectors, is called the scalar product of two vectors. defined in general in the following way: If the angle be- is and B, is 0, then the scalar product, S, equal to the product of the lengths of the two vectors and the cosine of the angle because d. This product is conventionally written as it is scalar product (S) If for the vector also called the dot product A • B. Thus we have = A B = AB cos d B we now choose one or other of the unit vectors of an orthogonal coordinate system, the scalar product of with the unit vector is just the component of A along A the direction characterized by the unit vector: Ax = A Ay = A i Thus the vector A can be A = + (A-i)i • j written as follows: (A-j)j This result can, in fact, be developed directly from the basic A statement that can be written as a vector sum of components along x-and y: A = Ax + Ay\ \ Forming the the unit vector A Now • (i i • 57 The i, = A x (i i) unit length = both sides of scalar product of 1 this equation with we have i) + and A„(i i) i) = (j • 0, and the values of resolution of vectors because these vectors are are and t/2, respectively. all of Thus ; we have a more or automatic procedure for selecting out less and evaluating each component If one were to take in turn. no this above development further, the would seem perhaps pointlessly complicated. more apparent if is interested in relating different coordinate vector in given one Its value becomes the components of a Consider, for systems. example, the second set of axes (x\ y') shown in Fig. 2-1 1 (a) they are obtained by a positive (counterclockwise) rotation from The vector the original (x,y) system. A then has two equally valid representations: AIf + AJL we want = AJV + A u '¥ A„j to find A» scalar product with A,' = • • i') + = i' cos 6 • j terms of A„(i at Fig. 2-1 l(a), Looking i AS in A x and A v we just form , the throughout. This gives us i' i' i') we = see the following relationships: cos f | - 6 j = sin Hence A z = A z cos ' + A„ sin Similarly, A y = AA-f) ' = + AJl'f) /*xCOs(;; +B + ) /*„COS0 Therefore, Ay = ' -A x sin d + A y cos 6 This procedure avoids the need for tiresome and sometimes awkward considerations of geometrical A projections of the vector onto various axes of coordinates. The same approach can be useful into nonorthogonal components. if a vector is to be resolved Consider, for example, the shown in Fig. 2-1 (b). The axes are the lines Osi and Os 2 and the components of the vector A in this system are OE and OF. If we denote unit veetors along the coordinate direetions by ei and e 2 we havc situation 1 , , 58 Space, timc, and motion A = siei we form we havc If s\ + + S2*2 = A zi + A„\ the scalar product throughout with, let us say, ei, J2(e2 • ei) = A x (i ei) + /i„(j ci) • Given a knowledge of the angles between the various axes, is a linear equation involving the two unknowns Si and s 2 this . A second equation can be obtained by forming the scalar product throughout with e 2 instead of ei, and it then becomes possible to solve for Si and s 2 separately. It is to recognize that the base vectors e! important, in this case, and thogonal, so that the scalar product (e, e 2 are • no longer or- does not vanish. e2 ) The use of oblique coordinate systems of this kind is, however, rather special, and as a rule the resolution of a vector along the three independent directions of an orthogonal coordinate system is the reasonable and useful thing to do. VECTOR ADDITION AND THE PROPERTIES OF SPACE You may tion, 1 be tempted to think that the basic law of vector addi- and the fact that the final result is independent of the order which the combining vectors are taken, is more or less obvious. Let us therefore point out that it dcpends crucially on having in our space obey the rules of Euclidean geometry. If we are dealing with displacements confined to a two-dimensional world as represented by a surface, This it is essential that this surface be flat. not just a pedantic consideration, because one of our is most important two-dimensional reference frames — surface is curved. As and flat, earth's long as displacements are small compared to the radius of the curvature, our surface poses — the is for practical pur- our observations conform to Euclidean plane geometry, all is well. But if the displacements along the surface are sufliciently great, this idcalization cannot be used. For cxample, a displacement 1000 miles eastward from a point on the equator, followed by a displacement 1000 miles northward, does nol bring one to the same placc as two equivalent displacements 59 (i.e., again on grcat circles intersecting at right angles) taken in This scction can be omitted without loss of continuity. Vector addition and the properties of space Fig. 2-12 Successiue displacements on a sphere are not commutative if the sizes of the displacements are not small compared to the radius of the sphere. the opposite order; This means, in it effect, misses by about 40 miles! (See Fig. 2-12.) that the correctness of vector addition matter for experiment and that tests for departures from it is a can make deductions about the geometrical properties space in which we operate. For example, we could take be used to of the the results of sensitive measurements on the difference between the two possible ways of making two successive displacements on a sphere and use the data to deduce the radius of the sphere. When we acknowledge that the space of our ordinary experience is really three-dimensional, then, of course, we look at from a different point of view. We recogon the surface of a sphere can actually be scen as displacements in a three-dimensional world that obeys the foregoing analysis nize that displacements a merely two-dimensional Euclidean geometry rather than in world that appears, within to be non-Euclidean. this point itself, a very intercsting speculation suggests say that our space of three dimensions Is it possible that the result is itself: But at Can we rigorously Euclidean? of adding displacements along the three basic coordinate dircctions is dependent to some minute on the order of addition? If this were discovered to be the case, then we might proceed by analogy and introduce a fourth spatial dimension, associated with some characteristic extent radius of curvature, such that our non-Euclidean space of three dimensions could be described as Euclidean of four dimensions. 60 Space, tirae, and motion in a "hyperspace" Some distinguishcd scientists, beginning with the great Kari Friedrich Gauss ("the Prince of mathematicians") ' have sought by direct observation to test the validity of Euclidean geometry, by measuring whether the angles of a closed triangle add up to (For example, in the non-Euclidean space repre- exactly 180°. sented by the surface of a sphere, the angles of a triangle add up to more than 180°.) No departure from a Euclidean character for three-dimensional space has been detected through such observations. The concept that space may, however, be "curved," and that this curvature might be revealed if one could only carry out observations over sufficiently great distances, occupies an important place in thcoretical cosmology. You may have read about the curvature of space another in connection— Einstein's theory of gravitation—which describes local gravitational effects in terms of a modification of geometry the space surrounding a massive object such as the sun. We not pursue this topic here, although we shall touch on briefly at the end of our account of gravitation (Chapter in it in shall very general 8). TIME In the preceding sections of spatial displacements. we have developed the basic analysis To describe motion we must link such displacements to the time intervals during which they occur. Before considering this as a quantitative problem, let us very supplement the remarks that we made at the beginning of the chapter concerning the actual nature of time. This is, of course, a hugc subject that has engaged the thoughts and speculabriefly tions of men— philosophers, scientists, throughout history, and continues to do and humanists alike— so. We shall not presume do more than to cxamine one or two aspects of the problem from the standpoint of physical seienec. The sense of the passage of time is deeply embedded in to every one of us. We know, in some elemental sense, what time is. 'Kari Friedrich Gauss (1777-1855) was one of the outstanding mathemaIn the originality and range of his work he has never been surpassed, and probably never equaled. He delved deeply into astronomy ticians of all time. and geodesy and was perhaps the first to recognize the possibility of a nonEuclidean geometry for space. See E. T. Bell's essay about him in The World of Mathemalics (J. R. Newman, ed.), Simon and Schuster, New York, 1956. 61 Time — ' But can we say what it is? The distinguished Dutch physicist H. A. Kramers once remarked: "My own pet notion is that in human thought the world of particularly, the those to which To gencrally, most important and most if one begin to see that tion of time may in physical science fruitful concepts are impossible to attach a well-defined meaning." it is nothing, perhaps, does this apply Nevertheless, and tries to more cogently than to time. analyze the problem, one can perhaps not entirely elusive. Even though a definihard be to come by, one can recognize that our it is concept of the passage of time is tied very directly to the fact that things change. In particular we are aware of certain recurrent events or situations — the beats of our pulse, the daily passage We of the sun, the seasons, and so on. treat these as almost subconsciously though they are markers on some continuous line But that already exists— rather like milestones along a road. all we have the rest is in terms of direct an knowledge intellectual construction. is the set of markers; Thus, although it may be valuable to have an abstract concept of time continuously flowing, our first-hand experience of a device called "a clock." is only the observed behavior In order to assign a quantitative measure to the duration of some process or the interval between two events, we simply associatc the beginning and end points with readings on a clock. use of a recurrent It is phenomenon not essential that the clock devices such as water clocks and graduated candles, or parallels such as continuously but we do ask that make — one need only consider ancient modern weakening radioactive sources provide us with a means of marking off it some recognizable way, and most such make use of repetitive phenomena of some successive intervals in devices do, in fact, kind. How do we know that the successive time intervals defined by our chosen clock are it is truly cqual? ultimately a matter of faith. The No fact is that clock is we don't; perfect, but we have learned to recognize that some clocks are better than others — better in the sense that the segments into which they divide our experience are his wristwatch and watch, however, is 'Physical Sciences and Press, Princeton, 62 N. J., tells more nearly equal. us that our pulse found to be Human is A doctor observes irregular; the wrist- itself irregular when compared Values (a symposium), Princeton University 1947. Space, time, and motion more critically against a crystal-controlled oscillator; the oscil- wandcrs noticeably when checked against a clock based on atomic vibrations. Whether or not the uniform flow of time, as lator an ultimate abstraction, has any physical meaning, fact is we approach that We flow existed. observing its the the remarkable measurement of time as if this consistency and reproducibility, so that we can quote its steady evaluate the behavior of any given clock by measure of a time or a time And specified range of possible error. in using it interval with a when we proceed then, from individual measurements to general equations involving we introduce time, the symbol t and treat as a continuous r variable in the mathematical sense. UNITS AND STANDARDS OF LENGTH AND TIME Most of our discussion of motion will be in terms of unspecified positions and times, represented symbolically by It r, /, and so on. should never be forgotten, however, that the description of actual motions involves the numerical measures of such quan- and the use of tities Our choice universally accepted units and standards. of acceptable standards of both distance and time is the result of a continuing search for the highest degree of consistency and reproducibility and present state in such measurements. The evolution of this process is briefly summarized below. Length The current Standard of length — the meter — was introduced, along with the rcst of the metric system, in the drive for and cultural order that developed of the 18th century. sent 1 The meter was -7 ten-millionth (10 ) in France scientific in the latter half originally intended to repre- of the distance from pole to equator of the earth along a meridian of longitude. But it proved impossible to construct any sufficiently precise Standard basis of this definition, and the meter on the was then defined as the length of a particular metal bar kept in Sevres, near Paris, and finally as a multiple —most line of the wavelength of a charaeteristie spectral recently (since 1960) as being a distance equal to 1,650,763.73 wavelengths of orange-red light from the isotopc krypton 86. Although the use of the extremely small 63 Units and standards of length and time units of length represented by light waves means that the meter can be defined with immense precision — to part in 10 1 8 or better — there is the disadvantage that an object as long as a meter cannot be directly measured, in a one-step process, in terms of light waves from ordinary sources. on observing if is that the measurements depend the distance in question becomes more than about development of lasers of over 100 m. some It wash out 1 ft. The has completely transformed this situation, up interference effects have been observed and at The reason optical interference effects that begin to to path lengths thus seems quite probable that the meter will future date be defined in terms of an optical wavelength obtainable from a laser source, perhaps one of the characteristic spectral lines of neon in a helium-neon laser. Time The process of defining a Standard of time involves a feature that from the establishment of a Standard of length. This is that, as Allen Astin has remarked: "We cannot choose a particular sample of time and keep it on hand for 1 reference." We depend upon identifying some recurring phesets it significantly apart nomenon and assumirrg that it always supplies us with time same length. The Standard of time the second intervals of the — the assumed constancy of the — was originally based on earth's rotation. as being equal to 1/86,400 of a mean solar day It was first — i.e., the average, defined from noon to noon or midnight to midnight at a given placc on the earth's surfacc. This is an awkward definition, because the length of the day, as measured from noon over 1 year, ol the time to noon, is not a constant; it varies because the earth's speed and its distance from the sun are continuously changing during one complete orbit. A logically more satisfaetory definition of the second can be based on the sidereal day— the time for any given star to return to the same position overhead. If the earth's rotation were truly uniform, the length of every sidereal day would be the same. In faet, it has gradually come to be rccognized, thanks to 'Allen V. Astin, "Standards of Measurement," Sci. Am., 218 (6), 50 (1968). people would, however, argue that even a solid, tangible bar as a length Some Standard 64 is equally vulnerable on philosophical Space, time, and motion and logical grounds. the extraordinary precision of astronomical raeasurements extended over thousands of years, that neither the length of the on solar year nor the earth's rate of rotation its axis is exactly constant, the latter in particular being subject to minute but The year has been found abrupt variations. to be lengthening about £ sec per century, so that in 1956 the second was redefined as being equal to 1/31,556,925.9747 of the tropical at the rate of (A year 1900. tropical year is defined as the interval of time between two successive passages of the sun through the vernal We equinox. "latehes on" poses.) It shall not attempt here to deseribe just to a how one second of the year 1900 for calibration pur- should be recognized that the variations being discussed here are fantastically small, as implied by the ability is to define the year in terms of the second to 12 significant figures. Finally, in 1967, the use of atomic vibrations to specify a time Standard was adopted by international agreement ; it defines the second as corresponding to 9,192,631,770 eyeles of vibration in an atomic clock controlled by one of the charaeteristie frequencies associated with atoms of the isotope cesium 133. Quite apart from the practical challenge of defining units and standards of time with the maximum attainable precision, there are some questions of fundamental time as defined by keep step at all celestial motions, controlled by gravitation, epochs with time as defined by atomic vibrations, atom? controlled by electric forees within the in Does interest involved. 1938 by P. A. M. It was suggested 1 Dirac that the constant of universal gravita- — with a "time constant" tion might be slowly changing with time on the order of the age of the universe itself, i.e., about 10 10 our astronomical and atomic standards years. If this were would, in the long run, be found to reveal diserepancies. true, This matter of units and standards is one that most of us do not bother our heads with. We think we know well enough what is meant by a meter and a second; and a ruler or a wateh is usually near at hand. But perhaps the above diseussion may help to suggest that the detailed story of sures are defined, redefined, is how these basic and made more and more a quitc faseinating business — especially, mea- precise perhaps, for time, to Dirac, a British theorelical physicist, was one of the leaders in the development of quantum thcory around 1926-1930. He was awarded the Nobel prize for this work. 65 Units and standards of length and time ' which astronomical observations over the centuries have contributed data of a refinement that almost passes belief. SPACE-TIME GRAPHS The primary data the description of any motion will be a set in of associated measures of position and time. for example, be a tabulation single stroboscopic i n Such data might, an astronomer's logbook, or a photograph such as Fig. 2—1. In general the statement of position at any instant will require the use of three coordinates, corresponding to the three independcnt dimensions of space. may I n many circumstances, however, the motion be confined to a planc, requiring two coordinates only, or to a single line, so that a single positional coordinate sumces. In this last case, and especially line, it is often Fig. 2-13 if the motion is along a straight extremely convenient to display the motion in Example of a space-lime graph for a one-dimensional motion. the (From PSSCfilm, "Straight Line Kinematics," by E. M. Hafner, Edueation Deiselopment Center Film Studio, Newton, Mass., 1959.) Time 'For further reading see, for instance, An Introduction to the Physics of Mass, Length and Time, by N. Fealher, Edinburgh University Press, Edinburgh, 1959. On the matler of time in parlicular, which probably holds the greatest books and articles are recommended: J. T. Fraser (ed.), The Voices ofTime, Gcorgc Braziller, New York, 1966; T. Gold and D. L. Schumacher (eds.), The Nature of Time, Corncll University Press, Ithaca, N. Y., 1967; G. M. Clemence, "Siandards ofTime and Frequency," Science, 123, 567 (1956); Lee Coe, "The Nature of Time," Am. J. Phys., 37, 810 (1969); Richard Schlegel, Time and the Physical World, Dover, New York, interest, the following 1968. 66 Space, time, and motion — ' terms of a space-time graph in which, as a rule, thc time is regarded as an independent variable, plotted along the abscissa, and the position such a graph. way is It plotted along the ordinate. Figurc 2-13 shows has the great merit that it conveys directly, in a that a numerical table cannot, a complete picture of One can immcdiately motion. minimum from the distance identify points of a given maximum and origin, regions of time in which the motion temporarily ccases altogether, and so on. VELOCITY The central concept in the quantitative description of that of velocity. a It is The vcctor. motion is quantitative measure of velocity, in the case of one-dimensional motion, is one of the we can extract from a space-time graph such as Fig. 2-13. Our way of designating velocities miles per hour, meters per second, and so on is a constant first pieces of information that — reminder of thc fact that velocity is these separatc mcasures of spacc and you that although physicists of measurcment of based on Has it ever struck have invcntcd names for the units a deriued quantity, time. of physical quantitics, they have all sorts never introduced special names for units of velocity? men have done though, with their unit the knot, equal to it, nautical mile per hour.) In nature otherwise, for although we have not that directly rccognizablc as a is length or time, magnitude c It = we do find a itself, ± things 1 seem to be quite as yet idcntified anything fundamental natural unit of — thc fundamental unit of velocity (c) of the velocity of light in (2.997925 (Seafaring 0.000001) X has become customary in empty space: 10 8 m/sec high-energy partiele physics to express velocities as fraetions of c. And in connection with high-speed (light, the in a comparable way, Mach number is used to express the speed of an aireraft as a fraelion or multiple of the speed of sound in air. But this does not alter the fact that 'Such graphical representations of correlated quantities often provide a much vivid insight into a situalion ihan do numerical tab- more immediale and ulations or algebraic formulas, and cven a rough graph, sketehed freehand, can be a great aid to thinking aboul a situalion. A facility in drawing and interpreling graphs as cxpressions of physical relalionships is well worth devcloping. 67 Velocity our basic description of velocities is in terms of the number of units of distance per unit of time. The measurement of a velocity requires at least two measurements of the position of an object and the two corresponding measurements of timc. Let us denote these measurements by and (ri, Ij) and (r 2 , t Using these we can deduce the magnitude 2 )- of what we can direction loosely call the average velocity between those points: T2—U —_ Vav t% However, — this /l average velocity Sometimes we may interesting quantity. versus linc, s (let / not, in is most cases, a very find that a us assume a onc-dimensional motion) graph of s is a straight so that the value of v deduced from any two pairs of values and t is the same. But there is a What can we do about problem: much more basic and general defining and evaluating the an arbitrary instant in a nonuniform motion such as that represented by Fig. 2-13? The next section is devoted to this question, which is of fundamental importance to the whole of the velocity at mathematical analysis of motion. INSTANTANEOUS VELOCITY Richard P. Feynman for speeding at tells the story of the lady who is caught 60 miles per hour and says to the police ofncer: "That's impossible, sir, I was traveling for only seven minutes." 1 The lady's objection does not convince us (or the police omcer); we understand that what is at issue is not the persistence of a uniform motion for a long time but the property of the motion as measured over a time interval that might be arbitrarily short. In order to talk about this in specific terms, imagine that alongside a straight section of a point, P, we have placed road in each direction from a chosen a set of equally spaced poles, say at Feynman with entertaining and instructive can be found in The Feynman Lectures on Physics, Vol. I (R. P. Feynman, R. B. Leighton, and M. Sands, eds.), Addison-Wesley, Reading, Mass., 1963. These lectures, of wonderful freshness and originality, range over the whole of physics and provide rich and exciting fare for everyone, whether a beginner or a veteran in the subject. Feynman, one of the most outstanding physicists of our time, was awarded the Nobel prize in 1966 for 'This story, further embellished by details, his 68 fundamental contributions to quantum Space, time, and motion field theory. Fig. 2-14 Arrange- menl for inferring the instantaneous velocity of a car as it passes the point P. 5-meter intervals (see Fig. 2-14). On clock with a sweep second hand. The A and run continuously. pole, so as to each pole movie camera is an electric clocks are synchronized is placed opposite each photograph the pole, the clock face, and the road. A car comes along the road; the cameras are set running and photographic records are developed. of a race. Each film the front pole. will is more or For a given pair let m) and let the Then the ratio As/A/ range of distance Ax and after the central point P. differcnce of time readings be is the average velocity over the centered on P. Now motion of the car has been extraordinarily be fitted by a smooth curve values of As. We = is We the value of As/A/ at A/ (a) = Eual- taneous velocity ds/dt by extrapolation to of a graph of against As. {b) Evalualion of the instantaneous velocity ds/dt by extrapolation WAt = of a graph ofAs/At against At. 69 it can baekward, and the value of could equally well plot the values of intervals A/, as ualion ofthe instan- As m As/At erratic, the points our measure of the instantaneous velocity at the central point P. 2-15 Unless the that flattens out for the smaller extrapolate As/A/ against the time Fig. construct a graph, as 2-1 5(a), of As/A/ as a funetion of As. in Fig. As/A/ for As exactly in line with the the separation in distance be called As (a multiple of 10 shown less assemble the records. Take the data for the poles in pairs, equal distances before called A;. the the photofinish contain one frame in which, let us say, bumper of the car We now It is like Instantaneous velocity is shown in Fig. 2- 15 (b); again the same, even though the The "slope" of a graph ofone physica! 2-16 Fig. guantity against another is is not primarily geometrical; defined by the ratio of the changes measured in whatever U each unils are appropriale. looks somewhat different. And here, in physical what we mean by evaluating the limit of the average when the range of position or time over which As/At is graph itself terms, is velocity, A Q and Aq, evaluated shrunk to zero. is The process described above corresponds — matical process of determining a derivative displacement with respect to time. Using the Standard calculus the context of the above discussion almost notation, which in speaks for we write itself, to the mathe- in this case, of instantaneous velocity o = —=— dt lim (2-5) A«-oA/ The synonym quantity ds/dl, a for the limit of As/At, is, in mathematical parlance, the first geometrical terms, reprcsents the slope of (a tangent to) the graph of s versus t it derivative of 5 with respect to at a particular value of In t. t. [A note concerning the mcaning of the word "slope" graphs of physical data Fig. — 2-16 is in is order. — The graph e. g., in as in a display of the numerical measure of one physical quantity plotted against the numerical measure of another, using and dictated by convenience no physical significance to the measured by a protractor. By "slope" scales that are entirely arbitrary Thus alone. there is, in general, inclination of such a line as we mean simply the ratio of the change in the quantity represented on the ordinate to the corresponding change in the quantity represented on the abscissa— e.g., (Q 2 — <2i)/(<72 — <7i)- In numerator and denominator are each a pure number this ratio, times a unit, and the slope then expressible as the quotient of is these numbers, labeled with its own characteristic units — e.g., metcrs per second.] same If the ideas as above are applied to changes with time of the displacement in space, we arrive at a general definition of instantaneous velocity as a veetor: .. v = hm Ar — — = dt dt In Fig. 2-17(a) Clearly, (2-6) Ar—O Ar if wc we indicate what the evaluation of Ar as in Fig. 2-17(b), the instantaneous velocity veetor 70 entails. display the path of an object as a continuous curve, Space, time, and motion is tangent Fig. 2-17 diagram (a) Vector to define a small chaiige ofposition. (b) Instan- taneous velocity vector, tangent to the path. to this curvc; this With this is implicit in the definition. departurc from straight-line notion, we shall draw attention to the distinetion that is made technically (but not always faithfully observed) bctwcen the words velocity and speed. The speed thus, by is the magnitude of the vector velocity v. definition, a positive, scalar quantity. The The spced is o that figures in our analysis of straight-line motion has, in faet, represented the velocity; it may which of course direetion of Even if is motion we take on negative as well as positive values, all the information needed to specify the in the one-dimensional case. are dealing with motion along a single straight line, the use of the vectorial deseription of position and velocity will be necessary Fig. 2-18 if displacements are referred to an origin not (a) Vector positions in a straight-line motion, referred to an origin not on the line. diagram for a curvilinear path. 71 Instantancous velocity (b) Corresponding on the line itself. because in fact, a very instructive thing to consider It is, motion from a straight-line this point of view [see Fig. 2-18(a)], helps us to scc the one-dimensional situation in it We larger context. can become displacement vcctor Ar the position vcctor in general, in a different direction from when the latter is referred to an arbitrary makcs the transition from the description of r, This then origin O. is, its accustomed to the idea that the a rectilinear path to the description of an arbitrary curved path scem [Fig. 2-18(b)] although may it less abrupt. It also emphasizes the fact that be very convcnient, in the motion, to choose an origin on the line may necessary and case of straight-line itself, this is certainly not not always be possible. RELATIVE VELOCITY AND RELATIVE MOTION Sincc the vector velocity the time derivative of the vector dis- is placement, the velocity of one object relative to another is just Thus if one the vector difference of the individual velocities. object is from object R = The and another object at r! r2 to object 2 1 Vm 1 , at r 2 , the vector distance R given by is - n rate of changc of to object is R then the velocity, V, of object 2 relative is and wc have — - ——— dt dt dt i.e., V = v2 - ~7) (2 vi This relative velocity V is the velocity of object 2 in a frame of reference attached to object 1. In diseussing frames of reference earlier in this chapter, we pointed out how the choice of some particular frame of reference may be advantageous because it gives us the clearest what is going on. Nothing could illustrate this better than the practical problems of navigation and the avoidance of collisions at sea or in the air. Imagine, for example, two ships picture of that at some instant arc in the situation shown in Fig. 2-19(a). The veetors v, and v 2 represent their velocities (which we take to be constant) with respect to the body of water in which they both 72 Space, time, and motion Fig.2-19 (a)Paths of two ships moving at consiam velocity along courses that intersect. ship B (b) Path of relative to ship A, showing that they do not collide even lhough their paths cross. move. The paths of motion from the the ships, extended along the directions of points initial Will the ships collide, or The answer distance? B, intersect at a point P. to this question ocean frame, but stiek to the A and they pass one another at a safe will if we is not at all elear if deseribe things we from the standpoint of one of the two ships the analysis becomes very straightforward. Let us imagine that we are standing on the deck of the ship marked A Putting ourselves in that frame of reference means giving ourselves the velocity Vi with respect to the water. But from our standpoint it is as if the water, and everything else, were given a velocity equal and opposite to vj. Thus to every motion as observed in the ocean frame we add the veetor — Vi, as implied by Eq. (2-7). This automatically, and by definition, . brings A to rest, as it were, and shows us that the velocity of the ship B, relative to A, and —V], the ships as is shown is obtained by combining the veetors v 2 in Fig. can see the whole picture. indicated miss by A by veetor distance between So now we line shown, as B follows the straight several successive positions in the diagram. the distance to the line of V. is The 2-1 9(b). unaffected by this change of viewpoint. AN, The time equal to the distance at the perpendicular distance which BN divided this elosest It will from A approach oecurs by the magnitude of V. Thus seems to sweep aeross /4 's bow, more or less sideways. If you have had oceasion to observe a elose encounter of this sort, especially if it is out on the open water with no landmarks in B sight, 73 you will know Relative velocity that and it can be a curious experience, quite relative motion disturbing to the intuitions, because the observed motion of the other ship seems to be unrelated to the direction in which it is pointing. PLANETARY MOTIONS: PTOLEMY VERSUS COPERNICUS Some and of the most fascinating problems in the study of motion, in particular of its relative character, have arisen in man's attempts to elucidate the motions of the heavenly bodies, including our the first own earth, through space. Observational astronomy, of the exact sciences, has yielded data of marvelous accuracy for several thousand years. But the question has always been how to interpret these data. main features of the problem. The first thing to recognize Fig. 2-20 (a) Palh of Venus omong the stars during a 6- month period, showing reoersed (retrograde) motion at one stage. (b) Similar set ofob- servalions on Mars. {.Both E. M. diagrams after Rogers, Physics for the Inquiring Mind, Princeton Uniuersity Press, Princeton, N. J., 1960.) IA Space, time, and motion is Let us consider some of the that naked-eye astronomy is, almost exclusively, the study of directions rather than distancei These unaided observations reveal nothing about the distances of the The great Greek astronomers Aristarchus (third and Hipparchus (~150b.c.) did make reasoned stars. century B.c.) moon estimates of the distances of sun and successfully), (the latter very but the only direct clues to the distances of the ' planets are through such quantitative evidence as changes of apparent brightness with time, suggesting that whatever the distances of the planets from the earth may be, they undergo systematic variations. Thus, as the practice, the positions is still of astronomical objects are defined in the of their directions only (this being all a telescope) and can be described as first we need if place in terms they were points on the surface of a sphere of large but arbitrary radius — with sphere its own In these terms the primary data It is —the celestial center at the earth, and with a polar axis and an equator defined by the earth's are of the type them with to find shown axis of rotation (cf. Fig. 2-2). on the motions of the planets in Fig. 2-20. already a tribute to the genius of the early astronomers that they were able to visualize such strange-looking paths as the projections, on the celestial sphere, of orbital motions of In particular, the belief took hold that the orbits various kinds. must be combinations of circular motions. This is not the place to go into a detailed account of the problem; some outstandingly 2 Instead, we shall simply focuson fine accounts exist elsewhere. an idealized presentation of the two main models: an earthcentered (geocentric) or a sun-centered (heliocentric) solar system. The most intuitively reasonable picture of the universe, in terms of everyday experience, is undoubtedly one that places the earth at the center of everything. Not one of us, without benefit of hindsight, could interpret his first impressions in any other way, and the ancient descriptions, such as the biblical one in Genesis, are entirely justifiable in these terms. Alexandrian astronomer Ptolemy (—150 a.d.) picture into It who was the built this a quantitative model of planetary motions and 'See the problems at the end of this chapter, and also Chapter 8. M. Rogers, Phyiics for the Inguiring Mind, Princeton University Press, Princeton, N.J., 1960, Chaps. 12-18, or E. C. Kemble, 2 See, for example, E. Physical Science, lis Structure and Developmenl, Mass., 1966, Chaps. 1-5; or any of a MIT Press, Cambridge, number of excellent books on elementary astronomy. 75 Planetary motions: Ptolemy versus Copernicus Fig. 2-21 in the Apparent matian of a planet as explained (a) Ptolemaic syslem. The planet epicycle whose center the earth, E. {b) C follows a P moues on Copernican explanation ofobserved The epicycle of (a) Is seen as being a of the earth' s own motion around the sun. motion. described earth's daily rotation, the be l The motion of a 1 planet, is approximated by imagining that a point closely fairly an approximately year, around the earth, E, as however, is compound. It can motion of the sun circular path, with a period of Figure 2-21 (a) Setting aside the effects of the illustrates the essential features. center. reflection work, the Almagest. in his great it the circular path around C around a circular path, and that the planet, P, another circular path with respect to C as center. This travels uniformly travels in extra circle motions, is called an epicycle; the combination of these two they are in the same plane, gives rise to a complicated if path that can have backward loops as shown. Precisely the same result would be obtained circles. If it we onto the celestial this motion from the earth, and sphere or to any other constant obtain almost the kind of variation of angular posi- We can come even closer by the plane of the epicycle out of the plane of the primary tion that path, is shown which in Fig. 2-20. carries the the celestial sphere, 76 the roles of the two (Verify this.) distance, This we interchanged we now imagine viewing projecting tilting if is sun eastward through the constellations around known as the ecliptic. Space, time, and motion circle a little. To the motion of a particular planet fit it is necessary to choose appropriate values for the ratio of the radii of the two and also for the times to make one complete circuit a noteworthy fact that in two cases (Mercury and circles of each. It is Venus) the period of the primary circle is exactly 1 sidereal year (i.e., the period for one complete orbit of the sun around the ecliptic), and in the other three cases (Mars, Jupiter, and Saturn) the period of the epicycle is Indeed, this can be sidereal year. 1 taken as the crucial clue to what we now regard as the truer picture. Suppose that we now place the sun Fig. 2-21 (b), and make the earth same radius the planet, P, travels at the center, as in around a travel circle that has Then the as the epicycle in Fig. 2-21 (a). around another simple circle, if of radius equal to that of the primary circle in Fig. 2-21 (a), the relative positions velocities of and We before. E and P can be made precisely the same as have drawn the positions of E and P in the The reason diagrams to display this exact correspondence. the appearance of the sidereal year in one or other of the two for two component motions of a planet in the Ptolemaic model is now very clear, and much of the arbitrariness of the whole description disappears. It is this new, heliocentric, description with which we name of Copernicus. He presented it in great detail principal work, De Revolutionibus Orbiwn Celestium associate the in his ("On the Revolutions of the Celestial Spheres") published in Actually, the 1543, the year of his death. the sun occupied the central position years earlier by Aristarchus, that the sun was much suggestion that who from other observations knew bigger than the earth (although he seriously underestimated just no first had been made about 1800 how huge and distant it is). There is was developed interesting, by the record, however, that the heliocentric theory in quantitative detail before Copernicus. It is way, to note his clear understanding of the relativity of motion. Here "For is a translation of all change his in position own which statement of the principle: is seen is due to a motion either of the observer or of the thing looked at, or to changes in the position of both, provided that these are different. things are is moved equally same things, For when no motion perceived, as between the object seen and the observer." 'From Arthur New 77 relatively to the Berry, A Short History of Astronomy, Dover l Publications, York, 1961. Planetary motions: Ptolemy versus Copernicus Given the persistent intrusion of the sidereal year into the Ptolemaic scheme, may seem it surprising that the heliocentric much picture did not prevail at a when earlier stage, especially had been recognized by Aristarchus about 400 years its possibility before Ptolemy's day. It must be remembered, however, that we have been presenting a greatly oversimplified model of the system, and the ancient astronomers were legitimately solar worried over discrepancies between thcse idcalized models numerous auxiliary introduce were driven to nicus and Both Ptolcmy and Coper- the precise, hard facts of observation. circular motions to obtain even approximate agreement between theory and observation. The Copernican scheme, in the form in which Copernicus himself developed it, was not in fact notably less Not until the arbitrary or less complex than the Ptolemaic. theory could free itself of the of cirele as the basis motion was a fundamental solution to be all celestial finally attained, although the introduetion of dynamical considerations of foree and the laws of motion — the laws — transformed the context within which the observations were interpreted. these questions in Chaptcr 8 and, more We shall fully, in come back Chaptcr to 13. PROBLEMS 2-1 Slarting from a point that can be laken as the origin, travels 30 miles northeast a course that heads in a SSW straight line, a. ship and then 40 miles on making a countcrclockwise Find the x and eastward, y northward) and its direetion (a angle of 247^° with a reference linc drawn eastward). y coordinates of its final posilion (x distance from the slarting point. 2-2 The scalar (dot) produet of ABcosOah, where 6 mi (a) By expressing nents, show that cos d a u (b) = is + A - B, is equal to the relation between rectangular and spherical show two points (R, and (R, 6\, <p\) cos0i2 = cos0i cos 62 + compo- A,B Z polar coordinates [Eq. (2-4)], radii to veetors, the veetors in terms of their Cartesian A Z BX + AyBy AB By using two the angle between the veetors. that the angle 0i2 02, <P2) sin 0i sin between the on a sphere $2 cos (^2 — is given by <pi) (Notc that the distance between the two points as measured along the 78 Space, timc, and motion them is equal to Rd 12, where 0i2 is This can be used, for example, to calculate great circle that passes through expressed in radians. mileages between points on the earth's surface.) 2-3 New Calculate the Cartesian coordinates of (a) N, 74° W) and Sydney, Australia (41° (34° S, 151° E). York, U.S.A. Take an origin of coordinates at the earth's center, with a z axis through the north x axis passing through the equator at mean radius is 6370 km. pole and an The earth's the zero of longitude. Find the distance along an imaginary straight tunnel bored (b) through the earth between Compare (c) New York and Sydney. the result of (b) with the shortest practicable route between these points by a great-circle You flight. can either calculate using the result of Problem 2-2(b), or measure this, directly it on a globe with the help of a piece of string. 2-4 (a) Starting from a point on the cquator of a sphere of radius R, a partiele travels through an angle a eastward and then through an angle /3 along a great circle toward the north pole. If the of the point is that coordinates are its final + x2 Verify that y 2 + z = R2 2 tiele if it first travels of the same par- final position through an angle a northward, course by 90° and travels through an angle /3 then changes along a great circle that out eastward. starts Show (c) that the straight-line distance points of the displacements in (a) and (b) As 2 = 2« 2 (sin/3 - sin Using the above (d) (1) is As between the end given by a cos P) 2 result, (p. 60) that there is a difference of position - Find the coordinates of the (b) initial x — R, y = 0, z = 0, show R cos a cos /3, R sin a cos /3, and R sin 0. taken to correspond to check the statement in the text of about 40 miles in the end points a displacement of 1000 miles eastward on the earth along the equator, followed by a displacement 1000 miles north, and (2) a displacement of 1000 miles north, followed by a displacement of 1000 miles starting out eastward on another great circle. The approximation 1000 miles. cos 6 s» 1 — (Put 2 d /2 Ra = Rp = will be found useful.) 2-5 If you found yoursclf transported to an unfamiliar planet, what methods could you suggest To To (a) (b) 2-6 The verify that the planet is spherical? find the value of radius of the earth its radius? was found more than 2000 years ago by Eratosthenes through a brilliant piece of analysis. andria, at the day at noon, He 79 also the Nile, the sun's rays knew Problcms mouth of and observed were at 7.2° to the He lived at Alex- that on midsummer vertical (see the figure). that the people living at a placc 500 miles south of ^-AleKandria ^ 7.2° Equator / / / 500 milesi\ sun Alexandria saw the sun as being directly overhead at the same date and time. From this What was the earth's radius. 2-7 It information, Eratosthenes deduced the value of his answer? has been suggested that a fundamental unit of length is repre- a nucleon diameter, and that a represented by the time it would take a sented by a distance about equal to fundamental unit of time light signal (i.e., Express the radius of the universe and the age a nucleon diameter. of the universe A 2-8 in particle fiecting walls at is the fastest kind of signal achievable) to travel across terms of these units, and ponder the is confined to motion along the x axis between re- and x = x = Between these two a. freely at constant velocity. Construct a space-time (a) If the walls upon each = i>2 A particle that starts at x collides with an = —v/2. Construct and after collision (a) For the case that moves upon reaching the magnitude of the velocity is —foi). at t = with velocity identical particle that starts at velocity it changes sign but not magnitude. reflection reduced by a factor/(i.e., 2-9 limits graph of its motion are perfectly reflecting, so that either wali the particle's velocity (b) If results. x = xo +v (along x) = with at / a space-time graph of the motion before the particles collide elastically, exchanging velocities. (b) 2-10 A tion is r(f) = For the case that the particle given by (a) xi moves along x = particles stick together the curve = Ax upon impact. such that its x posi- Bt. Express the vector position of the particle in the form + yy (b) Calculate the speed v path at an arbitrary instant 2-11 The refraction of considerations. 80 y 2 We (= ds/di) of the particle along this /. light may be understood by need to assume that Space, time, and motion purely kinematic light takes the shortest (timewise) path between two points (Fermat's principle of least time). Referring to the figure, medium point 2, B it medium 1 be vi and go from point A in Minimize with respect to x. the speed of light in let Calculate the time i?2- takes light to as a function of the variable x. to Given that = vi where the and c/rti n's are V2 known = c/ri2 as indices of refraction, prove Snell's law of refraction: = «i sinfli /»2sin02 A 2-12 At 12:00 hours ship port. the It is is 10 km east and 20 km north of a certain steaming at 40 km/hr in a direction 30° east of north. same time ship B is 50 km east and 40 At km north of the port, and is steaming at 20 km/hr in a direction 30° west of north. Draw (a) a diagram of this situation, and find the velocity of B relative to A. (b) If the ships continue to is their closest distance to 2-13 The distance from the above velocities, what A to B is /. A plane fiies occur ? it a straight course V relative from A air. Calculate the total time taken for this round trip speed v to is (a) B and move with one another and when does back again with a constant speed if to the a wind of blowing in the following directions: Along the line from A to B. (b) Perpendicular to this line. (c) At an angle d Show that the to this line. time of the round trip is always increased by the existence of the wind. 2-14 A ship is steaming parallel to a straight coastline, distance offshore, at speed V. A coastguard cutter, whose speed out from a port to intcrcept the ship. 81 Problems is v ( D < V) sets q^* *£ -o= -f: Port (a) Show that the cutter point a distance D(V 2 — must start out before the ship passes a v 2 ) 1 ' 2 1o back along the coast. (H ini: Draw a vector diagram to show the vclocity of the cutter as seen from the ship.) (b) Tf the cutter starts and when does out at the latest possible moment, where reach the ship ? 2-15 With respect to the "fixed Bulge rotates to keep pace with it stars/' the earth rotates once on its — one sidereal day that is how the sidereal day is defined. (a) The length of the year is about 366 sidereal days. By what axis in moon- amount sidereal is the mean solar day (from day ? (b) The moon completes one 27.3 sidereal days. That is, in this orbit with respect to the stars in time the line from earth to turns through 360° with respect to the Êarth's daily responding high tides on is longer than moon. (The high bulge at a fixed direetion with respect to the Show 2-16 that the daily lag is elose to 50.5 (a) The orbital radii of min tide moon—see (60 min = cor- solar day 1 is an ocean the figure.) ^ solar day). Venus and Mars are 0.72 and the radius of the earth 's orbit. moon The time between stars. successive days (24 hr) because of this motion of the rotation to noon) longer than the noon 1.52 times Their periods are about 0.62 and 1 .88 Using these data, construct diagrams by which the apparent angular positions of Venus and Mars change times the earth's year. to find how with time as seen from the earth, assuming that the orbits of planets lie in (b) the same plane. With respect to the Compare your ecliptic (the orbit) the planes of the orbits of all three results with Fig. 2-20. plane of the earth's Venus and Mars are tilted own by about and 2°, respectively. Consider how the apparent paths of Venus and Mars are affected by this additional feature. 3.5° 2-17 (a) What methods can you suggest for finding the distance from moon (without using radar or space fiight) ? the earth to the Moon 82 Space, time, and motion (b) The astronomer Hipparchus, more than 2000 years ago, found the distance of the moon as a multiple of the earth's radius by observing the duration of a total eclipse of the (see the figure). The rays from about ±5°, and the (the same days to l£ hr. moon itself Use and by the earth subtends an angle of just about ^° as the sun within about circle the earth, moon the sun have a spread of directions of 2%). The moon takes about 29 the duration of the total eclipse these data to obtain the moon's is about distance. 2-18 The astronomer Aristarchus had the idea of comparing the distances of the sun Moon and the moon from the earth by measuring the when the moon was exactly half angular separation 6 between them full (see the figure). Using our present knowledge of what these distances are, criticize the feasibility of the method. Aristarchus 6 = 87°. What result would this really is on earth - angle were uncertain by ±0.1°. Problems found Calculate what the angle and what error would be introduced Observer 83 imply? in the distance if this At present it is the purpose of our Author merely to investigate and to demonstrate some of the properties of accelerated motion (whatever the cause of this acceleration may be). galileo, Dialogues Concerning Two New Sciences (1638) Accelerated motions ACCELERATION from THn purely of motion descriptive point of view, the central feature velocity is — the instantaneous rate of change of But we must dig a position with time. deeper to get to the little quantity that proves to be the crucial one in relating motion per motion as governed by forces (dynamics). se (kinematics) to This is acceleration, the rate of Again we change of velocity with time. develop the basic ideas shall context of straight-line motion. in the first instance in the If the instantaneous velocity linear function of time, as in Fig. 3-1 (a), then found to be a conclude that the instantaneous acceleration is the same is we at all times and equal to the slope of the graph as evaluated from any two points on ! it. on the other hand, the values of the instan- If, taneous velocity define some sort of curve, as then we must obtain ticular value of by f in Fig. 3-1 (b), the instantaneous acceleration at a par- a limiting process: • • instantaneous acceleration a dv — —— A? a,_o Ad i- = lim .. ,. (3-1) di Thus the acceleration 'The units in which is the first derivative of the velocity with this is expressed will velocity divided by whatever unit of time the acceleration of a car is be defined by our chosen unit of found convenient. For example, is expressed most effectively and vividly in m.p.h. more analytical treatment of motions it is almost essential to use the same time units throughout, e.g., expressing velocities 2 Otherwise, when we see a symbol in m/sec and accelerations in m/sec such as i/, we have to stop and ask ourselves which of the different units it is measured in, and that makes for confusion and error. per second. But in the . 85 3-1 (a) Oraph ofv versus t for a Fig. uniformly acceleraled motion. (b) Motion with varying acceleration. In the indicated time interval Al ihe acceleralion is negative. respect to time. we however, If, wish to tie our definition of acceleration to a priinary record of position against time, then wc can write it as the second derivative of s with respect to /: (3-2) In general, we must be ready to take into account a variation of velocity in direction as well as magnitude. This then requircs us to considcr the acceleration explicitly as a vector quantity. Just as so we previously considered now we can show vectorial changes of position, the instantaneous velocity vectors at two neighboring instants, as in Fig. 3-2, and can proceed to a statement of the instantaneous vector acceleration, a: .. Av d dv r '-ôAt-di-dfi As reason far as kinematics why we should by itself is stop here. is no good and evaluate concerned, there We could define the rate of change of acceleration, but in general this does not represent information of any basic physical interest, and so our discussion of mechanics is based almost exclusively on the three and acceleration. you have some prior quantities displacement, velocity, You may fcel, cspccially with calculus, that v(/ + if we have gone to excessive lengths in our dis- A/) 86 Fig. 3-2 bot/i magnitude and direction. Small change of a velocity that is changing in Accclcratcd motions familiarity cussion of instantaneous velocity and acceleration. mistake about But make no The notion these are very subtle concepts. it; an object could both be at a certain point and moving past was one that perplexed some of the best minds of that that point antiquity. Indeed, it is the subject of one of the famous paradoxes of the Greek philosopher, Zeno, who contended that if an object l was moving it could not be said to be anywhere. If you want to test your own mastery of these ideas, try explaining to someone how an object that is at a certain point with zero velocity move away from (i. e., instantaneously at rest) can nonetheless that point by virtue of having an acceleration. It really isn't trivial. THE ANALYSIS OF STRAIGHT-LINE MOTION Given a detailed record of position versus time motion, the procedures that the associated we have shows an example of in Fig. 3-3, this. The and acceleration. variations of velocity sequence of diagrams a straight-line in described enable us to find going from top to bottom, But how about the converse of this process: given the acceleration as a function of time, to infer the graphs of velocity and displacement? velocity doing The basic definitions of and acceleration suggest the appropriate procedure for From Eq. this. (3-1) we see that the change of velocity Av in a short time At is given, at least approximately, by the equation Av = aAt Of course, "approximately" that the smaller we choose is not good enough, but we recognize At to be, the accurate statement of the change of In Fig. 3-4(a) graphical presentation. celeration versus time ; it more nearly we show cuts aeross the curve of a versus a graph of ac- t. From strip, the there top of which it is a short step to concluding that the over-all change of v between tributions. t is obtained by summing (We also negative.) 'On a all such rectangular con- We 87 The then imagine that the widths At are object with arbitrarily high precision. motion made quanlum mechanies and the velocity of a quite different basis, the uncertainty principle of analysis of straight-line is and hence the change of v, expresses our inability to measure both the position moving two given must, of course, recognize that wherever a negative, the area represented by a At, is Av an then becomes apparent that a At can be read as the area of a narrow rectangular values of is Again we resort to a v. (a) Fig. 3-3 Sel ofre- lated graphs showing the time dependence of (a) posilion, (b) velocity, and (c) acceleralion. vanishingly small so that the this limit, with the area This is u2 sum of all the strips coincides, in under the smooth curve of a versus t. then written mathematically as a definite integral: — 01 = wherc we write sidered as a (3-3) aU) di a(t) to show specific function evaluated up to some that the acceleration of time. indefinite time zero of time at which the velocity — 00 = is /, v Most starting . is to be con- often this integral is from some choscn Thus we put (3-4) fl(0 di Notice, then, that our integral, starting from knowledge of the acceleration as a function of time, gives us only the change of velocity during the time t 88 = (or at some /. Information about the value of v at other specific time) Accelcratcd motions must be supplied sep- (b) (a) Fig. 3-4 (a) Graphical integration o/a graph of acceleration versus time to find change ofvelocity. of velocity-time graph (b) Graphical integration to find displacement. arately; v requires is a typical example of a constant of integration that some knowledge of the initial conditions — or of the any one value of t. specific value of v at In like manner, given the curve such as that of Fig. 3-4(b) of v against / (which may represent the initial data or may itself have come from the above integration of the acceleration function) we can proceed to find the distance traveled. It is represented by the area contained between the velocity-time curve, the and the ordinates at S2 - si = two given values of r axis, t: (3-5) v(t) dt up to most usual to evaluate the integral from / = an arbitrary time, and again a constant of integration the must be supplied position s o at t = Again it is — — s Very — Si) = (3-6) v(t) dt often, of course, it is possible to choose Sq = 0, but one should never forget that the area included under the velocitytime graph gives us only the change of position. The a = first simplest applications of these kinematic equations, for or a = constant, are undoubtedly familiar to you. case the velocity-time graph in Fig. 3-5(a). 89 The If is the acceleration analysis of straight-line In the simply a rectangle, as shown is constant (but not zero), motion Fig. 3-5 (a) Veloc- ity-time graph for the special case ofzero acceleration. (b) Velocity-lime graph for a constant (positice) acceleration. is given by is The magnitude of appropriate. Fig. 3-5(b) celeration v a = slope = — the constant ac- do This conforms to the definition of acceleration in Eq. (3-1). If we took a as given, then we would obtain this same result by integration, according to Eq. (3-4), — V DO The area i / dt Jo - (3-7) at in Fig. 3-5(b) that represents the distance traveled can be thought of as made up of the two shaded regions as shown. Hence s — so Combining s — so = vot + the last = Dot \{o — Uo)l two equations, we + get \at 2 result of evaluating the which we can recognize also as the integral in Eq. (3-6) with v(l) — so = / (dq + at) dt = = v Dot + + at: (3-8) %at ./o It is sometimes convenient to remove the time, by o2 90 = u combining Eqs. (3-7) and 2 + 2a(s - so) Accelerated motions all explicit (3-8). reference to This gives us (3-9) For ease of reference, we repeat these equations below ( o lv 2 Kinematic equations (valid only for constant a){ s = = = + at v vq (3-10a) + 2a(s +o + 2 s t — useful, it (3-10b) so) \at 2 Although these mathematical expressions motion are tidy and extremely as a group: for (3-1 Oc) accelerated should be remembered that a truly constant acceleration is never maintained indefinitely. that everyone learns to solve For example, the problems fail under gravity, using a constant acceleration correspond to the is Later there. less may velocities the error we on free do not because air resistance causes the ac- facts, become celeration to g, really as the velocity increases. For low not be big enough to worry about, but shall be dealing with situations in it which the some mathematically well-defined way Thus the emphasis will shift away from Eq. (3-10) and toward the more general statements expressed in acceleration varies in with position or time. Eqs. (3-4) and (3-6). real problems in Another very important factor in solving Whether or is the digital computer. kinematics not the acceleration i s described by a mathematically convenient function, the actual technique of getting numerical answers to problems on motion will — e.g., the path of a rocket or a satellite be the summation of small but finite contributions, corresponding to the strips of Fig. 3-4. The program for solving problems in motion is then represented not by mathematical integrals, but by equations such as the following: +&t) = v(t) +a(/)Ar s(/+A/) = s(l) + v(t)At v(t s(t) = so +!>(') A' There are many problems in motion that can be handled as one-dimensional problems, even though the space in which dynamical processes go on prime reason for is this is that a space of three dimensions. it is A often feasible to resolve the and acceleration into their components in a rectangular coordinate system and then proceed to work with the separate components. Under these conditions, as we have mentioned before, it is not necessary to make use of vectors of position, velocity vector notation as such, even though 91 The anaiysis of straight-line we know motion that we are dealing with directional quantities. becomes It sufficient to axis of reference along the given line of the choose an motion and adopt a convention that selects one direction along this axis as positive and the opposite direction as positive made we must of a particular problem Which we choose negative. arbitrary, but having is stick to it. Directed quantities final coordinate, x, of a particle — e.g., the unstraight always be taken as measured along the positive direc- will line tion of the axis. matically of the — moving along a that are to be found as a result of a calculation known as a choice for the purpose tells In other words, origin. comes out the answer If actually inadvisable, because it it is negative, this auto- on the negative side not necessary (and may be us that the final position is can lead to confusion) to inject preconceived ideas as to which sign or direction a quantity will be found to have; the mathematics A Example. = x 10 will particle starts out at m with an initial velocity v = — 5 m/sec 2 = 8 sec. at acceleration a and position do in the = x it / 15 for you. = from the point m/sec and a constant Find direction. its velocity t We have = = o(0 d(8) oo + at 15 + = -25 (-5)(8) m/sec Also x(t) x(8) Thus at = x + = 10 + = t v l + ial 2 (15)(8) + |(-5)(8) 2 8 sec the particle has passed back through the origin to a point on the far side and t 8 sec much is shown information sees at once t = sees how is in is traveling in the direction of in- The whole creasingly negative x. = - -30 m progress of the motion provided at a glance in these the velocity falls to zero at 6 sec becomes equal and opposite to how the maximum its diagrams. = 3 sec and at its initial One value. its initial and that the sign, original displacement (x reached the negative of how One value of x corresponds to the instant at which u reaches zero and reverses returns to / up to Notice the two graphs of Fig. 3-6. ) at t = 6 sec particle when v has value v Q so that the total area , under the velocity-time curve up to that instant is zero. Even though this is a very simple and straightforward example, it dis- 92 Acceleratcd motions m Fig. 3-6 (a) Veloc- 40 30 ity-time graph for specific values tial velocily of ini- 20 and con- 10> sianl (negatiue) acceleration. (b) Posiiion versus /-20 timefor the motion represented in (a). plays many features that are various details fit worthy of note. And seeing how the together will greatly strengthen one's grip of basic kinematics. A COMMENT ON EXTRANEOUS ROOTS Occasionally, in turning the mathematical handle in the solution of the kinematic equations, one cranks out extraneous roots that are contrary to the physical situation. How docs one recognize these "incorrect" answcrs and when can thcy be discarded? Many of thcsc cxtraneous answers have their origin in the problem we always state initial conditions which specify the situation at the moment we first begin to follow the motion of the particle at t = 0. Specifying the position and fact that in solving a velocity at t = does not tell us anything whatsoever about the past dynamical history of the particle. sequence whatsoever For example, from 93 rest at if an A comment how lndeed, it of no con- the particle attained these initial values. we think of the motion of a body initial is falling freely height h above the earth's surface, on exlraneous roots we have = v O and = y We may h. have held the body at its initial we may have thrown it upward so that it rises to a maximum height h. The motion of the particle subsequent to the instant of time (t = 0) when it was at position and released from it the height h with zero velocity The most i.e., for Eq. (3-10c). is identical in both cases. is the equation motion with constant acceleration Mathematically, this and must be solved t the time is frequent source of extraneous roots relating displacement to time in — or rest, is a quadratic equation as such if the displacement is given and to be found. Because conditions alone do not give us information initial about the earlier motion of the particle (unless additional relevant data are given), a root of this equation corresponding to a negative value of t may not be valid. To illustrate this, consider Suppose we the problem used as an example in the last section. ask for the value of = x(t) Putting i 2 10 x = - 6t + - 15? we 0, - £/ x = 0. We have 2 get the following quadratic equation: = 4 with the roots at which / t = 3 ± V 13 = 6.6 sec or — 0.6 sec. Figure 3-6(b) makes quite apparent the origin of these two roots and shows how the negative root follows from an extrapolation of the graph backward into the region of times prior to may be quite unjustified. held the particle at fired it + 10 m off in the positive We might, if direction with But that "Oh yes, I and then its initial velocity of 15 m/sec." If that were the case, the solution sec 0. = from the origin until x = t asked, say: t x = Oat* = —0.6 would be a complete fabrication; one would be forced to recognize that should not, it simply did not correspond to One reality. however, discard extraneous roots without asking, in the way we have just done, whether there is first a clear physical reason for doing so. The matter of extraneous roots has had an interesting consequence in the history of physics. In quantum mechanics, developing a relativistically correct equation led to two values for the total energy of tive values were an electron : positive and negative. Nega- initially rejected outright as having no physical what meaning can one attach to a kinetic energy less than zero? However, at a later time, P. A. M. Dirac investigated more carefully the nature of these negative energy significance. 94 After all, Accelerated motions Fig. 3-7 Idealized parabolic trajectory for motion under gravity in the absence of air resistance. states and was led to a highly successful theory of electrons the existence predicted of positrons and which other "antimatter" particles. TRAJECTORY PROBLEMS One is IN TWO DIMENSIONS of the most famous and widely studied problems in motion that of free near the earth's surface. fail It provides an illustration of the fact that, in the rectangular-coordinate system represented by horizontal and vertical directions, the two orthogonal components of the motion are completely independent (provided that air resistance is negligible— see p. 225). The path of an object may be treated as two, separate motions occurring simultaneously, and each not present. Systems (1632), We initial may be first recognized this shall consider the speed v analyzed as Galileo, in his Dialogue on the at an angle fact. if the other were Two Chief World 1 motion of an object hurled with as shown in Fig. 3-7 from a height h 'The accomplishmenls of Galileo Galilei, born in Pisa in 1564, the year of Shakespeare's birth and Michaelangelo's dealh, are often citetl as the beginning of modern science. Galileo's publication on astronomy, Dialogue on The Two Chief World Sysiems, incorporated the Copernican model and led to confiicts with church authorities. While technically a prisoner of the Inquisition, Galileo turncd to the studies of mechanics and published (1638) surreptitiously in Holland the results of his investigations "Discourses and Mathematical Demonsirations Concerning Two New Sciences Pertaining to Mechanics and Local Motion," commonly referred to as Two New Sciences (translated, Dover Publications, New York). These books, written largely in the form of imaginary conversations, have a surprisingly modern flavor, and impress one with Galileo's insight and intellectual sophistication. are well worth reading. 95 Trajectory problems in two dimensions They above a level plain. motion is As you know, an approximation to the actual obtained by assuming that the horizontal component of velocity, vz , remains constant and that the vertical component of velocity, vv , g (= 9.8 subject to a constant acceleration of magnitude is m/sec 2 ) downward. the motion This approximate description of works well provided that the of air resistance effects are unimportant, which generally speaking requires compact (dense) objects and fairly low velocities. Later we cases in which the resistive effects are important idealized picture of the Let us now How 1. long What What 2. 3. We see shall is how to answer the following questions: the object in flight? R is the range is the velocity 3-8 — = - t'o ground? upward (y) we shall take the positive coordinate and to the righl (x). components are then as shown The values of the components, and in Fig. 3-8. Anatysis oflhe motion o/ Fig. 3-7 in terms of separale horizontal p»t striking the final coordinates, the initial velocity the acceleration Fig. (the horizontal distance traveled)? upon choose our origin of coordinates at the starting directions to be and and our present- motion becomes seriously inadequate. point of the particle, and initial shall consider and vertical components. cos e *- . wmm, Horizontal < ! ox "l Component 'o COS W,, „ cos «„ Vertical v„ sin o «e-0 u H (unknown) ai (always) Component li (minus because it is below the origin of the coordinate system g (acceleration due to gravity) (a) 96 Accelerated motions (b) [A note about to gravity 32 ft/sec The in order here. g. acceleration due In this book we shall denotes the positive quantity equal to 9.8 m/sec g is, is to represent the scalar magnitude of this vector. use the symbol g That signs represented by a vector, is 2 This . After a coordinate system has been chosen, and vector A. or convention in which represents the positive scalar magnitude of any A the symbol in accord with the usual is 2 if the y com—9.8 m/sec 2 ). In positive, then the upward vertical direction is taken to be ponent of the vector g becomes —g (i.e., a y = some situations it may be convenient to choose the downward In this case the direction as positive. set We equal to +g. y component of g must be must of course be consistent, within a given about what we mean by the positive coordinate calculation, we direction, but are completely free to take whichever choice please; the actual content of the final answers cannot on we depend this.] We Figs. 3-7 1. now return to the trajectory problem, as depicted in and 3-8: How long We know the object in flight? is the initial velocity, the initial and final values of y, and the vertical acceleration. Therefore, as applied to motion in the except the time of flight y = vo y t + \ay t y direction, if we take Eq. we know (3-10c), everything / 2 i.e., -h = We (do sin 8 )t + solve this equation \{-g)t 2 and take the positive root as the physically relevant one. 2. What is the range RP. Apply Eq. (3-10c) to the horizontal-component problem: x = v 0z t + ha ** 2 i.e., R = (do cos 0o)t In this culation 97 we + = (do cos 0oV substitute the specific value of t 1. Trajectory problems in two dimensions obtained in cal- What 3. the velocity is The components of the value of Vz — o vx = do cos Ba Then from may v striking the = vv = oy azt 4- ground? be found from Eq. (3-10b) using obtained from calculation / d upon + i»o» 1. Oyt "o sin do + (— g)' components we have these Magnitude of Direction of v o = tan d = v : (v z 2 + vu 2 ) l/2 ov — ox (In this case tan downward below tion pointing we can calculate the represents a direc- be negative, because will the horizontal.) magnitude of o 2 directly Alternatively, from Eq. (3-10b) as follows: = = o 2 vy 2 «Or v 2 = oo 2 cos 2 0o 2 + 2a u y y = 2 o sin 2 6 + 2(-*)(-A) Therefore, v2 The = o 2 + oy direction of v = sin d 2 = o 2 + 2gh can then be calculated from the relation — v FREE FALL OF INDIVIDUAL ATOMS Atoms moving of in a vacuum cast sharp traveling in straight lines. On shadows and give evidence it must surely the other hand, be true that atoms and molecules, as samples of ordinary matter, are subject to the usual free fail under gravity at the earth's surface. The effect is not very noticeable because evaporated atoms have high average speeds—about the same as a rifle bullet but it is measurable. How far would a beam of atoms — (initially moving horizontally) traveling a horizontal distance illustrates the lem of the problem. last section, It fail vertically L under gravity while at such speeds? would be just Figure 3-9(a) like the trajectory except that this time we take prob- the horizontal distance as given and then solvc for the vertical distance y. If an origin is taken at the point O as shown, from which the atoms 98 Accelerated molions — £^— . ^ / *- c Fig. 3-9 tory ofaloms vacuum with o (a) (a) Trajec- ^^^^. in an inisial horizontal velocity. The vertical displacement is greatly gerated. exag- {b) Para- bolic trajectory of atoms in an atomic (b) > beam ihat must pass Ihrough the and B slils Lt >3 Oven A to reach the detector lm- —*j« D. out horizontally with a speed start v, lm - D H we have Horizontal component: X = GQ xt + %ax t 2 Therefore, L = vt Vertical y y = = component: v 0y t + \ayt~ + K-*)' 2 2 which upon substitution for y = gives us -g L Suppose that we apply a this result to speed of about 500 m/sec. 1 / beam of atoms with a In traveling a horizontal distance of m, the time of flight would be 1/500 sec (2 msec) and we should have y The =— 2 2 \500/ X 10 -5 m deviation from a straight-line path is thus extremely small, it has been studied with only a few hundredths of a millimeter. Despite the small size of the 99 Frcc fali effect, of individual atoms precision in an experiment by Estermann et ment was as shown schematically Fig. in 1 al. Their arrange- Atoms of 3-9(b). cesium or potassium were evaporated out of an "oven" at about 450°K inside a vacuum system. Since the atoms emerge from the oven slits a variety of directions, the in A and B beam was on the same horizontal were about 0.02 mm wide. collimated by two level, as shown. The The beam of atoms was a horizontal hot-wire detector D, also about 0.02 slits detected by mm across. 2 any atom that reaches the detector must upward component velocity at A in order to negotiate the slit system. However, from a point midway between A and B where the beam is horizontal, the trajectory is just like that shown in Fig. 3-9(a). The detector is moved vertically across the beam and the ion current, produced by atoms that strike the wire, is recorded. As Fig. 3-9(b) shows, have a small initial In the absence of any gravitational deflection, the intensity distribution across the beam should be trapezoidal as in Fig. 3-10(a) because the central region would be bordered by "penumbra" regions that are a consequence of the two-slit system. Some shown in Fig. 3-10(b). obvious feature of these graphs is that they reveal most The a wide spread of speeds in the atoms of a beam. Some atoms are results are moving so fast that they are scarcely deflected at all; others are moving so slowly that their deflection is many times greater than the most probable deflection (which corresponds to the maximum of the intensity distribution). The complete curve must reflect a characteristic distribution of speeds of the atoms in the oven at a particular temperature. We shall not consider the detailed on the shape of the intensity A deflection of the peak. pattern but will fix attention comparison of the graphs for cesium obvious that potassium atoms (atomic and potassium makes mass = 39) move on it the Estermann, O. C. Simpson, and O. Stern, Phys. Reo., 71, 238 (1947). same time on the free fail of thermal neutrons— see L. J. Rainwater and W. W. Havens. Phys. Rev., 70, 'I. Similar experiments were reported at about the 136 (1946). K. atom, striking the hot wire, becomes ionized by losing nearby electrode at a negative potential with respect to the wire will collect these positive ions and the resultant current flow can be detected with a sensitive electrometer or galvanometer. If a thin straight wire 2 A neutral an electron. Cs or A it acts as a detector of width equal to its own diameter. Some atoms the halogens) tend to capture electrons and form negative ions; the sign of the potential of the electrode can be adjusted accordingly. is used, (e.g., 100 Accelerated motions Oven *• Detector current Slit/f Vertical line Graph showing detector along which detector travels response for various positions along vertical axis (a) 10 _ 8 8 A 'E 3 .O 6 i G <0 f "S l/l = c 4 4 4) *- « +-" /\ 2 c *-, ' \ r 7 Widthof 1 Width of V detector o \_; >>!-_ . 03 0.4 Deflection, mm 0.2 0.1 0.1 0.2 0.3 0.4 Def lection, mm (b) IPflJPI^^H^^^JIP^flU^^^^^^H F/g. 3-10 (a) Magnified detail of geometrical image formed by atoms moving in siraiglu lines through Iwo slits. The intensity is proportional t o the area ofthe source that can be "seen" by the detector in any given position. on the deflection of beams ofcesium and potassium atoms. [After 1. Estermann, O. C. Simpson, (6) Actual data and O. Stern, Phys. Rev., 71, 238 U947).] average much faster than cesium atoms (atomic mass comparable temperatures. This is an expression of the the molecules of different gases at the equal average kinetic energies —a further discussion at this point. = fact that same temperature have result that As 133) at we quote without for the actual magnitude of the thermal velocities, let us look at the peak of the cesium curve. It is displaced relative to the center of a gravitation-free by about 101 Free fail 0.11 mm. Now in the arrangement of individual atoms shown in beam Fig. 3-9(b), A B = BC = if L, then an analysis of the trajectory will show (see Problem 3-10) that atoms of speed v will be displaced downward by a distance y given approximately by y v* Hence 1/2 L'L = lm and y = Substituting the values = for the approximatc speed, v 1.1 X 10~ 4 m we get, 300 m/sec. This very beautiful and delicate experiment was used as a test of the theoretical velocity distribution of atoms at a given temperature. ' We cite it here as a nice illustration that atoms, like baseballs or earth satellites, follow action of gravitational forces. vacuum it is, laws of free a moving through the demonstrate This is Since the motion takes place in a more justifiable applicatibn of the idealized than are the more usual problems of objects in fact, fail curved paths under the air. You may wonder if it is possible to the free fail of individual electrons in a similar an immensely more difficult way. problem, because whereas atoms, being electrically neutral, experience only the gravitational acceleration, electrons are exposed to stray electric forces that completely swamp gravitational effects unless extraordinary all precautions are taken. Nonetheless, some experiments have been attempted on this problem, although the interpretation of the results is a rather complicated OTHER FEATURES OF MOTION IN affair. FREE FALL moving object near the earth's surface, the horizontal component of velocity always remains constant and the vertical acceleration always has the same value, g (downward). It may be interesting to point In the idealized description of motion of a freely out that, under these assumptions, every trajectory associated with the same value of the constant horizontal velocity, v 0x , forms part of a single parabola a kind of universal curve (see — on which one can mark in the beginning and end points of any particular trajectory, as shown. There is nothing profound Fig. 3-1 1) 'A similar experimenl, using velocity-selected atoms, has been reported by J. C. Zorn, Am. J. Phys., 37, 554 (1969). N. B. Johnson and 102 Accelerated motions Fig. 3-11 "Universal parabola" that embodies all possible parabolic trajectories for a gicen downward acceleralion and a given horizontal component of velocity. The heavily marked pari ofthe curve beiween poinis C and D corresponds lo the atomic beam trajeclory ofFig. 3-9(b). about this, but it can be useful in helping one to see any in- For example, it between the two atomic-beam paths dividual trajectory as part of a larger scheme. makes very clear the shown in Fig. 3-9. relation A closely related feature is the way in which the total velocity vector changes during the course of the motion. trated in Fig. 3-12. Suppose that the sented by the vector v i. is Then the obtained by adding to v shown. i a illus- initial velocity is repre- vertical (downward) vector a At as Similarly, every vector representing the instantaneous motion has a vertical line drawn from the end of W\. the fact that the horizontal same is velocity v 2 , at a time At later, velocity at a subsequent stage in the the This its end point on This result embodies component of every such vector has value. Yet another aspect of this same free-fall problem is illusby the venerable demonstration of the hunter and the trated Fig, 3-12 Array of successive velocity vectors for a motion in which the acceleralion is constant and vertically downward, as in free fail under gravity in the absence of air resistance. (This type of diagram is known as a hodograph.) 103 Other features of motion in. frce fail aA/ Fig. 3-13 bullet Ihe placements in monkey lets monkey. from The Classic monkey-shooting demonsiration. and monkey undergo egual gravilational disegual limes and are doomed to meet if ihe ihe The hunter aims directly at the a limb (see Fig. 3-13). makes no gun fired. go as soon as he sees This is monkey as it hangs really a mistake, because it allowance for the fact that the bullet follows a parabolic path as shown. But the monkey makes a compensating mistake. lets go of the limb as Seeing the gun aimed directly at him, he soon as he sees the monkey begin falling at due to the time of the same it It and— less justifiably— then follows that, in what- takes for the bullet to travel the horizontal distance from the gun to the bullet and the bullet instant (ignoring any delays transit of the light flash the reaction time of the monkey). ever time Thus the flash of the gun. and monkey vertical line of the receive the monkey's descent, both 2 same contributions, \gt , to their displacement as a result of the gravitational acceleration alone. Thus the bullet's trajectory crosses the line of the monkey's fail same time by the monkey with dirc consequences to himself. Note that this result is indcpendent of both the speed of the bullet and the at a point that is bound to be reached at the — value of g; it requires only that the bullet would, in the absence of gravity, go straight to the monkey's original position. remarkable, on the face of it, of the basic analysis of accelerated motion. 104 Accelerated motions Quite yet easily understood in terms Fig. 3-14 (a) Small displacement CP1P2) in a uniform (b) Velocily veciors at the beginning circular motion. and end of the shorl element ofpath. (c) Vector diagram for the evaluation of the change ofvelocity, At. UNIFORM CIRCULAR MOTION Probably the most interesting direct application of the vector definitions of velocity motion if in a circular and acceleration path the center of the circle to the is some constant at is speed. chosen as an origin, the vector always has the same length and simply changes uniform to rate. and r, its The instantaneous magnitude v is [Fig. 3-14(a)]. directions of r A0 = is is v A/, velocity constant. is its from P x From to Pz The angle A0 between therefore given by r direction at a always at right angles this we can For during a short time, calculate the acceleration. distance traveled problem of In this case, readily A/, the along a circular arc the two corresponding oAt Imagine that the bisector of drawn [Fig. 3-14(b)] and consider the changes in velocity parallel and perpendicular to this bisector. Initially the velocity has a component v sin(A 0/2) away from O, and v cos(A0/2) transversely. Subsequently it has a component v sin(A0/2) toward O, and again i; cos(A0/2) transversely in the same direction as before. Thus the change of velocity is of magnitude 2v sin(A0/2) toward O. Figure 3-14(c) shows how this same result comes from considering a vector 105 Uniform circular this angle motion is diagram in which Av + v(r), gives v(/ As A0 is M). vanishingly small, sin(A0/2) becomes in- made is defined as that vector which, added to A0/2 distinguishable from itself (in radian measure). Thus we ' can put = 2usin(A0/2)-x;A0 |Av| = But &d v M/r, so = |Av| we have M/r o2 Hence the magnitude of the acceleration (Uniform circular motion) and direction its circular path This is = given by — (3-12) radially inward, regardless of whether the is being traced out clockwise or counterclockwise. centripetal the called is |a| is acceleration (literally, "center- seeking") asscciated with any circular motion. The need for a dynamical means of supplying this acceleration to an object is an essential feature of any motion that is not strictly straight, because any change in the direction of the path implies a of Av perpendicular to v VELOCITY AND ACCELERATION The IN component itself. POLAR COORDINATES result of the last section, and other results of more general application, are very nicely developed with the help of polar coordinatcs in the plane. The use of this type of analysis is particularly appropriate of some kind— e.g., the starting point = the origin represents a center of force sun, acting on an orbiting planet. to writc the position vector r as the is the scalar distance r r if and the unit vector e r : (3-13) re r We now consider a change of The product of its the change of r with time. This can arise from length, or from a change of its direction, or from approximation is equivalent to (sin 6)19 -» 1 for approximaiion often. For a diseussion of it, Geometry, 3rd ed., sec, for example, G. B. Thomas Jr., Calculus and Analylic Addison-Wesley, Reading, Mass., 1960, p. 172. Mathematically, g-,0. 106 We this shall be using this Accclcraled motions Fig. 3-15 (a) Vector change ofdisplacement, Ar, during a short time At in form (b) a uni- circular motion. Changes in the unit vectors t, and e« during At, showing how Ae, et is and Ae« parallel to is parallel {but opposite) to e,. For the present we a combination of both. shall limit ourselves to circular motion, in which the length of r remains constant. change of which is a short time At r in Ar = shown in Fig. 3-1 5(b). direction of this Its drawn at magnitude, as is Thus we can put Fig. 3-15(a), is equal to r A0. from The in the direction of the unit vector e$ is right angles to e r as The then as shown in Fig. 3- 15 (a), almost the same as Fig. 3-14(a). change (Ar) clear is rAdee we then have Dividing by A/, and letting At tend to zero, the result (Circular motion) If we v = — = dB — dt dt r (3-14a) e» dt designate dd/dt by the single symbol w, for angular velocity (measured in rad/sec), we have (Circular motion) v = The derivation of the above fact that the unit vector e r is length is by expression of = ce» (3-14b) result embodies the important changing with time. Although its definition constant, its direction changes in accord with the direction of with r = wree its r itself. rate of In fact, we can obtain the explicit change as a special case of Eq. (3-14a), 1 d d6 (3-15a) ues In an exactly similar way, as Fig. 3-15(b) shows, a change of 8 implies a change of the other unit vector, positive, as direction of 107 Velocity shown, —e r ; it it is ej. If the change of can be seen that the change of given by the equation and accclcration in 6 is e$ is in the polar coordinates -** = | (e.)- -we (3-15b) r This possible time dependence of the unit vectors in a polarcoordinate system is a feature that has no counterpart in rectan- i, j, and k are defined same directions for all values of the position vector r. Once we have Eqs. (3-14a) and (3-14b) we can proceed to gular coordinates, where the unit vectors to have the calculate the acceleration by taking the next time derivative. we If r limit ourselves to the case of uniform circular motion, and to both are constant, so we have (Uniform circular —d (e») , a = ... motion) cor o = -co2 re r = . er di r (3-16) Thus the by Eq. (3-12) result expressed together with its correct direction. restricting ourselves to label this acceleration of magnitude a r we can put specifically as a radial acceleration If, still out automatically, falls we If , motion in a circle, we remove the condition that the motion be uniform, then the acceleration component vector a has a transverse (3-14b), (Arbitrary circular radial circular component of a motion do dd — e« — o — e di dt = a „ . motion) The Starting also. from Eq. we have the is (since dd/dt by a transverse component, same = a«. v/r as = ,. ,_ (3-17) r we obtained «), but it is for uniform now joined Thus we have — {_-r=r— = — • = _ 2 r ae = do do) di dt (where co =r d-—6 = (3-18) dfi dd/dt) PROBLEMS 3-1 At building. t = an object At the time /o is released a second from object rest at the is top of a tali dropped from the same point. (a) 108 Ignoring air resistance, show that the time at which the Accelerated motions objects have a vertical separation t —+ = glo / is given by 1T 2 2 do you interpret this result for / < £«*o ? (b) The above formula implies that there is an optimum value of f o so that the separation / reaches some specified value at the earliest possible value of /. Calculate this optimum value of fo, and interpret How the result. 3-2 Below are some careful measurements taken on a stroboscopic photograph of a particle undergoing accelerated motion. The distance is measured from the starting point, but the zero of time first is set at the position that could be separately identified: Distance (cm) Time in (in strobe flashes) photo 0.56 1 0.84 2 1.17 3 1.57 4 2.00 5 2.53 6 3.08 7 3.71 8 4.39 Plot a straight-line graph, based fitted by the equation s 3-3 A = these data, to to) child's toy car rolling across Taking a constant acceleration. the car %a(t on — at is (a) x = What 3 m at = / 1 2 , and show a sloping floor is x = sec, are the acceleration that they are find /o- known to have at / = 0, it is observed and at x = 4 m at / = 2 sec. and initial that velocity of the car? to (b) Plot the position of the car as a function of time up / = 4 sec. (c) When is the car at x = 2 m? an illegal water-filled window. Having lightning reflexes, he observes that the balloon took 0.15 sec to pass from top to bottom of his window a distance of 2 m. Assuming that the balloon was re3-4 The faculty resident of a dormitory sees balloon fail vertically past his — leased from rest, how high above the bottom of his window was the guilty party? 3-5 The graph on the next page versus time in a straight-line motion. 109 Problems is an actual record of distance Find the values of the instantaneous velocity (a) and 65 (b) Sketch a graph of (c) From celeration 3-6 at / = 25, 45, sec. had y as a function of / for the (b) estimate very roughly the times at its greatest positive and negative whole trip. which the ac- values. In 1965 the world records for women's sprint races over different distances were as follows: 60 (a) Make an m 7.2 sec 100 yd 10.3 sec 100 m 11.2 sec accurate graph of distance in meters versus time in seconds. (b) The graph will show you that the data can be well fitted by assuming that a sprinter has a certain accelcration a for a short time t and then continues with a constant speed distance x in terms of a, t, and /. (c) Find the numerical values of scription given in (b). l>, If this description Set v. a, is up the equation for and t that what correct, fit the de- is the distance in meters traveled by the sprinters before they reach their steady velocity? Two cars are traveling, one behind the other, on a straight road. Each has a speed of 70 ft/sec (about 50 mph) and the distance between them is 90 ft. The driver of the rear car decides to overtake the 2 car ahead and does so by accelerating at 6 ft/sec up to 100 ft/sec 3-7 (about 70 110 mph) after which he continues Acceleratcd motions at this speed until he is 90 ft ahead of the other How car. far does the overtaking car travel along the road between the bcginning car were in sight, coming in and end of this operation? If a third the opposite direction at 88 ft/sec (60 mph), what would be the minimum safe distance between the third car and the overtaking car at the beginning of the overtaking operation? (If you are a 3-8 how note of driver, take In Paradise Lost, Book I, large this distance is.) John Milton describes the fail of Vulcan from Heaven to earth i n the following words: . To Noon he from Noon fell, A to . . M om from dewy Eve, setting Sun Summer's day and with the Dropt from the Zenith like a falling Star .... (It was ; this nasty fail that gave Vulcan his limp, as a result of his being thrown out of Heaven by Jove.) in this trip, which was we assume that the acceleration had throughout, how high would Heaven be (a) Clearly air resistance mostly through outer space. the value (9.8 g m/sec 2 ) according to Milton's data ? can be ignored If What would have been Vulcan's velocity upon entering the top of the atmosphere? (b) (Much harder) One really should take account of the fact that the acceleration varies inversely as the square of the distance Obtain revised values for the altitude of from the earth's center. Heaven and 3-9 A the atmospheric entry speed. particle moves Below are values of in a vertical its x plane with constant acceleration. (horizontal) and y (vertical) coordinates at three successive instants of time: /, 2 X 4 X sec x, 10- 2 10~ 2 m y, m 4.914 4.054 5.000 4.000 5.098 3.958 Using the basic definitions of velocity and acceleration (u x = Ax/At, etc), calculate (a) The x and the time intervals (b) The v components of the average velocity vector during -2 sec and 2 10~ 2 to 4 X 10 -2 sec. to 2 X 10 X acceleration vector. [similar to Fig. 3-9(b)] shows a parabolic atomicvacuum, passing through two narrow slits, a distance L apart on the same horizontal level, and traveling an addiVerify that the atoms tional horizontal distance L to the detector. 3-10 beam (a) The figure trajectory in arrive at the detector at a vertical distance v 111 Problems below the first slit, such * — * L Oven C *. L/ i V ± that y » >> « gL 2/v 2 where u is , (You can assume the speed of the atoms. L.) A beam of rubidium atoms (atomic weight 85) passes at the same level, 1 m apart, and travels an additional distance of 2 m to a detector. The maximum intensity is recorded when the detector is 0.2 mm below the level of the other slits. What is the speed of the atoms detected under these conditions? Compare with the results for K and Cs shown in Fig. 3-10(b). What was the (b) through two slits component of velocity initial vertical 3-11 maximum over level ground (1638), stated that range of a projectile of given obtained at a is first slit? book Two New Sciences (a) Galileo, in his the theoretical at the initial speed of 45° to the horizontal, firing angle ± 5 (where 5 can be and furthermore that the ranges for angles 45° any angle < 45°) are equal to one another. Verify these results if you have not been through such calculations previously. (to the horizontal) (b) Show that for any angle of projection the maximum at the 3-12 Wall same A height reached by a projectile instant is half what it perfectly elastic ball is thrown against a house and bounces back over the head of the thrower, as shown in the figure. When above the ground and 4 leaves the thrower's hand, the ball is 2 m from the wali, and has i>ox - i>o» = How 10 m/sec. thrower does the ball hit the ground? (Assume that g 4m 3-13 A man a with at on a smooth stands He the horizontal. hillside that (a) Show that, down s 2v = (b) sin(0 if air resistance it m far behind the = 10 m/sec 2 .) makes a constant angle throws a pebble with an an angle d above the horizontal (see the at a distance j initial speed vo figure). can be ignored, the pebble lands the slope, such that + a) cos 8 g cos 2 <x Hence show value of s •Tmn* is that, for given values of uo obtained with — (Use calculus 112 would be gravity were absent. if t-'o (1 + = 45° — a/2 and is sina) g COS2 £* if you likc, but Accelerated motions it is not necessary.) and a, the biggest given by 3-14 A baseball out of the stadium and is hit observed to pass over is from home plate, at a height of 50 ft. The ball leaves the bat at an angle of 45° to the horizontal, and is 4 ft above the ground when struck. If air resistance can be ignored (which it the stands, 400 ft what magnitude of the ball's numbers? (g = 32 ft/sec 2 .) actually cannot), implied by these 3-15 A stopwatch has a hand of length 2.5 cm initial that speed would be makes one complete revolution in 10 sec. What (a) the vector displacement of the tip of the is tween the points marked 6 sec and 8 sec? hand be- (Take an origin of rectangular coordinates at the center of the watch-face, with a y axis passing upward througjt What (b) marked 4 the point = I 0.) are the velocity on the sec and acceleration of the tip as it passes dial ? 3-16 Calculate the following centripetal accelerations as fractions or (« 10 m/sec 2 ): The acceleration toward multiples of g (a) on the earth's axis of a person standing the earth at 45° latitude. The The (b) (c) acceleration of an electron on acceleration of a point of 26 in. diameter, traveling at 25 3-17 A particle r cos r sin 0, A (the first moves in the rim of a bicycle wheel mph. a plane; position can be described by its rectangular coordinates (x, y) or by polar coordinates x = a proton at Bohr atomic model). The (d) the earth. moving around 10 6 m/sec in an orbit of radius 0.5 X a speed of about 2 orbit of the moon toward acceleration of the and y = r (r, 0), where sin 6, and Calculate a z and a y as the time derivatives of r cos where both r and B are assumed to depend on /. (a) respectively, (b) Verify that the acceleration components in polar coordinates are given by ar = az cos 9 = — ax sin a» + 6 au + sin 6 a„ cos 6 Substitute the values of a z and 0j, from (a) and thus obtain the general expressions for a r and a» in polar coordinates. 3-18 x = A particle oscillates 0.05 sin(5/ — tt/6), along the x axis according to the equation where x (a) What are (b) Make a drawing to its is in meters and / in sec. and acceleration at t = 0? show this motion as the projection of a velocity uniform circular motion. (c) Using the position through the 113 (b), find how long it is, after the particle passes through x = 0.04 m with a negative velocity, before it passes again same point, this time with positive velocity. Problems // seems clear to me that no one ever does mean or ever has meant by "force" rate ofchange of momentum. C. D. broad, Scientific Thought (1923) Forces and equilibrium as we said at the beginning of this book, Newton's great achievement, in creating the science of mechanics, was to develop quantitative relationships between the forces acting and the changes in More the object's motion. declared that the main task of mechanics forces from observed motions. But was to learn about does not this on an object than that, he alter the fact that the idea of force exists independently of the quantitative laws of motion and comes periences — the muscular We puli. shall initially effort involved in applying a It push or a begin from this point of view, and rather than plunge at once into dynamics, in balance. from very subjective ex- we shall first take a look at forces has become rather unfashionable to do this be- you probably already know, the accepted units for the absolute measurement of force are defined in terms of the motions that unbalanced forces produce. We are bound to come to that, cause, as and in doing so we shall come to the heart of mechanics. Never- theless, the quantitative veloped in notion of force can be (and was) de- — the study of objects at another context equilibrium. Indeed, our basic knowledge of the two most important forces in mechanics Coulomb rest, in static — the gravitational force force between electric charges through laboratory observations of using techniques that are still — was static and the obtained largely equilibrium situations, important and broadly applied. In discussing forces in equilibrium, we shall begin with experiences that are familiar and seemingly quite straightforward. Later, after considering some of the problems of motion, we shall recognize that tacit assumptions and unsuspected subtleties are 115 we involved; be equipped to return to the problem of shall then equilibrium with deeper insights and a broader view. But, as said earlier, that kind do not of development IN we we but proceed by easy try to handle everything at once, stages to extend the range FORCES very good physics; is and sophistication of our ideas. STATIC EQUILIBRIUI Let us consider a very simple physical system In Fig. 4-1 (a) and straight situation shown its 4-1 (b), and to hold the human, in the Fig. 4-1 A (a) resting stale. What an arrow in the conditions have to be satisfied to shape shown in Fig. 4-1 (b)? Putting aside subjective aspects of the situation, one can say that force is C But what does are in balance. not disembodied; it is Schematic diagram of an archer's bow in (b) Bowstring drawn back between the two segments of string. (r) at midpoint, Arrow in the process ofbeing launched. (d) Forces applied at the center of the bowstring in situation (6). Forces and equilibrium this applied to something. regu/ring applied force along the bisector ofthe angle 116 into the back it straight condition, perhaps launching bow bow must the forces at the point mean? the there, a force process [Fig. 4-1 (c)]. hold the state; the string is that force and the bowstring snaps in Fig. Remove — an archer's bow. bow in its resting know that to bring the We taut. be supplied. toward we show Fig. 4-2 (a) Balanc- ing of equal forces in rangement. (b) Balaiicing o O simple egual-arm ar- S S S ^> S fe l> f l ofunequal forces in accordance with the law ofthe (a) lever. (b) One cannot imagine a force in the absence of a physical object on which In this case the object in question exerted. it is is some small part of the bowstring in the immediate neighborhood of point C. This directions CA and piece of string little CB and to a third force, supplied ACB. of angle We is exposed to pulls along from the adjoining portions of the by the archer, along can draw a separate diagram, as 4-1 (d), showing the piece of string at C in string, the bisector Fig. in equilibrium under the — a force F from the archer, and two forces, which by symmetry are of equal magnitude T, from the string. action of three forces These combine latter along the line of and opposite to give a force, equal to F, symmetry. But how do we know that the equilibrium requires that the bowstring supply a net force equal and F? You might say opposite to back up that this is obvious, but this intuition with a real experiment. It will we can involve one assumption: that identical objects, equally deformed, supply pulls or pushes of cqual size. can remain at rest if For example, a small loop of string pullcd in opposite directions by two identical coiled springs extended And by equal amounts. of equal and opposite forces is this balancing the most elementary of all equilibrium situations. Problems such as that of the archer's to you. But let bow will not be new us point out that the analysis of them involves our ability to assign numerical magnitudes to individual forces compare one force with another. How can we do that? Archimedes showed the way, when he discovered the law of the lever over 2000 years ago. Equal forces balance when applied and to at equal distances on either side of the pivot Unequal forces balance F2 h is satisfied. suggest the ment. ' establish 117 not so. Forces we need to have a believed that he could obtain the See Problem 4-5. in static = but the second has to be based on experiit Archimedes is [Fig. 4-2(a)]. Considerations of symmetry alone would first result, To O [Fig. 4-2(b)] if the condition F\l\ equilibrium result way of obtaining by pure logic, but this Fig. 4-3 Basic arrangement of a "steel- yard" for weighing an unknown (X) with the help of a Standard weight (S) that can be moved along the horizontal arm. multiples of a given force. One method would be to construct a number of identical coiled springs. We could first verify that, when stretched by equal amounts, they would balance one another individually in an equal-arm arrangement. Then we could attach several, pivot, same distance on one the all at and balance them with a single spring, or multiple, at the appropriate distance on the other in Fig. 4-2(b). (Note that such a procedure tions about the way in side of the some entails way, we can use it to measure an no assump- Archimedes' time, at the very and unknown justified force in terms The technique has been used of an arbitrary Standard. objects (see Fig. 4-3), shown which the force varies with extension for an individual spring.) Once the law of the lever has been in this different side, as least, for the since purpose of weighing basic principle, which allows us to its balance a small force against a large one, using suitably chosen lever arms, is exploited in many familiar mechanical devices. UN1TS OF FORCE For the purposcs of librium, we do not magnitudes of force, unit, this introductory study forces. The of forces in equi- necd to worry about the absolute strictly introduction of an arbitrary unit of and the specification of other would really suffice. However, forces as multiples of this it will often be convenient to express forces in terms of their customary measures, so before going further we shall state what these measures are. (You have probably met them The all before, in unit of force that and throughout the book, 118 we is Forces and equilibrium any shall case.) most frequently employ, here the newton (N). This is a force of a such a size that of 1 The all kg ; it it can give an acceleration of mass to a MKS system. detailed basis of this definition will be discussed in Chapter 6; we need that for the moment is the recognition that this does The magnitude of size. such that the gravitational puli exerted on a mass of is near the earth's surface is m/sec 2 represents the basic unit of force in the uniquely define a force of a certain unit 1 is represented by the earth's puli most appropriate A about 9.8 N. this kg 1 force of about 1 on a medium-sized apple N — may result in view of the old tradition (which well be true) that the fail of an apple provided the starting point of Newton's profound thoughts about gravitation. CGS In the force is the dyne (dyn), defined as the force that can give an acceleration of and = 1 g 1 dyn 10 1 -3 cm/sec 2 to a mass of (In saying this we have kg, = 10- 5 lished, so that F (centimeter-gram-second) system, the unit of Since g. 1 1 cm = 10~ 2 m, the relationship N we take the relation F = ma as already estab- changing m and a by given factors implies changing The dyne by the product of those factors.) is thus an exceedingly small force, about equal to the earth's gravitational puli on a mass of 1 mg — as represented, for example, by roughly a £-in.-square piece of this page. The only is other force that the pound. we shall have occasion to mention Unlike the newton and the dyne, the pound is on the measure of the earth's gravitational puli on a Standard object. As soon as one recognizes that the gravitational puli on any given object changes (or at least was, originally) based directly from one place to another, this has to be adjusted. However, detail, we can still definition loses its exactness setting aside such difnculties of pound say that the nitude to about 4.5 N. weighed out on a spring is a force equal in mag- Every time we buy something that scale, we tional units of force such as the practical implications and of that in is are accepting the use of gravita- pound. more We shall considcr the detail later (Chapter 8). EQUILIBRIUM CONDITIONS; FORCES AS VECTORS The static equilibrium of a given object entails two distinct conditions: 1 119. . The object shall not be subject to any net force tending to Ecjuilibrium conditions; forccs as veclors Fig. 4-4 (a) An object in ekuilibrium uuder the action of three nonparailel forces. (6) Two equally acceptable vector diagrams showing the eguilibrium condition 2F = 0. (b) (a) move bodily; it The 2. it is to twist or rotate The in what we call translational equilibrium. object shall not be subject to any net infiuence tending first what we it; it is in call rotational equilibrium. condition involves, in general, the combination of forces acting in different directions, as in our initial example of the bowstring. It has been known since long before Newton's much more than that they time that forces are vectors. This says have characteristic directions; another in the same way tional displacement. over a vertical peg. to it and as the prototype vector quantity, posi- Imagine, for example, a ring that fit nitudes says that they combine with one it Suppose that it a loosc that these strings are pulled by forces of relative 3, and 4, 5, as defined by corresponding periment will show that the ring remains the peg is removed, if mag- numbers of identical springs equally extended [see Fig. 4-4(a)]. when is has three strings attached Then in equilibrium, ex- even the directions of the forces correspond exactly to those of a 3-4-5 triangle. This means that the forces, represented as vectors with lengths proportional to their magnitudes, form a closed triangle, which can be drawn in two different ways, as shown in Fig. 4-4(b). that the three force vectors add Fi +F2 + It is a basic 120 we say property of vectors that the order of addition forces applied to the one In formal terms, to zero: F3 = immaterial (Chapter addition in up many 2). same Thus, object, if we have a it is different ways; Fig. large possible to represent their 4-5 gives an example. The form essential feature is that, in every case, the force vectors Forces and equilibrium is number of — Fig. 4-5 (a) Seueral forces acting at the same poin:, (b) The force veciorsform a closed polygon, showing equilibrium. (c) Equioalent veclor diagram lo (b). a closed polygon Since forces may (i.e., they add up to zero) be applied space, the force polygon and the is in any direction if in equilibrium not necessarily confined to a plane, single statement that the force vectors will in general add up to zero be analyzable into three separate statements pertaining to three independently chosen directions though exists. three-dimensional — usually, al- not necessarily, the mutually orfhogonal axes of a rectangular coordinate system. Geometrically, one can think of this as the projection of the closed vector polygon onto different planes; regardless of the distortions of shape, the projected polygon remains a closed When figure. written out in algebraic terms, the projection involves a statement of the analysis of an individual force vector into components, or resolved parts, along Thus the the chosen directions. condition of equilibrium first equilibrium with respect to bodily translation — can be written as follows: Vector statement: F = Fi +F2 + F3 + •• = (4-la) Component statement: Fz = FU + F2l + F 3l + Fu = Fi„ + F2u + Fa, + F = Fu + F2l + F3l + z It is • • • = = = (4-1 b) worth noticing the relationship between the process of adding force vectors and the process of resolving an individual force vector into force F, in the its components. xy plane, resolved I n Fig. 4-6(a) we show a into the usual orthogonal components: F = Fj 121 + F„j Equiiibrium conditions; forces as vectors Fig. 4-6 Re sol u t ion ofaforce (a) inlo orthogonal com- The components of F, added vectorially ponents. (b) theforce —F, giue zero. (c) Resoluiion of F to inlo nonorthogonal components. We represent F by a stitute— for F triangle, in —F, itself. and its components by dashed lines, components are a replacement a sub- full line to emphasize that the — In Fig. 4-6(b) which the veetors we show a elosed veetor foree Fx and F„\, added to the foree veetor \ represent an cquilibrium situation: FA + Fvi + (-F) = We can then recognize that any other pair of veetors that give when added to — F are a complete Thus we have at our disposal an infinite F zero equivalent to solving a given foree into components. variety of ways of reFor example, the two forees Fm > and Fv > in Fig. 4-6(c) could represent the itself. way of analyzing F into components in an oblique coordinate system; the faet that F,< may happen to be larger than F itself does not invalidate this way of resolving In a similar vein, as in Fig. 4-7(a), Fig. 4-7 (a) if any we can The F 2 and F 3 , is balanced by the equil(b) The same combination of ibrantFB . forees Fi, F2, and F3 has the resultant FK . 122 combine to give zero, regard one of them (which we have combination offorees Fi, the original foree. set of forees Forees and equilibrium labeled here as the others. A ¥E) as the equilibrant force equal completely equivalent to the vector it is their resuJtant, shown as ¥ R the balancer) of (i.e., and opposite to sum of FE is all then a force these other forces: in Fig. 4-7(b). A familiarity with these simple relationships can be a considerable help in problems of combining or resolving forces and can strengthen one*s understanding of equilibrium situations in general. ACTION AND REACTION IN THE CONTACT OF OBJECTS We have already used a kind of principle of uniformity to prescribe a way of reproducing a force of a given magnitude or making a known multiple of the springs, stretched or force. We assume that identical compressed by equal amounts, define equal This assumption leads to consistent conclusions in the Picture now a simple exanalysis of equilibrium situations. do with a pair of identical periment that any two people can forces. They stand facing each other, and one of them, A, agrees to be the active agent. His goal is to push (via the spring that he holds) on the end of the spring held by B, the passive partner, who tries to avoid pushing back. The size of the push springs. that each person exerts spring is is measured by the amount by which his compressed. You know the result of the experiment, of course. Regard- less of the maneuvers made by A and B, the compressions of the two springs are always the same only the amount of both com; pressions together can be varied. direct This displays, in a particularly manner, the phenomenon that is expressed by the familiar statement: Action and reaction are equal and opposite. general form In its this says that, regardless of the detailed form of the contact or of the relative hardness or softness of the two objects involved, the magnitudes of the forces that each object exerts on the other are always exactly equal. Note particularly that, from the very way they are defined, these two forces cannot both act on the same object. This may seem a trivial and obvious remark, but many calculations in elementary mechanics have come to grief through a failure to recognize it. The production of a force of reaction in response to an applied force always involves deformation to some extent. You push on a wali, for example, and that is a conscious muscular act; but how does the wali know to push back? The answer is that it 123 Action and reaction in the contact of objects yields, however imperceptibly, and it is as a result of such elastic deformations that a contact force exerted by the wali comes into No existence. gives a little force until matter how it has done so. There what happens when we sit down chair and what happens when we But in may rigid a surface sefem, it always under a push or a puli and cannot supply a contrary is no basic in a sit difference between comfortably upholstered down on the one case the springs are soft a concrete ftoor. and yield visibly by distances of an inch or more, whereas in the other case the springs atoms in a tightly packed solid and a deformation by a small fraction of an atomic diameter is enough to produce the support required. are essentially the individual structure, ROTATIONAL EQUILIBRIUM; TORQUE We shall now consider the second condition for static equilibrium of an object, assuming that the has been will a satisfied. Whether the object now depend on whether way first is, condition, in fact, in as to produce a resultant twist. on by two equal and opposite Figure 4-8 illustrates the forces. If, forces are along the line joining the points forces are applied, the object is An forces remain tion Fig. 4-8 is finally bound A and B (a) Rota- equal, opposite forces applied at different points of an extended (b) Equal, opposite forces applied in a way that does not give rotational eouilibrium. (c) Iffree to rotate, the object mouesfrom orientation (b) to an equilibrium orientation. 124 acted at which the truly in equilibrium; to twist. is If [e.g., as it has no shown in the directions of the unchanged as the object turns, equilibrium orientashown in Fig. 4-8(c). How do we con- reached, as tional equilibrium with object. object as in Fig. 4-8(a), the tendency to rotate. In any other circumstances is 0, or not the forces are applied in such problem with the simplest possible example. Fig. 4-8(b)] the object £F = equilibrium Forces and equilib.rium 4-9 Alt the forces acling Fig. on a pivoted bar (pf negligible weight) in rolalional equilib- F,+F, rium with one force applied on each side ofthe pivot. knowledge a quantitative criterion for struct out of such familiar rotational equilibrium? The law of the lever provides the clue. Look again at the situations shown in Fig. 4-2. In particular, consider situation (b). The balancing of the forces F and F 2 with respect to the pivot The product of the at O reauires the condition FJi = F 2 / 2 force and its lever arm describes its "leverage," or twisting ability; the technical term for this is torque. The torques of F t t . and F2 with respect to direction — that O F2 due to Let us counterclockwise. are equal in magnitude but opposite in is call clockwise and that due to F\ one negative; then the condition of balance can be expressed another way: The total toraue is one of them positive and the other is in equal to zero. Although the situation as described above is extremely simple, there is more to it than meets the eye, because a further force is exerted on the bar at the position of the pivot of magnitude F, be 2 if the first Now, is as to be sure, this third force exerts about the pivot point O itself. consider the torques about, let it ; must be condition of equilibrium Thus the complete array of forces satisfied. Fig. 4-9. +F However, what if is to shown in no torque we choose to us say, the left-hand end of the some distance d to the left of the point of application of F x ? Then clearly, with respect to this new origin, the force at O is bar, supplying a counterclockwise torque is — but it turns out that this by the sum of the torques due exactly balanced to Fj and F2 (both clockwise, notice, with respect to the new, hypothetical pivot), provided that the condition F\l\ new this result is of its correctness. to you, take a What it says is = F2 / 2 momcnt is satisfied. If to convince yourself that, if the vector sum of the on an object is zero, and if the sum of the torques about any one point is zero, then the sum of the torques about any forces other point So far is also zero. we have of parallel forces. that a force 125 F is limited ourselves to the balancing of torques Now let us make things more general. Suppose applied at a point P, somewhere on or in an Rotational equilibrium; torquc K l F sin + \Sf COS ip (J* ' /v ( (a) Rgi 4-/0 (a) Force F applied at a vector distance r along and perpendicular lo F by finding effeclive lever its some other (c) r. arm, be The first I. Let the vector distance from thing to notice define a plane, which diagram. into components point O, which might be the position of a real pivot, or just an arbitrary point. r. F Evaluation of lorgue of Consider the torque produced by F about object [Fig. 4-10(a)]. F would be if O were indeed a to be the plane of the has become second it real pivot point, the effect and F It lie. therefore makes excellent itself, regarded sense to associate this direction with the torque torque? We to to produce rotation about an axis perpendicular to the plane in which r as a vector of O and F between them that r is we have chosen Experience, so familiar that nature, tells us that of sin^) (b) front a pivot point. (6) Resolution of Z' =r some Now, what about sort. the magnitude of the can calculate this in two ways. The between If the angle r. indicated components along and in Fig. 4-10(b), is to resolve the force into perpendicular to first, and F r is <p, these components are Fcos<p and F sin v», respectively. The radial component represents a force directed straight through O and The transverse com- hence contributes nothing to the torque. ponent, perpendicular to Another way of seeing r, gives a torque of this result is, to recognize that the effective lever arm be formed by drawing the perpendicular along which F Then acts. magnitude rFsin of the total force ON the torque due to F from is O F can to the line just the same as were actually applied at the point W, at right angles to a if it lever arm of We length / equal to r sin shall introduce the single ' <p. symbol M for the •Leonardo da Vinci envisaged the effeclive lever arm of and called it "the spiritual distance of the force." 126 <?. as suggested in Fig. 4-10(c), Forces and equilibrium magnitude a force in this way Fig. 4-11 A and B. and F. {a) (b) Cross product. C, oftwo arbilrary vectors, M, as the cross product oft Torque vector, (c) Silualion resulling in a torque vecior opposile in direclion to thal in (b). Then we have of the torque. M = rF (4-2a ) siri <p This equation does not contain the necessary information about the direclion of the torque, but a compact statement in vector algebra, invented specifically for such purposes, is is Given two vectors A and B, the cross product to be a vector perpendicular to the plane of A This at hand. the so-called cross product or vector product of two vectors. C defined is and B and of magnitude given by C = AB sin where is 6 the Iesser of the two angles between other being 2v — d). There are, of course, A and B (the two opposite vector normal to the plane of A and B. To establish a unique convention we proceed as follows. Imagine rotating the vector A directions through the (smaller) angle [see Fig. 4-1 fingers of the right tion, keeping the until it lies along the direction of This establishes a sense of rotation. l(a)]. hand are curled around in the sense of thumb extended (do it!), B If the rota- the direction of the along the direction of the pointing thumb. The shorthand mathematical statement, which is understood to emcross product body all is these properties, is then written as C = A XB 127 Rolational equilibrium; torque : carefuily that the order of the factors Note is crucial; reversing the order reverses the sign B X A = -(A Using is X this vector notation, the torque as a vector quantity completely specified by the following equation: M = r XF (4-2b) Figure 4- 11 (b) and of B) two (c) illustrates this for different values in each case a right-handed rotation about the direction in <p; which M points represents, as you can verify, the direction in which F would cause rotation write down and the the vector sum of second condition Then, to occur. of cquilibrium we can finally, the torques acting all on an object, — equilibrium with respect to rotation—can be written as follows: YM = ri X If the object Fi + X F2 + r2 on which a as an ideal particle (i.e., rotational equilibrium set r3 X F3 + • • • = (4-3) of forces acts can be regarded a point object), then the condition of becomes superfiuous. Since all the forces are applied at the same point, they cannot exert a net torque about this point; and if the condition £F = is also satisfied, they cannot exert a net torque about any other point either. If one wants to put the same value of r condition r X £M = this in more formal terms, one can say that applies to every term in Eq. (4-3), so that the reduces to the condition (EF) = and so embodies the condition £F = o for translational equilibrium; the equation for rotational equilibrium adds no new information. FORCES WITHOUT CONTACT; WEIGHT Figure 4-12 depicts a pair of situations that look so simple, and embody results that are so familiar, that paused to wonder about the you may never have relation between them. In Fig. 4-1 2(a) an object is shown attached to two spring balances that puli on it horizontally with equal and opposite forces— a clear case of static equilibrium. In Fig. 4-1 2(b) the same object hangs 128 Forces and equilibrium Fig. 4-12 —#—E>^ in equi- Objecl (o) librium under equal and -~n=\ opposite horizontal forces (from spring balances). {b) (g) Objecl in eauilibrium under ofa the action single uerlical contaclforce (from a spring balance) and an incisible influence (gracity). (c) The reading on the spring balance in (b) defines the measured weight (W) of the objecl. Byinference the gravitational force F„ equal and opposite to (C) (b) is W. single spring balance that pulls vertically from a Experience us that this tells is yet as far as visible connections are concerned, How an unbalanced situation. This may seem like a trite or even At very profound one. upward on this time do we trivial we it is quite clearly dilemma? resolve the question. In fact, examine shall it has much it is a only from it the standpoint of our knowledge of static equilibrium. shall see that it. also a case of static equilibriutn, Later we wider ramifications. coming to situation (b) of Fig. 4-12 for the first time, with a background of experience in other kinds of You have learned from this previous exstatic equilibrium. Try to imagine always corresponds to having the perience that equilibrium vector sum of the applied forces equal to zero. having tangible and via strings, springs, situation in and so on. which only one force your confidence You are used to visible evidence of these forces being applied, Now is i n you fuli in the general validity see an equilibrium view, so to speak. But of the equilibrium conditions is so strong that you say: "Even though there is no other contact, there must be another force acting on the object, equal and opposite to the force supplied by the spring." And so you postulate the force of gravity, pulling downward on the object. But your only measure of it is the reading on the spring balance that counteracts this gravitational force. It is in terms such as these that we have come to postulate and accept the existence of a downward gravitational force exerted on every object at or near the earth's surface. In order to hold an object at rest relative to the earth, we must apply a certain upward force on force, as shall call 129 it. The magnitude of this equilibrating measured for example on a spring balance, is what we the weight of an object. This is an important definition; Forces without contact; weight we shall restate it : The weight of cm object be defined as the will magnitude, W, of the upward force that must be applied to the object to hold of what process that in an object as the gravitational force on connection with the supporting spring the object begins to now measurement Notice especially that we are not question. defining the weight of If the an example is an operalional dcfinition; we describe the actual to be followed to get a practical is of the quantity is This at rest relative to the earth. it called is fail freely, then by our definition the object weightless, although there is no suggestion that the tational force on the object has been changed in any (We process of breaking the connection. lessness further in we If Chapter it. broken, so that is way graviin the shall diseuss weight- 8.) return to the situation in which the object held is stationary relative to the earth by the puli of the spring, then our picture of the forees aeting on the object Fig. 4-l2(c). On the assumption that this equilibrium situation, then we can say is as is shown in indeed a static that the spring force W and the gravitational force F„ are equal and opposite, so that the magnitude of W does provide, under these cireumstances, a measure of the magnitude of F„. that, in not synonymous. much But keep it firmly in mind our usc of the terms, weight and gravitational force are By maintaining this distinetion better equipped to handle dynamical we shall be problems involving gravity later on. PULLEYS AND STR1NGS The use of in it pulleys and strings to transmit forees has more physics than one might think and contains some nice applications of the principles of static equilibrium. over a cireular pulley of radius R Consider a string passing [Fig. 4-1 3(a)]. Let the string be pulled by forees F i and F 2 as shown. If this is to be a situation of static equilibrium, the pulley must be in both translational and rotational cquilibrium. Let us consider the rotational equilibrium first. If the pivot is effectivcly frictionless, the only torques on the pulley are supplied by the forees Fi and F2 . Since both of these are applied tangcntially, they have equal lever arms of length exactly at its center). magnitudes of Fj and T, which 130 we can thus R (assuming that the pulley is pivoted Therefore, to give zero net torque, the F2 must both be equal to the same value, — the strength call the tension in the string Forees and eqiiilibrium Fig. 4-13 (o) Ten- sions in the segments of a string where it meels a slationary circular pulley must be in rotational equilibrium. (b) Set of three forces involced in the static eguilibrium of a pulley. (c) Pulley as a deuicefor (O (b) (a) applying a force of gicen magnitude in any desired direction. of the puli that would be found to be exerted by the string at any point along its The length. pulley thus enables us to change the direction of an applied force without changing its magnitude. However, to satisfy the condition of translational equilibrium, the pivot must be able to supply a force F 3 equal and opposite to the vector sum F2 of F, and along the bisector of the angle ments of the 2Tcos(6/2) its This force must therefore lie between the two straight seg- magnitude must be equal to [see Fig. 4-13(b)]. If the tension at is and string, . one end of a string passing over a pulley supplied by a suspended object, then a force equal to the weight, W, librium [Fig. 4-13(c)]. of the object is F of magnitude needed to maintain equi- This means that a pulley-string system W in any desired can be used to supply a force of magnitude direction. Figure 4-14 illustrates the typical kind of arrangement that exploits these features in a simple experiment equilibrium of three concurrent forces. To to study the the extent that the pulleys can be treated as ideal, the magnitudes of the forces Fi, Fig. 4-14 (a) Simple static equilibrium arrangement involving three nonparallel forces (string tensions) applied at the pomt P. (b) Vector diagram showing the equilibrium condition. (a) 131 Pulleys and strings (b) F 2) and F 3 are equal to the respective weights W\, W W and 2, 3 of the suspended objects. PROBLEMS The ends of 4-1 two men who a rope are held by puli on it with equal and opposite forces of magnitude F. Construct a clear argument to show why 4-2 may defined F, not 2F. is a well-known fact that the total gravitational force It is object the tension in the rope on an be represented as a single force acting through a uniquely point— the "center of gravity"— regardless of the orientation of the object. (a) K For a uniform bar, the center of gravity (CG) coincides with Use the geometrical center. this fact to tional torque about the point P considered as arising from a single force (a) H show [see part (a) two individual forces of magnitudes that the total gravita- of the figure] may be W at the bar's center, or from Wx/L and W(L — x)/L acting at the midpoints of the two scgments defined by P. l- (b) If at H-. -I' one end a bar or rod has a weight W, and a small weight [see part (b) of the w is hung simpler of the above figure], use the two methods to show that the system balances on a fulcrum placed atPif x = (b) 4-3 LW/UW + Diagram »v). (a) represents a rectangular board, of negligible weight, with individual concentrated weights mounted at corners. its 1 3 H'. H», > iy O w, 'W, (b) (a) (a) To find the position of the CG of system, one can this proceed as follows: Choose an origin at the corner O, and introduce x and y axes 132 as shown. Imagine the board to be pivoted about a horizontal axis along y, and which an upward force W calculate the distance x (= wi +K-2-f-H'3 Forces and equilibrium + from w4) this axis at will keep the system Next imagine the board rotational equilibrium. in be to pivoted about a horizontal axis along x, and calculate the correspond- Then the ing distance y. (b) An center of gravity, C, board from two corners that matter) in is at the point (x. y). experimental method of locating the in succession (or and mark the To each case. CG is to hang the any other two points, for direction of a plumbline across the board verify that this board to be suspended from O is consistent with (a), imagine the and in a vertical plane [diagram (b)] show by direct consideration of the balancing of torques due to W2, h% and m>4 that the board hangs in such a way that the vertical line from O passes through C (so that tan d = y /2). 4-4 You have just finished a 20-page boyfriend) and you first want to mail thing in the morning. fortunately it is it You letter to your at once, so that it giri friend will (or be collected have a supply of stamps, but un- 2 a.m., the local post office is closed, and you haven't and a and you happen to have learned somewhere that the density of nickel is about 9 g/cm 3 The ruler itsclf balances at its midpoint. When the nickel is placed on the ruler at the 1-in. mark, the balance point is at the 5-in. mark. When the letter is placed on the ruler, centered at the 2-in. mark, the balance point is at Sîn. The postal a letter scale of your own. However, you do have a 12-in. ruler nickel, . rate is 20 cents per half-ounce (international airmail). postage should you put You on? (This problem is How much drawn from real life. might like to do a similar experiment for yourself, perhaps using a differcnt coin as your Standard of weight.) 4-5 (a) As mentioned in the text (p. 1 17), Archimedes gave what he believed to be a theoretical proof of the law of the lever. Starting the necessity that equal forces, F, at equal distances, /, from from a fulcrum must balance (by symmetry), he argued that one of these forces could, again by symmetry, be replaced by a force F/2 at the fulcrum and another force F/2 at 21. Show that this argument depends on the truth of (b) 133 what A Problems it is less purporting to prove. vulnerable argument is based on the experimental knowledge that forces combine as Suppose that vectors. parallel and F2 are applied to a bar as shown. Imagine that equal and opposite forces of magnitude / are introduccd as shown. This gives us two resultant force vectors that intersect at a point that defines the line of action of their resultant (of magnitude Fi + F2). forces F\ Show that this resultant intersects the bar at the pivot point for which Fili = 4-6 Find the tensions in the ropes F2/2. W The weight 4-7 A (a) way 0.5 kg elothesline that the sag a hung is at the by 8 cm. What is is tied When middle of the line, the midpoint is and a tree that away he then pushes transversely on the rope at midpoint of the rope is ; pushes with on the car? man A The a, down pulled were N (= 50 kg suflicient to to be 50 ft midpoint. its ft move pushed the midpoint of the rope another 2 that Instead, he ties when what force does ), begin to stranded is knows happens displaced transversely by 3 force of 500 If this car driver not strong enough to puli the car out directly. the rope tightly between the car (b) is the tension in the elothesline? in a diteh, but the driver has a length of rope. he apart, in such a wet shirt with a mass of (With acknowledgements to F. W. Sears.) (b) m between two poles, 10 negligible. is two configurations shown. in the equilibrium in each case. is in static how man this exert the car, ft, If the the and the would far the car be shifted, assuming that the rope does not streteh any further? seem Does this 4-8 Prove that a practical method for dealing with the situation? like on an three forces act if object in equilibrium, they must be coplanar and their lines of action must meet at one point (unless all three forces are parallel). 4-9 r Painters sometimes work on a plank supported end of each rope pulley the rope is attached to the plank. is such a plank, of weight 50 Keeping (a) side, what (b) 150 lb. around he lets 4-10 is the On the other side of the in mind maximum A that he must be able to move from side to tension in the ropes? Suppose that he uses a rope that supports no more than finds a firm nail on the wali and loops the rope One day he this instead of around the hook on the plank. go of the rope, An it breaks and he inn sign weighing 100 lb is is falls hung as But as soon as to the ground. shown Why? in the figure. lb, and The the held by a guywire that should not be subjected to a tension of more than 250 134 works on painter weighing 175 lb lb. supporting arm, freely pivoted at the wali, weighs 50 system by One loopcd around a hook on the plank, thus holding the plank at any desired height. £- at its ends long ropes that pass over fixed pulleys, as shown in the figure. lb. Forces and equilibrium * Slr Isaac Newton (a) poin t What (b) What the supporting 4-11 man the is minimum safe distance of the point B above A? A man the magnitude is arm at A under and direction of the force exerted begins to climb up a 12-ft ladder (see the figure). weighs 180 lb, on these conditions? the ladder 20 The lb. The wali against which the ladder smooth, which means that the tangential (vertical) component of force at the contact between ladder and wali is negligible. The foot of the ladder is placed 6 ft from the wali. The ladder, with rests is very the man's weight on will slip if the tangential (horizontal) force it, at the contact between ladder and ground exceeds 80 the ladder can the man 4-12 You want hang a picture to only available nails are at points £ at a certain place 1 ft of the edges of the picture (see the ift How lb. up far safely climb? to the left figure). You on a wali, but the and 2 ft to the right attach strings of the appropriate lengths from these nails to the top corners of the picture, 2ft M 2ft i as shown, but the picture will not balancing weight of some (a) If the picture with its possible balancing weight, i (b) In straight unless frame weighs 10 lb, and where would you put the point of intersection of the w hang two tension forces what it A is (a) is the least (Hint: Find the discussion, be yo-yo rests on a table (see the figure) and the free end of its gently pulled at an angle 6 to the horizontal as shown. What is the critical value of 6 such that the yo-yo remains stationary, even though geometrically if with the table.) 135 ? how would hang? (If you want to go beyond a qualitative prepared for some rather messy trigonometry.) string a in the strings.) the absence of the balancing weight, picture 4-13 you add kind. Problems it is free to roli? (This you consider problem may be solved the torques about P, the point of contact What happens (b) a yo-yo, test 4-14 a A for greater or lesser values of 0? (If you have your conclusions experimentally.) simple and widely used chain hoist pulley. diffcrential different diameter are rigidly of rotation. free pulley An is based on what is callcd In this arrangement two pulleys of slightly connected with a common, endless chain passes over these pulleys from which the load W is suspended (see the components of the ditferential pulley have tively, what downward puli applied to one radii fixed axis and around a figure). If the a and 0.9a, respec- side of the freely hanging part of the chain will (ignoring friction) suffice to prevent the load from descending if can be neglected? (a) the weight of the chain itself (b) (more realistic) one takes account of the fact that the freely hanging portion of the chain (PQR) has a 4-15 A force F with components x = 0, y = applied at the point nitude and direction of the torque, Fz — 5 3N, m, and M, of z Fy = AN, and Fz = is = 4 m. Find the mag- F about the direction in terms of the direction cosines angles that — the origin. i.e., (Express the cosines of the M makes with the axes.) 4-16 Analyze vertically weight w? total in qualitative but careful downward on the pedal of a terms how the act of pushing bicycle results in the production of a horizontal force that can accelerate the bicycle forward. (Clearly the contact of the rear wheel with the ground plays an essential role in this situation.) 136 Forees and equilibrium And thus Nature will be very conformable to herselfand very simple, performing all the great Motions of the heavenly Bodies by the attraction of gravity almost all the attracting and . . . and small ones of their Particles by some other repelling Powers .... newton, Opticks (1730) The various forces of nature THE BASIC TYPES OF FORCES 1 all forces arise from the interactions between difFerent objects. Once upon a time it must have seemed that these interactions were bewilderingly diverse, and one of the most remarkable features in the development of modern science has been the growing realization that only a very few basically distinct kinds of interaction are at work. we know of that 1 The following are the only forces at present: Gravitational forces, which arise between objects because of their masses. 2. in Electromagnetic forces, due to electric charges at rest or motion. Nuclear forces, which dominate the interaction between subatomic particles if they are separated by distances less than 3. about 10 It >An -15 m. may be excellent that evcn this degree of categorization will prove background to this topic is the PSSC film, "Forces," by J. R. Zacharias, produced by Education Development Center, Inc., Newton, Mass., 1959. The title of this chapter is borrowed from that of a set of popular lectures delivered in London by the great scientist, Michael Faraday, just over 100 years ago and available in paperback (Viking, make easy and New York, 1960). They rather delightful reading. 139 to be unnecessarily great; the theoretical physicist's dream would allow us to recognize find a unifying idea that is to all these one and the same thing. Albert Einstein later years on this problem but to no avail, forces as aspects of spent most of his and at the present time the assumption of several different kinds of forces seems meaningful as well as convenient. In the following sections we shall briefly consider these three primary types of forces, with examples of physical systems in which they are significant. standpoint of classical classification will It be useful, and from the mechanics very important, to add to our what we "contact forces" shall call — the forces Al- manifested in the mechanical contact of ordinary objects. though these forces are merely the gross, large-scale manifestation of the basic electromagnetic forces between large numbers of atoms, they serve so well to describe most of the familiar phenomena interactions in mechanical that they merit a category of their own. GRAVITATIONAL FORCES Ali our experience suggests that a gravitational interaction between material objects is a universal phenomenon. The an attractive interaction. It is always gravitational forces exerted earth on different objects near its by the surface can be compared, by using a spring balance for example, and these gravitational attractions are proportional, in every case, to the property of the attracted object that we call its plications of this familiar mass. (The content and the im- and seemingly simple statement be a matter for detailed discussion in later chapters.) The general law of gravitational interaction arrived Newton states that the force any other is F with which any will at by particle attracts proportional to the product of the masses of the particles, inversely proportional to the square of thcir separation, and directed along the line separating the found experimentally that this pendence on the velocity of the magnitudo of the force Departures from ihis theory of relativity. F particle It is on which it acts. ' The mass m 1 exerts on a velocity indcpendence are analyzcd in the general is discernible only if the effect of quantities of the speed of light) can be detected in the gravitational interaction. The various particles. measurable de- 3 that a particle of They are the order of o 2 /c 2 (where c 140 two force has no forces of natura of m2 2 can be written as G^f Fl2 = where m mass particle of (5-1) r l2 is the distance G and is from the center of mi to the center a constant of proportionality called the universal gravitational constant. This law of force holds for point masses and for uniform spheres of finite 2 G is 0.667 X 10- lo 3 /kg-sec m size. The value of the constant . Equation (5-1), and the preceding verbal expression of is the first example of a physical law. what it in this It is really says. book it, of the quantitative expression worth spending a few words to discuss Every mathematical statement of experi- no more than a statement of a and m 2 we simply mean the relationship between numbers. and 2 in terms numerical measures of the masses of particles mental relationships in physics is By m i 1 of some arbitrarily chosen unit. The concept of mass has been developed to aid us in wise the concepts of force and distance. numbers that we deal with. is one that our description of nature— like- Thus But it is always the the full verbal equivalent of Eq. (5-1) would be: The numerical measure of the force F with which any particle attracts any other is proportional to the product of the numerical measures of their masses and inversely proportional to the square of the numerical measure of their separation. mind whenever you read the mathematical statement of a physical relationship. They are almost never included explicitly, yet without them there is a Keep those italicized phrases in danger of reading more into such a statement than tains. Thus it really con- the customary type of abbreviated colloquial statement of the law of gravitation begins: "The force of gravitational attraction between two particles is proportional to the product ." What, one may ask, is meant by the product of their masses of two masses? What sort of a physical quantity is that? Even . . . very good seientists have been drawn, at times, into almost metaphysical arguments through the effort to read some special significance into the "dimensions" of such combinations. By reminding oneself what an equation such as Eq. (5-1) actually represents, The one can avoid any such confusion. experiment to measure G was performed by classic the British physicist 141 Henry Cavendish Gravitational forees in 1798. It involved a Fig. Sehemat ic diagram of a gravity 5-1 lorsion- balance experiment. measurement of the gravitational foree between of modest dimensions and arrangement —a solid lead spheres made use of an ingenious mechanical torsion balance — to detect this tiny attractive foree. Figure 5-1 shows the essential features of the arrangement. Two rod to small lead spheres are placed at the ends of a light rod. suspended horizontally by a very thin metal is its midpoint. Two The fiber attached larger lead spheres are positioned elose to the smaller ones, in symmetrical fashion, so that the gravitational attraction between these pairs of large and small masses An tends to rotate the rod in a horizontal plane. orientation is reached when equilibrium the twist of the supporting fiber provides a restoring effect that just balances the gravitational attraction. A beam of light reflected from a small mirror attached to the rod allows the tiny angular deflection to be amplified into a substantial movement of a spot formed by the reflected light a distant wali (the "optical lever" effect). arrangement of of the most The this type, A on torsion-balance incorporating an optical lever, is one sensitive of all mechanical devices. gravitational foree is astonishingly weak under the conditions of a laboratory cxperiment involving the interaetion of relatively small objects, and the detection and measurement of it is an extremely delicate operation. version of the Cavendish apparatus 1 For example, a modern uses two small suspended kg each. The center-to-center distance between a small sphere and a lead spheres, each of 15-g mass, and Manufactured by the Leybold Co. 142 The va-rious forees of nature large spheres of 1.5 Fig. 5-2 Globular cluster ofstars held together by graoitalionalforces (globular cluster M 13 in the constellation Hercules). from (Photograph the Hale Obser- vatories.) one large is about 5 cm. these conditions the gravita- Under tional force of attraction is only weight of a single than human hair is about 6 X 10~ 10 N. The about 10,000 times greater this! Although the gravitational interaction is intrinsically so very feeble, it plays the prime role in most astronomical systems, because (1) the interacting objects are extremely massive and (2) other forces are almost absent. An interesting example is pro- These are collections of vided by globular star clusters. stars more than 120 such clusters have been identified in our galaxy. Figure 5-2 shows a globular cluster containing perhaps more than a million stars. One can infer from the symmetry of this system that there is in a spherically symmetric distribution ; probably no net rotational motion to the cluster as a whole though individual stars undoubtedly travel in all directions. Direct experimental evidence on the actual motions of individual stars is very meager. If we consider a star "at rest" near the outer edge of the cluster, the net force due to (by symmetry) toward the center of the accelerate, reaching its maximum all other stars should be cluster and the speed at the center. star will The net considerations) will ap- on the star (again by symmetry proach zero as the star reaches the center of the cluster. Having gained considerable speed, the star will pass through the center force and gradually slow down due to the resultant force of attraction 143 Gravitational forces Fig. 5-3 Schematic diagram ofa group of globular clusters associated with our Galaxy. of all the other stars (which at toward the center of the all point diametrically opposite its times The cluster). is a net force directed star will finally reach a Thus a starting position. star could perform oscillations along a diameter passing through the Because the separation between center of cluster. much any greater than the diameter of stellar collision is is chance of a very small, even at the center of the cluster (although the photograph would not suggest Thcre stars is so very star, the this). clear astronomical evidence for other systems of a on a much larger scale. Our Galaxy is surrounded by a spherically symmetric "halo" of globular clusters (see the sketch shown in Fig. 5-3). Since the individual clusters maintain similar type but compact identity, each can itself be considered a "mass point," and together they form a sort of supercluster of globular their Direct observational evidence shows that clusters are clusters. traveling in all directions single cluster Galaxy and there is no reason to doubt that a can perform oscillations through the heart of our (just as single stars can oscillatc within a cluster), again with almost negligible chance of individual stars colliding with each other. These two similar examplcs tional attraction between masses illustrate cases is where gravita- the sole force that governs the motion. Usually gravitational forces are important only where at one body of astronomical that it is exercises least solely because the earth is size is involved — note in this category that gravity a major influence on our everyday lives. One can think of certain exceptions to this general rule, as for example the initial stages of the aggregation of neutral hydrogen 144 The various forces of naturc atoms under mutual gravitational attractions to form a proto- their galaxy. ELECTRIC AND MAGNETIC FORCES The on one another The basic law of forces that electrically charged particles exert are of fundamental importance in nature. by the nineteenth-century French Like physicist C. A. Coulomb, and known by his name. Cavendish, Coulomb used a torsion balance for his measurements. But whereas Cavendish simply measured the gravitaelectric force is that found 1 tional constant, taking the basic form of the force law as already explored the actual form of the law of electric known, Coulomb force. Coulomb's law charged particle at rest states that a will attract or repel another charged particle at rest with a force proportional to the product of the charges, inversely proportional to the square of their separation the charges are unlike If and directed along the 2 The force is attractive when and repulsive when they are alike in sign. line separating the two particles. one denotes by qi and q 2 the charges carried by the particles, 1 exerts on particle 2 is the magnitude of the force that particle given by *»-km (5-2) metric unit of electric charge in is coulomb more than is is a coulomb the X Coulomb's law has the value 9 A The The constant k form with the law of universal gravitation. identical in 10 9 huge amount of N-m 2 /C 2 electric usually found in isolation in nature. separated from a similar charge a repulsive force of about 1 3500 N C . charge —vastly One coulomb mile distant would experience (approximately 800 an electrically neutral droplet of water | nearly 1000 (C). in. in lb). Yet diameter contains of positive charge in the nuclei of the hydrogen and oxygen atoms, balanced by an equal amount of negatively charged electrons. (Check this for yourself.) 'Their experiments were performed at almost the same time, but apparently they acted quite independently of one another in their choice and development of the torsion-balance technique. 2 Note of" 145 that insertion, is tacitly Electric where appropriate, of the phrase "numerical measure assumed. and magnetie forces ' 5-4 Fig. Cakulaled irajectories ofprotons ofaboul 14 Gev kinetic energy approaching the earth 'm its magnetic egualorial place and deflected by its magnetic field. (After D. J. X. Monlgomery, Cosmic Ray Physics, Princeton Unioersity Press, 1949.) It is interesting to compare the sizes of electrical forces with those of gravity. Consider, for example, the gravitational torsion balance mentioned in earlier. If only one out of every 10 8 electrons each lead sphere were missing, the resultant imbalance of electric charge on the masses would produce an electrical force force. Through such examples immense strength of the electrical force one can appreciate the comparable to the gravitational compared to the gravitational force. Although gravity is always present, the electrical force is overwhelmingly the most significant agent in all chemical and and processes biological objects of everyday size domain). It in the interactions between physical (i.e., below those of the astronomical holds atoms together, provides the rigidity and tensile strength of material objects, in and is the only force involved chemical reactions. We have been discussing the forces electrical between Moving charged particles also exert on each other. But an additional force arises in this case which we call the magnetic force. It has the interesting property that it depends on the velocity of the charges and always acts on a given charged particle at right angles to the particle's motion. We shall discuss these magnetic forces more precisely stationary charged particles. electrical forces in Chapter An 7. illustration of the magnetic force is provided by the trajectories of protons (positively charged) ejected and approaching the earth. of protons approaching 146 The various in from the sun Figure 5-4 shows possible paths the equatorial plane, forces of nature when they pass in the vicinity of the earth's magnetic It is field. confidently believed that the earth's magnetic field itself arises as a result of charged particles in motion inside the earth, so basically this interaction between charges is an example of the electromagnetic in motion. Actually, from the standpoint of relativity theory, the magnetic force is not something new and different. Charges that are moving with respect to one observer can be stationary with one accepts the basic idea of relativity, one may expect to be able to relate a magnetic force, as observed in one reference frame, to a Coulomb force, as obThus, respect to another. if For the detailed working out of this example the volume Special Relativity in this series. served in another frame. idea, see for NUCLEAR FORCES Although electric forces are responsible for gether, they would, by atomic For we nuclei. electrically repelling themselves, know all holding atoms to- prevent the existence of that nuclei contain protons, one another and not stabilized by a comBut nature has supplied another pensating negative charge. force, as the sirong interaction, which binds together the known nucleons (protons and neutrons) stronger than the its of 10 Coulomb unknown until recently because For distances greater than about nuclear force quickly becomes negligibly extremely short range. -13 cm (= small, Although much force at sufficiently short distances, properties were relatively its in a nucleus. but 1 it F) this dominates over nucleons at shorter distances. of interaction, attractive pulsive for still force— that is, down all It is other interactions between an exceedingly complex type F and strongly re- to about 0.4 smaller separations. It is, in part, a noncentral in contrast to the gravitational and Coulomb not directed along the line joining the centers of the interacting particles. Somewhat analogously to the Coulomb force, which exists only between electrically charged interactions, it is particles, the strong nuclear interaction exists only certain class of particles, known among a as hadrons (Gr. hadros: heavy, bulky), which besides the nucleons themselves includes a number of lighter particles (mesons) and heavier particles (baryons), all -8 sec or less). of which are unstable and very short-lived (10 interactions nuclear associated with force type of Another 147 Nuclear forces is also known to exist but is not very well understood as yet. The range of this force is It is called the "weak even less than that of the strong interactions, and its strength is -15 about 10 as great as the other. This interaction." estimated to be only weak interaction is important only for certain types of nuclear process, such as radioactive beta decay. It is instructive to compare the magnitudes of the types of forces exerted between 15 two protons 10~ different m apart— i.e., separated by about one nucleon diameter: Type of interaction Gravitational 2 Coulomb 2 X 10-" X 10 2 Nuclear (strong) 2 X The magnitude in which it is of the weak nuclear effective, is but even so force, Approximate magnitude of the force, it is N 10» interaction, within the range 1 of the order of 10~ 3 of the electrical 23 by a factor of the order of 10 , greater, than the calculated gravitational force at the same distance between two particles. The nuclear forces have really no place in a discussion of Newtonian mechanics. Indeed, the very use of the word "force" in connection with nuclear interactions is open to question, for any statement about the force between two nuclear at best a remote inference. We cannot cite any particles is direct observa- comparable to those of Cavendish and Coulomb for Moreover, the subject of and electrical forces. tions gravitational nuclear forces, like everything else in subatomic physics, requires the ideas and techniques of quantum mechanics, which is a theory that has almost no use for the concept of force as such. Thus, when we talk of classical or Newtonian mechanics, concerned with situations really in we are which the only relevant kinds of interactions are electromagnetic or gravitational. FORCES BETWEEN NEUTRAL ATOMS It is tion, a fact of profound importance that, the electric in contrast to gravita- forces between individual particles may be repulsive as well as attractive, because of the existence of different signs of electric charge. possible the existence of mattcr that 148 The various forces of nature two This duality of charge makes is clectrically neutral in bulk and of atoms that are individually neutral by virtue of 1 having equal numbers of protons and electrons. On no the face of forces it, therefore, on one another at all two neutral atoms would exert when separated (if one excepts the usually negligible gravitational force). quite true. It would be the case if This, however, not is and negative point. We know, the positive charges in an atom were located at a single however, that the electrons and the nucleus are separated by a Also, the atom is not a rigid structure; certain small amount. and although the "center of gravity" of the negative charge due to the electrons coincides with the positive the atom is isolated, the approach of another distribution when nucleus particle can disturb this situation. characteristic force of attraction named after the great force increases Dutch much more proach one another— or different manifestation of this (to put atoms— a between neutral physicist J. van der Waals. rapidly than l/r message) the force much more One it a force This two atoms ap- as way that carries a subtly off with increasing distance in a falls 2 is rapidly than the attraction between two unbalanced charges of opposite sign. The basis of the van der Waals force is still, of course, the inverse-square law of force between point charges, but its quantum character cannot be calculated without the use of mechanics. neutral The varying as l/r final result is a force detailed 7 between atoms of the same kind. There is, however, another kind of force that comes into play between neutral atoms them together. This is when the attempt is made to squash a positive (repulsive) force that increases with decreasing separation even more rapidly than the van der The result is that the net force exerted by one atom on another, as a function of the separation r between their centers, has the kind of variation shown in Fig. 5-5. The force passes through zero at a certain value of r, and this Waals force. neutral value (r o) can be thought of as the i.e., one atomic diameter if the sum of the two atomic radii— atoms are identical. The repulsive component of F below r atoms behave to some approximation as hard that grows so sharply with further decrease of r spheres; one manifestation of this is the highly incompressible 'J. G. King has shown by direct invesligation of the degree of neutrality of a volume of gas that the basic units of positive and negative charge in ordinary matter cannot differ by more than about 1 part in 10" (private communica- tion). 149 Forces between neutral atoms — Fig. 5-5 Qualitative graph oflhe force between two neutral aloms as a function of the distance be- The dashed tween their centers. nonphysical line represents the idealization of aloms t hai act as completely hard spheres Ihal one anolher. attract The dashed nature of condensed matter. line in Fig. 5-5 indicates the result of idealizing the interatomic force to correspond to complete impenetrability for r attraction for r > r < and the van der Waals r This model can be used quite effectively . to analyze the deviations of an actual gas Since almost all the objects we from the ideal gas laws. deal with in classical mechanics are electrically neutral, this basic atomic interaction the electric force importance, as we between neutral particles — is of fundamental shall discuss next. CONTACT FORCES Many we of the physical systems shall deal with are the objects of everyday experience, acted friction, the strings we and push and cables, upon by such ordinary forces as and beams, the tension of Each of these forces involves what puli of struts and so on. naively call physical "contact" with the object under observation. Consider, for example, a book resting on a horizontal table. The book is supported by the sum total of countless electromagnetic interactions between atoms layers of the book and the these interactions purposes, however, all table. would be we can in the adjacent surface A submicroscopic analysis of prohibitively complex. For most lump ignore this complexity and can these interactions together into a single force that call we shall category but a a contact force. This is a rather artificial Broadly speaking, all the familiar forces of a me- useful one. chanical nature, including the force that a liquid or a gas exerts on a surface, are contact forces in this sense. the previous section 150 The various makes it Our discussion in clear that they are electric forces forces of nature Fig. 5-6 (a) Two charged conlact. the electrical repulsion. apparently in conlact. this spheres, obciously not in The weight of the upper sphere (b) Two is balanced by uncharged spheres, Oli a sufficiently magnified scale, appearance ofa conlact ofsharp geometrical boundaries would disappear. exerted between electrically neutral objects. of such forces when one smooth, hard object The development is pressed against another comes about through a distortion of the distributions of positive and negative electric charges. It is characteristic of such forces that their variation with the distance between the objects is much more rapid than the inverse-square dependence that holds for objects carrying a net charge. Thus the contact forces are, in effect, forces of very short range; they fail to when objects are more than about one atomic diameter apart. The fact, however, that they do have a systematic dependence on the distance of separation means that the notion of what we ultimately mean by "contact" is not clear-cut. negligible size no fundamental distinction between the situation represented by two charged spheres, visibly held apart by their electrical repulsion, and the same two spheres, uncharged, apparently There is in contact, as assumes a trical shown force on separation in Fig. 5-6. final position it it just balances (a) Quali- tatwe graph offorce versus separation for the charged spheres ofFig. 5-6{a). conlact i. e., is if in Fig. 5-7, For any smaller we display these variations we see a drastic difference. In the case of uncharged objects, the transition from negligible force to very large force is so abrupt that it supports our impression of a completely rigid object with a geometrically sharp The "soft" — the difference between F and slowly with (b) weight. its exceeds the weight. But of force with distance, as 5-7 Fig. In each case the upper sphere of equilibrium in which the net elec- W varies r. Comparable graph for the uncharged spheres ofFig. 5-5(6). The conlact is "hard" —the near equality of F and W occurs only Separation of centers, r over an extremely narrow range of (a) values of r. 151 Contact forces H Separation of centers. (b) t Do boundary and no interaction outside that boundary. an and that not forget, however, that this refined measurements would always reveal a continuous variation is idealization, sufficiently of force with separation. FRICTIONAL CONTACT FORCES The discussion the previous section has concentrated on in contact forces called into play by the simple pushing together of two objects. Such forces are then at right angles to the surface — we of contact of that word. call But them normal forces in the geometrical sense much importance and interest attaches to the tangential forces of friction that appear when the attempt is made to drag an object sidcways along a surface. Figure 5-8(a) depicts a block resting on a horizontal surface; its weight is supported by a normal contact force N. We now apply a hori- Suppose that the magnitude of P is gradually increased from zero. At first nothing seems to happen the block remains still. We know, from our analysis of equilibrium situations, that this means that a force equal and opposite to P is zontal force P. ; being supplied via the contact between the block and the surface. This is the frictional force, S. balance P, just crease if rection, we It automatically adjusts as the normal force deliberately pushed on the top of the block. minute deformations of the N down we can imagine charge distributions along the interface, sufficient to develop the requisite forces. (a) Block on a rough horizontal table, subjected to its maximum limiting value, certain value, the frictional force it may a horizontal puli, P. {b) Qualitative graph of the frictional force, S, as afunclion ofP. The condiiion can be S = P satisfied up to the point at which S = nN. After the eguilibrium bound that, is to be broken. 152 But then, S when P is broken is no longer ablc to keep step with it. The equilibrium is down, and motion ensues. A graph of S against the applied force P might look like Fig. 5-8(b). Once S has been brought to increased beyond a Fig.5-8 in- harder, in a vertical di- In both cases electric to itself would automatically The various forces of naturc even drop at first as P is — F v «(i')/ Fig. 5-9 (a) Force b on a sphere in a flowing fluid. (b) Total force o f fluidfriction is 1 mtr-i r^ ) Av -« mode ^ P ofseparate terms 1 that are respectively linear and quadratic in the relativeflow O celocity, v. further increased, although and to motion, it tends to remain at a fairly constant However, the whole regime value thereafter. ff may depend uniquely defined region which we can put JF = is in detail on the P > P, as represented by the 45° line on the fact that the limiting value of (n) — The only velocity. that of static equilibrium, throughout graph of Fig. 5-8(b). The other feature of interest normal force N, so that S corresponds £F is is the empirical roughly proportional to the —the their quotient coefficient of friction a property of the two surfaces in contact: is = 11N (5-3) The above discussion applies to the contact of two solid surfaces. If the contact is lubricated, the behavior is very dif- One ferent. is then dealing, in effect, with the properties of the contact between a fluid (liquid or gas) and a solid. The basic properties of so-called fluid friction can be studied by measuring the force exerted on a fixed solid object as a stream of fluid driven past it range of values of u, this fluid frictional resistance is is Over a wide at a given speed d [see Fig. 5-9(a)]. well described by the following formula: R(v) where The A first second is = Ao + Bv 2 and B are constants for a given object in a given fluid. term depends on the viscosity of the 153 fluid, and the associated with the production of turbulence. the ratio of the second term to the knows (5-4) first is proportional to that at sufficiently high speeds the fluid friction Frictional contaci forces is Since v, one dominated by turbulence, however small the ratio B/'A may be. The same consideration guarantees that at sufficiently low speeds the resistance will be dominated by the viscous term, directly proportional to o [see Fig. 5-9(b)]. CONCLUDING REMARKS In this chapter given a brief account of the three major we have types of physical interactions and have indicated the general To areas in which they are dominant. recapitulate: Nuclear forces are significant only for nuclear distances, the gravitational force is important only if objects of astronomical scale are involved, and nearly everything else ultimately depends on electromagnetic interactions. The study of physics is attempt to understand these interactions and sequences. In mechanics we essentially the their con- have, for the most part, the more all modest goal of taking the forces as given and considering various dynamical situations in which they enter. We shall, however, be discussing two classic cases— gravitation and alpha-particle — which Newtonian mechanics provided the key to the basic laws of force. The present chapter has provided a kind of preview, because it summarizes the state of our current knowlscattering in edge without entering into any detailed discussion of have come to know it. The real work lies ahead! how we PROBLEMS At what distance from the earth, on the line from the earth to the sun, do the gravitational forces exerted on a mass by the earth and the sun become equal and opposite? Compare the result with 5-1 the radius of the moon's orbit around the earth. 5-2 By what of normal its angle, in seconds of are, will a plumbline be pulled out vertical direetion 10-ton truck that parks 20 ft by the gravitational attraction of a Do you away? think that this effect could be detected ? 5-3 In a Cavendish-type apparatus (see the figure) the large spheres g. The length of the arm con- are each 2 kg, the small spheres each 20 necting the small spheres is 20 cm, and the distance between the centers The torsion it is 5 cm. of a small sphere and the big sphere elose to constant of the suspending fiber is 5 deflection of the suspended system 154 The various is forces of naturc -8 m-N/rad. The angular deduced from the displacement X 10 Scale ^20 g change in direction of the which the mirror It is is (Remember that the through turned.) when observed that effect Deduce the value of G shifted is their initial other side (dashed lines) the on the position of the spot of light (a) moved from the large spheres are positions to equivalent positions mean m away. reflected light is twice the angle of a reflected spot of light on a scale 5 by 8 cm. according to these data, ignoring the on each small sphere of the force due to the more distant of the larger spheres. on (b) Estimate the percentage correction quired to allow for the effect of the The 5-4 more the result of (a) re- distant spheres. Cavendish experiment was done with a large-size original one wants to make the gravitational forces and torques as big as possible. However, this requires a strong, stiff wire to support the suspended masses. Much later (1895) C. V. Boys apparatus, as is natural apparatus, using thin fibers of fused quartz for the made a miniature suspensions. sensitivity if It is an interesting exercise to see how the attainable its size. Imagine two versions of the apparatus depends on of the Cavendish apparatus, in which the radii A and and separations of B, both using solid lead spheres, the masses in B, together with all down by maximum the length of the torsion fiber, are scaled with respect to A. We then design for a certain factor L each sensitivity in apparatus by using the thinnest possible torsion fiber that will take Now for a the weight of the suspended masses without breaking. torsion fiber of given material mum and supportable load its Using is torsion constant this and of circular cross proportional to is two , section, the where d is its proportional to d* /I, where information, compare the obtainable with the d2 different sizes maximum maxi- diameter, / is its length. angular deflections of apparatus. (Remember, the lengths of the torsion fibers also differ by the scaling factor L.) 5-5 155 The radius of the hydrogen Problems atom according to the original Bohr ' is 0.5 A. (a) What theory is the Coulomb force between the proton and What is the gravitational force? the electron at this distance? How (b) Coulomb far apart must the proton and electron be for the force to be equal to the value that the gravitational attraction has at 0.5 A? What familiar astronomical distance is this comparable to? 5-6 Suppose electrons could be added to earth and moon until the repulsion thus devcloped was of just the size to balance the Coulomb gravitational attraction. electrons that 5-7 What would be would achieve For a person the smallest total mass of this? living at 45° latitude, what fractional difference, during the day, minimum due to the moon gravitational forces is the approximate between the maximum and — the change resulting from the fact that the earth's rotation causes the person's distance from the moon to vary ? What is a manifestation of this kind of force effect in 5-8 nature? You know that the Coulomb both obey an inverse-square law. force and the Suppose that that the origin of the gravitational force is gravitational force it were put to you a minute difference between the natural unit of positive charge, as carried by a proton, and Thus the natural unit of negative charge, as carried by an electron. "neutral" matter, containing equal numbers of protons and electrons, would not be quite neutral in fact. What (a) fractional difference between the positive and negative elementary charges would lead to "gravitational" forces of the right How magnitudo between lumps of ordinary "neutral" matter ? could such a difference be looked for by laboratory experiments? (b) Is the theory tenable? 5-9 The (a) text (p. 148) quotes a valuc of the nuclear force for two nucleons close together but also suggests that to describe the nuclear interactions in terms of forces any way in (b) is not very practical. According to one of the earliest and simplest descriptions of the nuclear interaction (by H. attraction between F( ,) Can you suggest which a nuclear force as such could be measured? = two nucleons Yukawa) at large separation theoretical the force of would be given by _d e -'"o 15 m and the constant A is about where the distance r o is about 10~ 10" 1 N-m. At about what separation between a proton and a neutron would the nuclear force be equal to the gravitational force between these two particlcs? 5-10 Can you think of any systems or processes in which gravita- 156 The yarious forces of nature tional, electromagnetic, 5-11 As mentioned and nuclear forces all play an important role? of the in the text, the attractive (long-range) part force between neutral molecules varies as l/r 7 For a number of . molecules, the order of magnitude of this van der Waals force well is represented by the equation 76 7 Fvw (r)~ -10- A FV w where newtons and r is in Compare the magnitude of Coulomb force between two ele- in meters. the van der Waals force with the C]: mentary charges [Eq. (5-2), with<?i = ?2 = e = 1.6 X 10~ = about equal to the diameter is a distance r 4 A. (This (a) For of a molecule of oxygen or nitrogen and hence barely exceeding the closest approach of the centers of two such molecules in a collision.) For a value of r corresponding to the mean distance between molecules in a gas at STP. (b) 5-12 One of the seemingly weakest forms of contact force surface tension of a liquid film. One the is of the seemingly strongest is the However, when expressed in terms of a force between individual atoms in contact, they do not look so different. Use the following data to evaluate them in these of a stretched metal wire. tensile force terms: (a) If -3in.- 3 in. wide, a water film and a the weight of about Waterfilm is formed between a rectangular wire frame, freely sliding transverse wire (see the figure), 1 contractile force can be ascribed to the contact of the within a monomolecular layer along each side of the A that the molecules are 3 lg it g to prevent the film from contracting. across and film. atoms takes This lying Supposing closely packed, calculate the force per molecule. (b) A copper wire of 0.025-in. diameter was found to break when a weight of about 10 kg was hung from its lower end. First, calculate this breaking force in tons per square inch. If the fracture is assumed to involve the rupturing of the contacts between the atoms on the upper and lower sides of a horizontal section right across the wire, calculate the force per atom, assuming an atomic diameter of about 5-13 3 A A. time-honored trick method for approximately locating the midpoint of a long uniform rod or bar is to support it horizontally any two arbitrary points on one's index fingers and then move the fingers together. (Of course, just finding its balance point on one finger alone works very well, too!) Explain the workings of the trick method, using your knowledge of the basic principles of static equilibrium and a property of frictional forces: that they have a maximum at value equal to a constant /x (the coefficient of friction) times the component of force normal to the surface of contact between two objects. 157 Problems 5-14 A string in tension (a) is in contact with a circular rod (radius A0 over an arc subtending a small angle the force with which the string presses radially inward (and hence the normal force string) on r) that the pulley which the pulley pushes on the equal to TA9. is Hence show (b) This to T/r. AN with Show (see the figure). is that the normal force per unit length is equal a sort of pressure which, for a given value of T, gets bigger as r decreases. when a (This helps to explain why, string is around a package, it cuts into the package most deeply passes around corners, where r is least.) tightly tied as it the contact (c) If not perfectly smooth, the values of the is amount AF AN, where n is Deduce from this tension at the two ends of the arc can differ by a certain before slipping occurs. The AT is value of the coeffkient of friction between string equal to and rod. y. the exponential relation T(6) = T ep» where To is the tension applied at one end of an arbitrary arc (0) of and T(6) is the tension at the other end. (d) The above result expresses the possibility of withstanding a large tension T in a rope by wrapping the rope around a cylinder, a phenomenon that has been exploited since time immemorial by sailors. Suppose, for example, that the value of n in the contact between a rope string a dock and a bollard on of T corresponding 7" For is 0,2. = 100 lb, calculate the values and four complete turns of rope to one, two, three around the bollard. (It is interesting to note that T proportional to To- is This allows by having a rope passing around a continuously rotating motor-driven drum. The arrangement sailors to produce a big puli or not, at can be described as a force will, amplifier.) 5-15 In a very dclicate torsion-balance experiment, such as the Cavendish experiment, the stray forces due to the fluid friction of slow air currents pushing on the suspended system may be quite significant. To make this quantitative, consider the gravity torsion-balance ex- For suspended spheres of the periment described in Problem 5-3. size stated (r « 0.8 cm), the force due to a flow of air of speed v is given approximately by the formula [Eg. (5-4)] R (newtons) where u due to is in = 2.5 X 10- 6 <; X 10 -s,,2 equaled the gravitational force exerted on the (i.e., the force exerted on a 20-g sphere by a 2-kg sphere with their centcrs 5 The 5 m/sec. Calculate the value of v that would cause a force air currents that sphere in this experiment 158 + cm apart). various forces of nature (Hint: Do not bother to solve a quadratic equation for v. Just find the values of d for which the contributions to R, taken separately, force. The smaller of the two enough to 159 spoil the experiment.) Probiems would equal the gravitational values of u so obtained is clearly already To tell us that every species of things occult specific quality by which effects, is to tell us nothing. it But acts is endowed with an and produces manifest to derive two or three general principles of motion from phenomena, and afterwards to tell us how the properties and actions of all corporeal things follow from those manifest principles, very great step in would be a Philosophy, t/wugh the causes of those principles were not yet discovered. newton, Opticks (1730) — Force, inertia, and motion THE PRINCIPLE OF INERTIA the preceding chapters have treated matter, motion, and force Now we come to the central problem of as separate topics. Newtonian dynamics: affected We by forces? on the face of it, How really began. We saw much What can we simpler: is Chapter 4 in how will in the the study of static cquilibrium is zero. What For an object this, to net force conclude that undoubtedly have solved following diseussion in if is superfluous. on not, A on — and, as a kind order to keep an object many problems Do the net force at rest must bc maintained. After laws before rcading this chapter. at rest, the could be more natural than to zero, the object must remain of corollary to >You was ] turn this statement around and infer that, moving a It problem by Galileo that the science of dynamics net force aeting is is, say about the subjected to no forces? situations leads us to a basic principle: an object objects shall preface this with a question that motion of an object that analysis of this are motions of material all, our experience of Newton's assume that the in the use that account, wish to get down to business — and using them is very sound. The quantitative use of a physical theory is an essential part of the game; physics is not a spectator Where do the equations sport. But to gain real insight and understanding come from? What do they really say? one must also examine the basic assumptions and phenomena. And some of the greatest advances in physics writing equations — — have come about in just this way. Einstein arrived at special relativity by thinking deeply about the nature of time. And Newton, when asked once about how he gained his insight into the problems of nature, replied "By constantly thinking unto them." 161 ' Limitalions on in- Fig. 6-1 molion at the earth's ertial surface. An object starling oul horizontally in a Iruly straight line at A mighl end up on a In the process hilttop at B. would be bound to slow it down. shows that moving objects on the earth's surface do come to and a continuing rest if left to themselves, effort does have to be applied to keep an object moving steadily. But, as Galileo was the first to realize, an extrapolation beyond the range of ordinary experience possible; is in is expressed in his principle of inertia. simply asserted that an object would, Initially, this resistive forces, plane. it free of all if continue with unchanging speed on a horizontal Galileo himself recognized that this assertion For a a limited sense. truly flat to the earth's surface; therefore, if is horizontal plane true only is extended far enough, tangent it must be seen as going perceptibly up hill (Fig. 6-1), and objects traveling outward along it must ultimately slow down. Subsequently Isaac Newton stated the principle of inertia in a generalized form the Principia: "Every uniform motion law" of motion as presented in his "first body perseveres in its in a right line, unless it is that state by forces impressed which we probably upon all learn shows a say? The chanics, and Fig. 6-2 in it." our It is first state of rest, in or of compelled to change a familiar statement, encounter with me- practical illustration of it. But first thing we must recognize is what does it really that, as discussed in Chapter 2, every statement about the motion of a given object involves a physical frame of reference; we can only measure displacements and velocities with respect to other Thus objects. the principle of inertia is not just a clear-cut much around and make a statement about the behavior of individual objects; deeper than that. We can, in faet, turn it it goes statement that goes roughly as follows: There exist certain frames of reference with respect to which the motion of an object, free of all external forces, is a motion in a straight line at constant velocity (including zero). For Galileo's Two New tions, 162 own diseussion of these matters, see his Dialogues Concerning Crew and A. de Salvio, translators), Dover Publica- Sciences (H. New York. Force, inertia, and motion Fig. 6-2 A — motion al conslanl velocity a bullet photographed strobo- traveling at about 1,500 J1 /sec, scopically at 30,000 flashes/sec. (Photograph by Prof. Harold E. Edgerton, M. I. T.) A reference frame in which the law of inertia holds good frame, and the question as to whether a is called given frame of reference is inertial then becomes a matter for an inertial observation and experiment. Most observations made within the confines of a laboratory on the earth's surface suggest that a frame of reference attached to that laboratory all, it was on is After suitable. the strength of observations within such a frame that Galileo arrived at the principle of inertia in the first place! A more critical seruti ny shows that this is not quite good enough, — but we shall do that later. to look further afield For the moment we shall limit ourselves to introducing the main principle, which is not affected by the later refinements. Sir Arthur Eddington, who had a flair for making comments and we need that were both penetrating and witty, offered his own of the principle of inertia: "Every body continues of rest or uniform motion in a straight it doesn't." 1 In other words, he regarded 'A. S. Eddington, The Nalure o/ the Physical University of Michigan Press, 163 The Ann principle of inertia line, it version in its state except insofar as as being, in the last World (Ann Arbor Paperbacks), Arbor, Mich., 1958. an expendable proposition. (He was paving the way for a discussion of general relativity and gravitation.) This remark of Eddington's nevertheless draws attention, in a colorful way, analysis, to what the very foundation of is deviation from a straight-line path No Newtonian mechanics: Any is taken to imply the existence — no force and vice versa. It must be recognized that we cannot "prove" the principle of inertia by an experimental test, because we can never be sure that the object of a force. under test is deviation, truly free of all external interactions, such as those due to extremely massive objects at very large distances. Moreover, there is the far from trivial question of defining a straight line in a real physical sense: nor is it is certainly not intuitively obvious, an abstract mathematical question. it (How would you can be define a straight line for this purpose?) Nevertheless, claimed that the principle of inertia a valid generalization is it from experience; it is a possible interpretation of observed motions, and our belief in its validity grows with the number of phenomena one can correlate successfully with its help. FORCE AND INERTIAL MASS: NEWTON'S LAW The law of of an object inertia implies that the is "natural" state of motion a state of constant velocity. Closely linked to the recognition that the effect of an interaction between this is an object and an external physical system is to change the state of motion. For example, we havc no doubt that the motion of a tennis ball is affected by the racket, that the motion of a compass needle is affected by a magnet, and that the motion of the earth by the sun. "Inertial mass" is the technical phrase for that property which determines how difficult it is for a given applied force to change the state of motion of an object. Let us consider how this description of things can be made quantitative. is affected "Force" ciate it is an abstract term, but we have seen how to assocompressing springs, with practical operations such as and so on. We can readily study the effect of pushing or pulling on an object by such means, using The observations become parforces of definite magnitude. stretching rubber bands, ticularly clear-cut if otherwise move we apply a force to an object that would with constant velocity. A very close approximation to this ideal can be obtained by supporting a flat-bottomed object on a cushion of gas— for example by placing the object on 164 Force, inertia, and motion s 3 I I 4 I © (a) 6-3 Fig. (a) Strobo- scopic pholograpli of a uniformly accel- The eraled molion. time interval between Hght flashes was £ sec. Acceleratior (From PSSC 2 springs Physics, D. C. Heath, Lexington, Massa- o chuselts, 1960.) (b) Simple dynamical experimenls ihat can Acceleration be used as a basis for 1 block 1 block deoeloping Newton' second law. 1 = « a° spring (b) a horizontal table pierced with holes through which air is blown from below. It is then possible to puli horizontally on the object and make such observations as the following 1. A spring, stretched (see also Fig. 6-3): by a constant amount, causes the velocity of the object to change linearly with time tion produced 2. If a in a given object by a given force second spring, identical with the by the same amount, celeration force, is doubled. is — the accelera- is constant. first used side by side with the That is, if we take a known and stretch- first, the ac- multiple of a according to our criteria for comparing forccs in static equilibrium (see Chapter 4), then the acceleration produced in a given object It is directly proportional to the total force. thus becomes possible to write expressing the relation betwccn forces 165 Force and ineriial down F and simple equations accelerations a: mass: Newton's law ) = kF a or F= k'a k' describc thc inertial properties of the particular where k and Which of object. We convenient? 3. If we the above two statements of the results find the place on the answer first is more in another simple experiment: object a second, identical object, by given arrangements of the springs are rcduced to half of what was obtained with one object alone. We can express this most easily by choos- it is observed that all accelerations produced ing the second of the above equations, so that the inertial property is additive — the inertial constants k' of i.e., two different objects can be simply added together, and the acceleration of the combined system under a given force F= + (Ar'i k'2 )a is immediately given by l+2 i.e., 1+2 fci It is is + JG by such steps that one can be led to the equation that known as "Newton's (second) law": universally F = ma = m— di (6-1 where the proportionality factor m (identical with k' as defined above) is called the inertial mass of the object and F is the net force acting on it. Embodied in this basic statement of Newton's law is tities the feature that force and acceleration are vector quan- and that the acceleration is always in the same direction as the net force. An interesting historical fact, often overlooked, is that Newton's own statement of the basic law of mechanics was nol in the form of Eq. (6-1); the equation F = ma appears nowhere Newton spoke of the change "motion" (by which he meant momentum) and related this in the Principia. the value of force Instead, X time. In other words, of to Newton's version of the second law of motion was essentially the following: FAt = 166 mAo Force, inertia, and motion (6-2) We Chapter 9 that Newton's way of formulating the shall see in law grew, by inference, out of the particular kind of evidence available to him: the consequences of collision processes. Direct experiments of the kind illustrated in Fig. 6-3 were not possible with the limited techniques of Newton's day. (The only type of uniformly accelerated motion easily accessible to Newton was that of motion under gravity, but this, of course, did not allow any independent control over the value of F applied to a given object unless the use of an inclined plane, giving a driving force — mg sin regarded as 6, is fulfilling this purpose.) SOME COMMENTS ON NEWTON'S LAW and familiar as Eq. (6-1) is, it nevertheless contains an enormous wealth of physical concepts indeed, almost the whole First comes the assumption that basis of classical dynamics. Sitnple — quantitative measurements of displacements and time intervals lead us to a unique value of the acceleration of an object at a we remind If given instant. ourselves that displacements can only be measured with respect that this, like the principle of inertia, the choice of reference frame. frame in In fact, which the acceleration is measured For the kinds of basic experiments earth fills celeration vector an important the direction of the net force vector. result; it is an expression of the This fact that the combine in a linear Suppose, for example, that an object has two springs way. attached to it [see Fig. 6-4(a)]. x force Fi in the acceleration y. F /m t direction. along Acting alone direction. — that the acting together is just vectors Fi and F2 to is it acceleration caused — not by F2 i n the y predictable by the two springs what one would calculate by adding the form a resultant force F [Fig. 6-4(b)] and mass [Fig. 6-4(c)]. The observed equal to F/m. This result provides the dynamical basis for the "independence of motions" that purely kinematic effect It tells on the object a would produce an would produce an acceleration F-i/m applying this single force to the acceleration a exerts 1 Spring 2 exerts a force x. it Spring Acting alone then a matter of experiment It is pure logic 167 inertial frame. the feature, already emphasized, that the acis in accelerative effects of several different forces along an is illustrated in Fig. 6-3, the this role. Next comes is we see cannot be separated from we tacitly assume that the to other physical objects, in the trajectory we discusscd as a problem of Chapter us that the instantaneous acceleration of an object Some comments on Newton's law is 3. the ' Fig. 6-4 (a) Object pulled by two springs in perpendicular directions. (b) Resultan! force calculated according (c) Obserced acceleralion ofthe objecl agrees With lhal due lo the net force cector to Ihe laws ofveclor addUion. as found in (b). consequence of a linear superposilion of the applied forces or of the accelerations that they would individually produce. If this resiilt did not hold, the prediction and analysis of motions as produced by forces would become vastly more complicated and difficult. The Let us add a word of explanation and caution here. linear superposition of instantaneous components does not mean that we can always of acceleration automatically proceed to whole course of development of the y component of an object's motion without reference to what is happening in the x direction. To take an example that we shall calculate, let us say, the consider in magnetic more detail later, if a field, the charged particle component of is moving in a force in a given direction depends on the component of velocity perpendicular to that direction. In such a case, we have to keep traek of the way in which that perpendicular velocity component changes as time goes on. one may In the case of an object subjected to a single force, be tempted to think that celeration is in the same it is intuitively obvious that the ac- direction as the force. pointing out, therefore, that this velocity particles are involved is — sufficiently modified kinematies and dynamics of special Finally comes the It may be worth not in general true if high- fast to require the relativity. assertion that a given force, applied to a particular object, causes the velocity of that object to change at a certain rate a, the magnitude of which depends only on the See, for example, the volume Special Relatioity 168 Force, inertia, and motion in this series. Fig. 6-5 inertial Increase of i mass with speed, as revealed in experiments on high- speed electrons. Based on data of (open circles) Kauf- mann (1910), (filled circles) Bucherer (1909), and (crosses) Guye and Lauanchy (1915). (AflerR.S. Shankland, Atomic and Nuclear Physics, Macmillan, New York, 1961.) magnitude and direction of F and on a characteristic of the object remarkable result; Newton's law let ment, or is No! it its inertial us consider it single scalar quantity mass m. is all a very further. by any example by a stretched spring, has the Thus, according to conditions. does not matter whether the object traveling at high speed. Is this alvvays, It This asserts that the acceleration produced constarit force, as for same value under — is initially this state- stationary and universally true? — speeds that — the acceleration turns out that for extremely high speeds are a significant fraction of the speed of light produced by a given force on a given object does depend on v. Under these high-speed conditions Newton's mechanics gives way The to Einstein's, as described by the special theory of relativity: inertia of a given object increases systematically with speed according to the formula m(o) = (6-3) - oVc2)" 2 (1 shown in Fig. 6-5, with experimental data that substantiate it. The quantity m Q which is called the "rest mass" of the object, represents what we can simply call the inertial mass This relation is , in all situations to for any v « which classical c the value of preciably different from m m mechanics applies, because according to Eq. (6-3) is inap- . Another implication of Newton's law, as expressed by Eq. (6-1), is that the basic dynamics of an object subjected to a 2 2 given force does not depend on d v/dt or on any of the higher time derivatives of the velocity. 169 The absence of any such com- Some comments on Newton's law plication is in itself a remarkable result, which as far as we know continues to hold good even in the "relativistic" region of very high velocities. has, however, been pointed out that It if one considers physiological effects, not just the basic physics, the d 2 o/dt 2 (= da/dt) can be important. good feeling of a "smooth acceleration" in a existence and magnitude of We all car, and what we mean by know the close to being constant. A that phrase — to an acceleration that is rapid rate of change of acceleration produces great discomfort, and unit of da/dt is it has even been suggested that a be called a "jerk" — should be introduced as a quantitative measure of such effects! The conclusion is that we can draw from the above that Newton's law, although ultimately limited in discussion applica- its does express with insignificantly small error the relation between the acceleration of an object and the force acting on it for almost everything outside the realm of high-speed atomic tion, particles. SCALES OF MASS AND FORCE Granted that a given force produces a unique acceleration of a given object, we can then apply this same force to different Such observations can be used to establish quantitative scales for measuring both inertial masses and forces. In taking this step, we adopt Newton's law as the central feature of meobjects. Although we have hitherto used static situations to compare forces with one another, we now turn to dynamics for defining the absolute magnitudes of forces in terms of the motions chanics. This also means that instead of relying on static measurements to give us prior knowledge of the magnitude of a force, we accept the idea that Newton's law can be used as an they produce. from the observed acceleration that it produces. Because the measures of force and inertial mass are linked in the single equation F = ma, there is danger of circularity in our definitions. But we shall not delve into the subtleties of this problem; we shall simply present a analytical tool for deducing the force pragmatic method of establishing scales of measures for these quantities. Our observations permit given spring is us to assume that every time a stretched to the same magnitude of force on an 170 Force, inertia, and motion same elongation, it object attached to exerts the — for it we We can then observe a reproducible acceleration. of different objects, labeled with the same force ... 1, 2, 3, , : a x a 2 a3 F= , , can use these experimental results to define an scale because number at a time our spring stretched to the same elon- (i.e., gation) and measure the individual accelerations We take a them one puli , mass inertial we can put m\a\ = «12^2 = W!l fl3 W3<33 - • • Therefore, m\ One fl2 particular object (e.g., Standard unit mass of all m,) can be chosen A called a "kilogram." (arbitrarily) as a quantitative measure other inertial masses can then be obtained in terms of the Originally, the kilogram Standard object. was defined to be the 3 mass of 1000 cm of water at its temperature of maximum density (about 4°C), but it is now the mass of a particular cylinder of platinum-iridium alloy, kept at the International Bureau of Weights and Measures France. in Sevres, tended to be exactly equivalent to the small discrepancy was discovered it liter of water, but when a was decided metal Standard as being generally superior reproducibility, (This object was inin to switch to the terms of durability, and convenience.) If the inertial mass truly a property of the object alone, is then the ratios (6-4) must be independent of the particular force Repeating the procedure with a spring extended by a used. amount and hence with different finds experimentally that the The mass different accelerations, ratios are the fact that these experimentally same as determined mass ratios are independent of the magnitude of the force establishes the mass as a Our one before. inertial characteristic property of the object. quantitative scale for forces likewise stems from ton's law once the scale of so that in the MKS inertial system, as New- masses has been established, we mentioned in our first discussion of forces in Chapter 4, the unit force (the newtori) defined as the force that imparts to 1 m/sec 2 1 N kg an is acceleration of : = 1 kg-m/sec 2 Dynamical units of force 171 1 Scales of mass and in other systems of force measurement can an exactly similar way, A newton, as we noted in about equal to the gravitational force on an apple, on a mass of a few ounces. be defined Chapter i.e., in 4, is In describing the kinds of simple experiments on which the formulation of F = ma might be based, we introduced the result that the inertial property represented by may be tempted to experimental As test. commonsense feel that this is is, it is sum of to the a shade But sound. just to question here, we let them come mass of the the masses of the electron and proton? Why? less. is the inertial masses of a proton and that together to form a hydrogen atom. No; You Is the inertial electron, separately, atom equal additive. in principle, a legitimate we have measured imagine that and an is far as ordinary objects are concerned, this reaction to the problem recognize that there mass obvious and not worthy of an then Because in the formation of the atom, with the binding together of the proton and electron, the amount of mass escapes in the form of Conversely, if an object is made up of various parts equivalent of a tiny radiation. held together by cohesive forces, so that effort has to be supplied to separate into those parts, then the it individual parts For ordinary in is objects the difference atomic and nuclear systems it sum of the masses of the mass of the original object. is immeasurably small, but greater than the can become a significant feature of the total mass, and provides the basis of calculations on the energy of nuclear reactions, etc. Returning now to the observed additivity of for macroscopic justifies the objects, we can inertial masses see that this property fully procedure of making a set of Standard masses by constructing blocks of a given material with their volumes in simple numerical ratios ality — e. g., 1 This proportion- :2:5: 10 .... of mass to volume was taken by Newton himself as basic, embodying the concept of a constant density for The very in fact, a definition first sentence in the Principia is, a given material. — a definition that of mass as the product of density and volume has drawn heavy "How," they say, fire, because some critics regard it as circular. "can one define density except as the quotient of mass and volume?" But Newton had a picture of solid matter as built up of small particles packed together in a uniform manner, and it probably seemed to him more logical to take inner structure as primary. The calculation of the this mass of a lump of matter would then be in essence a matter of counting the number of particles that it contained and multiplying by the mass of a single particle. 172 Force, inertia, and motion THE EFFECT OF A CONTINUING FORCE Our main concern stantaneous effect in this chapter of a and the next is with the in- Let us, however, take a force. look first at a question that will be the subject of very extensive analysis later in the book. This question is: What is the effect of a force You is applied to an object and maintained for a while? no doubt aware that the answer to this question can be given in more than one way, depending on whether we consider the time or the distance over which the force is applied. To take the that are simplest possible case, let us suppose that a constant force applied to an object of mass F= we have motion m that at rest at time zero. is F is Then ma, defining a constant acceleration. The resulting thus deseribed by the most elementary versions of the is kinematic equations: d = at x = At time the object has traveled a total distance x, t caleulate Ft \ttfi two possible measures of the total effect and we can of F: = mat = mv 2 Fx = (ma)$at 2 ) = %mv two primary dynamical properties that we associate with a moving object: its momentum and its kinetic Thus we arrive at the energy. The effect of F as measured by the produet Ft is called its of F as measured by the produet Fx is of course work. The quantitative measures of Ft and Fx are impulse; the effect what we call newton-seconds and newton-meters. The former, which defines our measure of momentum, does not have a special unit named The latter, however, is expressed in terms of the in its honor. work or energy basic unit of in the MKS system — the joule. Thus we have impulse work —* momentum in > = kg-m/sec N-m = joules N-sec kinelic energy in THE INVARIANCE OF NEWTON'S LAW; RELATIVITY We have emphasized how the experimental basis of Newton's law involves the observation of motions with respect to an inertial 173 The invariance o\' Newton's law; relativity Fig. 6-6 a particle to Molion of P referred iwoframes that have a relatiue uelocity v. The frame of reference. actual appearance of a given motion from one such frame to another. will vary therefore, how It is worth seeing, the dynamical conclusions are independent of the particular choice of frame — which means that Newton's mechanics embodies a principle of relativity. The one inertial forces S', first point to establish frame, S (i.e., moves uniformly is that, if we have having a constant velocity relative to the instantaneous velocity u in S, and is v, given (see Chapter 2) by Thus if any other frame, also an inertial an object has the first is frame. This follows directly from the fact that S any a frame in which an object under no in a straight line), then is identified if the velocity of S' relative to then the instantaneous velocity of the object relative to S' if u and v are constant velocities, so also is and the u', object will obey the law of inertia as observed in S'. To up rectangular discuss the problem further, let us set coordinate systems in both frames, with their x axcs along the direction of the velocity v (see Fig. 6-6). Let the origins O' of the two systems be chosen to coincide y and instant, also, the moved a the coordinates of ing equations. work in law of P — 0, at O and which at a later time, x /, when the Then axis of S. two systems are related by the follow- appropriate, in view of Galileo's pioneer kinematics and especially of his clear statement of the inertia, that they should have become Galilean transformations.) 174 P distance vt along the in the (It is t z axes of S' coincide with those of S. Let a moving object be at the point origin O' has at Force, inertia, and molion known as the x' (Galilean transformalion: moves S' S relative to a constant speed u L' with | in the The last | r' - - vt = (o const.) =y = = z' -\-x direction) = X (6-5) z r of these equations expresses the Newtonian assumption of a universal, absolute flow of time, but it also embodies the specific convention that the zero of time is taken to be the same instant in both frames of reference, so that all the clocks in both frames agree with one another. We can then proceed to obtain relationships between the components of an instantaneous velocity as measured frames. Thus for the x components we have , ** dx' = = x — Putting x' u'z d = — , (x — two dx = u -d7 in the ' 'di and vi, ^ vt) = u, dt' - = dt, we have v dt The transformations of all three components of velocity are as follows: u'x u'y «'s = ux — = u„ = v (6-6) "z Finally, diflferentiating these velocity we have time, (for = v components with respect to const.) three equalities involving the components of acceleration: du'x du x dUy dUy du'z du z dt' dt dt' dt dt' dt Thus the measure of any a' = acceleration is the same in both frames: (6-7) a Since this identity holds for any two inertial frames, whatever their relative velocity, we say in classical mechanics. relativity in that the acceleration This result is Newtonian dynamics (and it 175 The illustrate an invariant ceases to hold good in the description of motion according to special To is the central feature of relativity). the application of these ideas, consider the invariancc of Ncwton's law: relativity Fig. 6-7 Two ferent views trajectory after it dif- of the of an object has been released from rest i with respect to a moving frame. ide released from rest n .V frame S'. Frame S Frame S' BI simple and familiar example of a particle falling freely under Suppose that at gravity. 5 t = 0, when the axes of the systems and S' are coincident, an experimenter from The rest in this frame. drops a particle in S' trajectories of the particle, as seen in S is observed to follow the expected trajectory according to the and 5', are plotted in Fig. 6-7. In each frame the particle kinematic equations with a vertical acceleration g. frame, the particle has an it S In the horizontal velocity and therefore initial follows a parabolic path, whereas in S' the particle, under the aetion of gravity, different frames falls down. straight Observers in these two would agree that the equation F = ma, where they use the same F, accounts properly for the trajectories for any particle launehed in any manner in either frame. The frames are thus equivalent as far as dynamical experiments are concerned frame — either in frame may be assumed stationary and the other motion, with the same laws of mechanies providing correct explanations from the observed motions. This example of the invariance of Newton's law INVARIANCE WITH SPECIFIC FORCE LAWS In this seetion we a simple is itself. 1 shall consider a little more carefully what is involved in a transformation of Newton's second law of motion. What do we mean by a transformation of this law? Unless we have an explicit F = ma can be regarded as only F from the observed motions. So let law of foree, a preseription for deducing us consider the foree provided by the interaetion between two objects. Suppose, for simplicity, that the interaetion by something like a stretehed is provided rubber band, so that the foree a funetion only of the distance r between the objects; can put F = f (r). For further simplicity, let us 'This seetion can be omitted without loss of continuity. 176 Foree, inertia. and motion i.e., is we assume that the motion object 2 by object F12 =/(*2- I , as measured in F 12 = — f(x? x\) now shall X2 S, can be written (6-8) = then stated as follows in 2, is (6-9) /M2fl2 rewrite this equation so that entirely in terms of Eq. (6-5) frame *i) Newton's law, as applied to object terms of measurements in S alone: We Then the force exerted on confined to the x axis. is measurements made in the it is frame expressed S'. From we have - = Xi (X2 + ~ X[) - vl) (x[ + ot) = x2 - x\ Thus = Fl2 f(X2 but according to — f(x'2 x[) is the terms of measurements = Fl2 assumed law of precisely the specification force, function the of the force F i2 in Hence we can put in S'. F\2 Turning now to the right-hand side of Eq. (6-9), the Galilean transformations give us a inertial mass F[ 2 We motion inertial is = a constant: = f(x 2 - x\) a'; and m2 = m'i. Newtonian dynamics the Thus we are able = m'2a 2 how the Newtonian law of invariant with respect to the particular choice of frame, provided that the Galilean transformations correctly describe the transformations between one frame and another. as long as A of displacements and times more complicated invariance. force law, involved only the relative positions and velocities it same property of on absolute positions however, the force depended e.g., if the force law were of the form of two interacting objects, would possess and to write (6-10) see here in explicit terms is in If, this — velocities F12 = f(xl - x?) then the form of the equation of motion would cease to be the same in all inertial frames. Nothing in our experience has vealed such a situation, which would 177 make Invariance with Specific force laws re- the laws of physics appear different in a laboratory and in a train or plane moving physical laws were different for If the at constant velocity. might be a clue to the uniqueness of different observers, this certain frames of reference. was, indeed, believed for a long It time that a unique reference frame must medium, pervading space, how exist, in the which the waves of in form of a light could be could light travel from the stars to the carried. (Otherwise, earth?) But Einstein showed how the equivalence of frames and the invariance of all all inertial physical laws could be pre- served, provided that the kinematics of Galileo and the dynamics of Newton were replaced by ncw formulations that merged into the old ones in the region of moderate or small velocities. NEWTON'S LAW AND TIME REVERSAL The subject of this section might be less accurately, stated more dramatically, although as a question: Is time reversible? Look two stroboscopic photographs in Fig. 6-8. The first shows an individual object moving vertically under gravity. Is it falling down or "falling up"? The second shows a collision between at the two objects. Which were the paths of the objects before collision? In both cases the answer must be "We A don't know." motion two sequences of events could be run backward would be impossible for the viewer to detect any violation picture of these and it of Newton's laws. The reason is that all velocities of a collection of particles can be reversed without violating Newton's law of motion. A time-reoersal operation (replacing t acceleration the unchanged. downward (Gravity, This direction.) is — by kinematical equations) changes every u to -v, but for example, it t in the leaves the remains in because acceleration involves Thus any conclusions about forces that we reach as a result of watching a dynamical process in reverse sequcnce are identical with what we would the second derivative with respect to time. conclude from the process itself. We do not see attractions apparently turn into repulsions, or anything like that. And it yet, when we see an ordinary motion picture in reverse, quickly becomes apparent from the behavior of inanimate — leaving objects aside the ludicrous effects of reversing human — that actions, which appear strange for quite different reasons Imagine, for direction. most physical actions havc a well-defined example, a sequence in which a glass shatters into small fragments on the floor. 178 Force, inertia, and motion from a table and If wc saw a motion falls t V-/ l j - U (b) (a) Fig. 6-8 (a) Strohoscopic photograph ofan object moving vertically under gravily. Which way is it moving up, or down? collision. lis (b) Slroboscopic time sequence completely reversible. is photograph ofan lo all intents and purposes (From PSSC Physics, D. C. Heath, Lexinglon, Massachusells, 1965.) 179 — elaslic Ncwton's law and time reversal picture in which the fragments gathered themselves together into a whole glass, which then jumped up onto would the table, this unbelievable— nature doesn't act like clearly be Yet a "micromovie" of the individual atomic encounters at every stage that. of the process ought to be perfectly time-reversible. Thus we are faced with a puzzle: Newton's law implies that the fundamental dynamical behavior of an individual particle when one takes a system of very reversible in time, but numbers of apparently the behavior ceases to be time- particlcs, The reversible. resolution of this mystery statistical analysis in the detailed — the subject known As long we as do not associated with time rcvcrsal we are dealing, as be here, with systems of only a few them found is of many-particle systems as statistical mechanics. sider is large shall particles, the problems and we shall not con- arise further. CONCLUDING REMARKS It will probably have become apparent to you during the course of this chapter that the foundation of classical mechanics, as represented by Newton's second law, respects subtle matter. The a is complex and law precise content of the matter for debate, nearly three centuries after the first version of it. in many is still Newton a stated "The Origin In a fine diseussion entitled and Nature of Newton's Laws of Motion," one author (Brian Ellis) says: is "But what of Newton's second law of motion? What the logical status of this law? mass? Or is it acceleration?" Consider fields it is to define Is it a definition of foree? Of an empirical proposition relating foree, mass, and 1 Ellis argucs that how Newton's it is something of second law is all of these: actually used. In unqueslionably truc that Newton's second law a seale of foree. How else, for example, can sure interplanetary gravitational forees? But questionably true that Newton's second law to define a scale is it is some is used we meaalso un- sometimes used of mass. Consider, for example, the use of the mass spectrograph. And in yet other fields, where foree, mass, and independently measurable, and Newton's second law of motion funetions as an empirical coracceleration are all easily relation between these three quantities. Consider, for example, 'Published in a colleclion of essays, Beyond the Edge ofCeriainly (Robert G. ed.), Prentice-Hall, Englewood Cliffs, N.J., 1965. Colodny, 180 Foree, inertia, and motion Newton 's second law in ballistics and rockTo suppose that Newton's second law of motion, or any law for that matter, must have a unique role that we can the application of etry . . . describe generally and call the logical status and an unfounded is unjustifiable supposition. Since forcc and mass are both abstract concepts and not objective realities, we might conceive of a description of nature which we dispensed with both of them. But, as one physicist (D. H. Frisch) has remarked, "Whatever we think about ultimate in reality convenient to follow Newton and it is of our observations into 'forces,' the description which are what make masses and 'masses,' which are what forces make accelerate, This would be just tautology were nomena can on the sameset of masses." more detail Now accelerate. not that the observed phe- it best be classified as the result of different forces acting in split there are, in fact, Ellis spells many and out this same idea various procedures by which the magnitudes of the individual forces acting on a given system may be determined electrostatic forces by charge and distance — by measurement of strain, magnetic and distance determinations, gravitational forces by mass and distance measurements, and so on. And it is an empirical fact that when all such foree measurements are measurements, elastic forces forces by current made and the magnitude of the resultant foree determined, then the rate of change of tion is momentum of the system under considera- found to be proportional to the magnitude of this resultant foree. And so it is that we obtain an immensely fruitful and accurate description of a very large part of our whole experience of objects in motion, through the simple and compaet statement of Newton's second law. PROBLEMS 6-1 Make a graphical analysis of the data represented by the stroboscopic photograph of Fig. 6-2(a) to test whether this is indeed acceleration under a constant foree. For further diseussion of these questions, the American Journal of Physics a perennial source. See, for example, the following artieles: L. Eisenbud, "On the Ciassical Laws of Motion," Am. J. P/iys., 26, 144 (1958); N. Austern, "Presentation of Newtonian Mechanics," Am. J. Phys., 29, 617 (1961); R. Weinstock, "Laws of Ciassical Motion: What's F? What's ml What's a?", Am. J. Phys., 29, 698 (1961). J is 181 Problem-, A 6-2 cabin cruiser of mass 15 metric tons drifts in toward a dock at a speed of 0.3 is 10 :i kg.) m/scc A man from the dock, and try to stop 6-3 a it. its engines have been cut. (A metric ton able to touch the boat is thereafter he pushes Can he A man (a) window after on the dock on bring the boat to rest before of mass 80 kg jumps ledge only 0.5 m down is about 2 cm. With what average force does (b) If the man jumps from a ledge bends it is 1 it m N to touches the dock? to a concrete patio from He above the ground. knees on landing, so that his motion his when with a force of 700 it neglects to bend arrested in a distance of this jar his m 1.5 bone structure ? above the ground but knees so that his centcr of gravity descends an additional his distance h after his feet average force exerted touch the ground, what must h be so that the on him by ground the is only three times his normal weight ? 6-4 An object of mass 2 kg tion of forces in the 20 N at 8 = tion. 47r/3. is xy plane: The acled upon by the following combina5 N direction 6 at 6 = = 0, 10 N = at 6 corresponds to the At t = the object is at the point j;=-6m and y = components v x = 2 m/sec and vy = 4 m/sec. 3 velocity object's velocity and position at t = jt/4, +x and direc- m and has Find the 2 sec. The graphs shown give information regarding the motion in the xy plane of three different particles. In diagrams (a) and (b) the small 6-5 Note! Parabolic N (a) ° (Vertical lines equally spaced) Direction ,of motion (b) / = (c) 182 Force, inertia, and motion dots indicate the positions at equal intervals of time. write equations that describe the force An 6-6 observer first 10~ 2 m/sec and to be components For each Fx and case, F„. measures the velocity of an approaching object 10 -2 m/sec. No then, 1 sec later, to be 2 X intermediate readings are possible because the observer's instruments second to determine a velocity. full of 5 what conclusions can the observer make about g, The size of the force that had been aeting? The impulse supplicd by the force? The work donc by the force? (a) (b) (c) 6-7 A has a mass If the object take a partiele of mass 2 kg along the x axis according to oscillates the equation x = where x 0.2 sin is in (b) 6-8 A -!) meters and What What (a) 5/ 1 / in seconds. is the force aeting is the maximum car of mass 10 3 kg is on the partiele at on force that acts traveling at 28 / = 0? the partiele? m/sec (a over little 60 mph) along a horizontal straight road when the driver suddenly sees a fallen tree blocking the road 100 m The ahead. the brakes as soon as his reaetion time (0.75 sec) allows rest 9 and comes to m short of the trec. Assuming constant deceleration caused by the brakes, what (a) is driver applies What the decelerating force? = (take g fraetion is it of the weight of the car 9.8 m/sec 2 )? -1 with had been on a downward grade of sin the brakes supplying the same decelerating force as before, with what speed would the car have hit the tree ? ^) (b) If the car 6-9 A partiele of mass m follows a path in the xy plane that is deseribed by the following equations: x = A(at — sin a*) y = A(l — cos at) (a) Sketch (b) Find the time-dependent force veetor that causes Can you 6-70 A suggest a if 183 is way the circle maximum is Pro b lem s /, which can support a used to whirl a partiele of mass is the this motion. of producing such a situation in praclice? piece of string of Iength tension T, What this path. m speed with which the partiele (a) horizontal; (b) vertical? maximum in a cireular path. may be whirled Part II Classical mechanics at work Does the engineer ever predict the acceleration of a given bodyfrom a knowledge of its mass and of the forces acting upon it Of course. Does ? the chemist ever measure the mass of an atom by measuring field offorce? Yes. strength of a field mass in Does its acceleration in a given the physicist ever determine the by measuring ha t field? Certainly. t the acceleration of a Why then, should known any one of these roles be singled out as the role of Newton's second law ofmotion? Thefact is brian that it ellis, has a variety of roles. The Origin and Nature of Newton's Laws of Motion (1961) Using Newton's law rr is worth used in reemphasizing the fact that Newton's law two primary ways: Given a knowledge of 1. we can calculate its forces acting on a body, all the motion. Given a knowledge of the motion, we can 2. may be infer what force or forces must be acting. may seem This like a very obvious and quite trivial separa- The first category represents a purely deductive activity using known laws of force and making clearly defined predictions therefrom. The second category includes the induclive, exploratory use of mechanics making use of observed motions to learn about hitherto unknown features of the intertion, but it is not. — — actions between objects. law in is Skill in the deductive use of Newton's of course basi c to successful analytical and design work physics and satisfaction. engineering and can bring great intellectual But, for the physicist, the real thrill comes from the inductive process of probing the forces of nature through the study of motions. It was in this way that Newton discovered the law of universal gravitation, that Rutherford discovered the atomic nucleus, and that the particle physicists explore the structure of nucleons (although, to be sure, this last fleld requires analysis in terms of quantum mechanics rather than Newtonian mechanics). If the law of force is also known, this can be used to obtain information about an unknown mass. The quotation opposite treats this as a third category. 187 SOME INTRODUCTORY EXAMPLES we shall have something to say about the way in which Newton arrived at his insight into gravitational forces from the study of planetary motions. But first we shall discuss how onc goes about calculating motions from given forces. There will be a certain lack of glamour about some of this initial work but that is an inescapable aspect of science, as of everyIn Chapter 8 — thing will else. And some pay rich dividends systematic groundwork at the beginning We later. shall restrict ourselves at first Thus we to cases in which the forces are constant. shall be able to use the kinematic equations for constant acceleration [Eq. (3-10)]. For convenience, we quote the equations again = = = Kinemalic equations s > d2 (Constant acceleration only) do + so + eot + do 2 here: at + 2a(s In any given problem our procedure will be W — 2 (7-1) so) first to identify from Fnct = ma, to calcan then use the kinematic the forces acting on an object and, all culate the resultant acceleration. We equations to describe the subsequent motion. latter step is merely an exercise of the situation Do lies in in mathematics; the real "physics" the analysis of what forces are present. not be misled by the apparently Taken troductory problems. Generally, this trivial nature of these uninteresting and inconsequential. But the method of analysis of salient importance— exactly the same approach more We sophisticated problems. in- at face value, they are, indeed, is is used in far purposely begin by choosing rather elementary systems to clarify the procedure, so the problems are trite. Examp!e we mean one to itself. this to No But keep your eye on the 1: Block that is on a smooth table. method— it is powerful. By a "smooth" surface incapable of exerting any force tangential such surface exists but some come close enough for be a useful idealization. Consider a block of mass frictionless surface. Wc first Question: m What pulled horizontally along a is the motion of the block? must dctcrmine what forces act on the block. To mentally "isolate the block" with an imag- assist the analysis, In this and succeeding cxamplcs we make the assumption that an object rest rclative to the earth's surface has zero acceleration. This mately true—see Chapter 12 for a 188 Usine Newton's law full diseussion. is at only approxi- Fig. 7-1 (a) BIock pulled horizontally on a perfeclly smooth surface. (b) Same block with a slring puli in g in an arbitrary direction. inary boundary surface [see Fig. 7-1 (a)] and showing all draw a sketch the forces that act from outside through this surface on the block. 1 They are = T = force of gravity N contact force the table exerts on the block; this force F„ = tension in the string (a contact force) normal to the surface, hence our choice of the We is N letter have introduced vector symbols here because clearly forces more than one direction are involved. The complete acting in statement of Newton's law for this case Since we assume ma. we conclude vertical direction, is that the block has that N thus T +N+ F„ no acceleration = in the and F„ are equal and opposite vectors; n other words, i N+ F„ = Expressing this a F of F„ is way, we can say that the magnitude diflferent equal to the magnitude A' of N, because the magnitude of a vector, without regard to positive quantity. The along y can then be stated sum of the forces on its direction, is defined as a condition of zero net force component the in the equation N— F„ = 0. The block—the "net" force—is thus (by and hence we have T = ma (with vector addition) equal to T, T both 'We with 189 and a horizontal). shall, in these all Some examples, treat each object as if it were a point forces accordingly acting through a single point. introductory cxamples particle, I The problem becomes somewhat more suppose that the puli of the string The initial components now gives us N+ Vertically Tsind N (The - F„ Tcosf? equation of the samc form and horizontal = = ma us the magnitude of tells is into vertical the following equations: Horizontally first it we substantial if not horizontal [Fig. 7-1 (b)]. vector statement of Newton's law as before, but the analysis of The is N'. = Fg - Tsmd component of vertical the tension in the string hclps to support the block, and so A' becomes equation less than Fg .) The second us directly the magnitude of the horizontal ac- tells ccleration. We notice that there is on a physical limitation this analysis. Unless the table can puli downward on the block, as well as being able to push upward on is necessarily upward, as is necessarily positive. it Thus, if in the Tsin Example 2: Block on a rough suppose that the block and we is shall, for simplicity, Fig. 7-2). Although cannot say what we can put 5 The value of equal to 9 < nmg. m A' The 7-2 < the satisfy table. As Example in 1, we take the puli to be horizontal (see frictional force 3\ Fa /* is If it is "dry fric- the coefficient of friction. in this case, and so, writing gravitational acceleration g, FB as we have inequality expresses the ability of the frictional itself, up to a Block pulled horizontally on a rough surface. 190 we cannot F„, pulled by a string with a force T, where fiN, cqual to times force to adjust Fig. is > certainly is about the properties of the tion," N not a difficult problem, we happen unless we have some information this will N its What happens then? the assumed conditions. shall weight, the force diagram, and the scalar to support shown Using Newton's law certain limit, to balance the force T. Thus the equation T+ T< If T > If We nmg: nmg: = JF ma, which defines the horizontal two separate statements: acceleration, leads to 7 - = ff = = (and hence a T — nmg = ma [and hence a shall not consider in detail 0) (T/m) — fig] what would happen if T were But one can see that this applied at an angle to the horizontal. will of modify the value of You 5F. N and hence, in turn, the limiting value should analyze this case for yourself. Example 3: Block on a smooth the block (Fig. 7-3): the force N surface and the gravitational force F„. bound directions, they are to We Two incline. normal forces act on to the (frictionless) These are in different have a nonzero resultant, and the we wished, introduce horizontal and vertical coordinates x and y and write equations for F = ma that would define the components az and a y of the block must accelerate. could, if vector acceleration a Ncosd — N It is clear, F„ = ma„ 6 = ma x siri however, that a in this case to is parallel to the slope, introduce the coordinate s and it is representing distance downward direction. Resolving along and perpendicular to 5, we have along the slope in F„ sin 6 N- F cos 6 the we m times the gravitational acceleration g, the write the magnitude of the gravitational force as equal to tions gives us = g sin Fig. 7-3 d Forces acting on a block on a perfectly smooth incline. 191 the forces = ma = If a simpler Some introductory examples first of these equa- Fig. 7-4 BIock at respect to resi with an accelerating elevator; the spring scale records its apparent weight. Example A block sits on a 4: BIock in an elevator. scale (a spring scale!) in package an elevator (Fig. 7-4). Question: What does the scale read as the elevator moves up and down? Mentally isolating the block, we recognize that only two forces act on it. They are: Va = the force of gravity (downward) N = the contact force (upward) exerted by the scale on the block The package scale records the magnitude of N, because, by the equality of action and reaction for objects in contact, the force exerted on the scale conditions, is is what we —N. The value of this reading, under any shall call the measwed weight {W) of the block. If the elevator acceleration a, la (and the block and scale with it) has an upward, then Newton's measured as positive w requires Thus, if velocity the elevator upward is stationary or on the block. positive acceleration (upward), then m(g + moving with constant or downward, the reading of the scale the gravitational force N= is But if is equal to the elevator has a we have a) In this case the weight of the block, as measured by the scale, is 192 greater than the gravitational force Using Newton's Iaw on it. One will often on the notice this effect personally, as an increased force of the feet, the elevator is picking up speed, going upward, and (b) is slowing down, going downward. Both of these involve acceleration Similarly, . soles riding an elevator in two situations: (a) when the elevator if is slowing when when it upward down in its upward motion or just beginning its downward motion, a is downward, so that the measured weight is less than F„. (Incidentally, the internal discomfort that one sometimes feels in an elevator can be linked directly to Newton's law. positive a acquires (upward) acceleration, — — stomach must literally sink a extra forces from the surrounding little If the elevator heart onc's and before they experience tissues to supply the acceleration called for. Example 5 : Two connected masses. Here is a simple example designed to illustrate the important point that one isolate, in one's imagination, apply F = ma to it alone. is free to any part of a complete system, and Figure 7-5 shows two masses connected by a light (massless) string on a smooth (frictionless) A surface. horizontal force, P, pulls at the right-hand mass. What can we deduce about the situation? we can imagine an isolation boundary drawn around both m, and m 2 and the string that connects them. The only First, P, and the common to both external horizontal force applied to this system total mass P = is rtii (mi plus m2 - Hence we have + mi)a This at once tells us the acceleration that we can imagine an masses. Next, the connecting string alone. T! and T 2 with which equality of aetion and — Ti and — 2 isolation In Fig. 7-5 the string pulls Two applied to its ends. connected blocks pulled horizontal/y on a perfectly smooth surface. Newton's law must apply to any part of the system that one chooses to consider. 193 Some is boundary surrounding we indicate the forces on the masses; by the reaetion, the string has forces equal to equal the mass of the string times Fig. 7-5 is introduetory examples The sum of its must Assuming these forces acceleration a. mass of the string to be negligible, this means that T] and T 2 would have the same magnitude, T, which we call the tension in the (This idealized result the string. what is worth noting of course, rather obvious; is, any that, in is would real situation, there have to be enough differencc of tcnsions at the ends of the string to supply the requisitc accelerative force to the mass of the string itself.) Finally, m x and m2 we can imagine drawing separately, isolation boundaries around and applying Newton's law to the horizontal motion of each: T = m\a P — T = mza Adding these equations, we arrive at the equation of But the total system once again. if we take motion of either equation alone, substitution of the already determined value of a will give the value of the tension Once again these problems. T in P and the masses. acknowledge the simple character of terms of we shall But if they are studied in the spirit in which they are offered— not for their own sake, but for the they exemplify the systematic use of Newton's law way in which — they will be found to suggest a sound approach to almost any problem that involves a direct application of MOTION IN F = mz. TWO DIMENSIONS Ali the examples in the last section dealt with motion along one dimension only. This is not a serious limitation because, thanks to the "independence of motions," any given instant situation at force and acccleration along separate coordinate simplest case of this vector. itself we can always analyze a terms of the components of in It is occurs when directions. the acceleration a is The a constant then very convenient to choose the direction of a as one coordinate; along any direction perpendicular to component is, by definition, zero, and the this the acceleration velocity component must be constant. Probably example of this procedure is the most familiar the analysis of motion under gravity can be ignorcd. We considered this as a purely kinematic problem in Chapter 3. Another example, with the additional feature that we have control over in the idealization that air resistance 194 Using Newton's law ' Accelerating Vertical Fluorescent deflectins screen plates anode (a) Controlgrid- Focusing anode Horizontal - deflecting Glasswallof tube plates Vertical deflection Fluorescent plates (simplified) screen Resultant deflection from central axis (b) Fig. 7-6 Diagram of (a) cathode-ray tube. the main features ofa simple diagram ofelectrical (b) Simplified connections and electron trajectory. the applied force, is the motion of an electron Let us consider ray tube. this as a beam in a cathode- dynamical problem. Figure 7-6(a) shows a sketch of a cathode-ray tube, and Fig. 7-6(b) A a schematic diagram of is well-focused beam strike beam principal features. from an electron gun passes between two pairs of deflection plates which, deflect the some transversely if appropriately charged, will away from the central axis so as to any particular point on the fluorescent screen. We shall call the central axis the z direetion, so that the direetions of transverse deflection, perpendicular to z and to one another, can be called x and y, just as The 'We shall electron make gun in a real oscilloscope. accelerates the electrons through a po- use of certain results concerning the effects of electric forees. familiar to you, see, for example, PSSC, Physics If these are not already (2nd 195 ed.), Part IV, Heath, Boston, 1965. Motion in two dimensions K tential difference eV (where e . This gives to each electron a kinetic energy the elementary charge) so that is gitudinal velocity it acquires a lon- component v 2 given by = eVo j»n?,fi (7-2) Therefore, After leaving the gun, the electron not subjected to further is acceleration or decelcration along the z direction, so ordinate continues to change at the rate vz We shall suppose that a potential difference between the upper and lower ^-deflection if an electron were it would acquire an energy eV on to travel all the by the transverse it in its z co- . V applied is plates. This means that way from one plate to another, consequence of the work done electric force Fy . If we suppose the plates and spaced by a distance d that is small compared length, the force Fv would have the same value at all to be parallel to their points between the plates, so that the gain of energy would also Fv d. be given by Hence we have =^d F * . Given Fv this value of between the plates, m md * How the electron will, throughout far vertically will the electron depend on the amount of time If the horizontal extent plates. zontal component of value v z , the time = passage its have an upward acceleration given by it be deflected? This will spends between the deflection of the plates is /, and the hori- the electron's velocity has the constant given simply by is ]_ Vz From Eq. we determine the transverse displacement y that occurs during the time this upward force acts: 196 (7-1) sy = vov t y = + + \a 2 cari yt md\vj Using Newton's law t But from Eq. = , (7-2), mvz 2 /e is 2V just , so we obtain IL (7-3) This expression gives the transverse displacement of the electron as it emerges from the deflection plates. The resultant displacement away from the central axis of the spot on the fluorescent screen may bc found from trigonometric considerations. The z and y components of the displacement of the electron while between the plates are given z = y = by o, W 2 from these equations shows that y is proportional and the trajectory is therefore a parabola. to z Once the electron leaves the region between the plates, there Eliminating t 2 , is no further force on it and it travcls in a straight line at angle equal to arctan (o„/o,), where v u component upon leaving the vy = aj = plates. is its an transverse velocity Now we have Y as the electron travels the eV — — % i v„ eVI VI vz mdo? IVad The additional transverse deflection distance D from the deflector plates to the screen is thus given by r-*?-g£ 2Vod (7-4) v, If D» /, as is usually the case in practice, most of the total transverse deflection is contained in Y, and to estimate the approximate = [sensitivity AU we can use Eq. (7-4) of an actual oscilloscope (spot deflection)/(deflection voltage).] We may note (7-4). sensitivity in passing a striking feature of Eqs. (7-3) and distinguishing characteristics of the particle as an electron have disappeared. Any any charge or any mass, could negatively charged particle, with in principle pass through the system and end up at the same spot on the screen. In a real oscilloscope, as Fig. 7-6(a) indicates, the deflector plates are not flat strictly apply. It and parallel; thus Eqs. (7-3) remains true, and (7-4) do not however, that these equations which the deflection depends on the accelerating and deflecting voltages, and also, in a less specific way, 197 indicate the way Molion two dimensions in in on the characteristic dimensions of the tube. MOTION IN A CIRCLE The problem of circular motion is often presented as though it were separate from the kinds of applications of Newton's law that we have discussed so far. It may therefore be worth emphasizing that F = it involves a completely straightforward use of wa. The only special feature of the acceleration is that the radial is component uniquely related to the radius of the path and the instantaneous speed. Consider we suppose that around a circle of radius r at a Such a motion can be set up, for the simplest case, in which first an object of mass m is traveling constant speed v (Fig. 7-7). example, by tethering a puck, by means of a taut string, to a peg on a very smooth horizontal surface (e.g., an air table) and giving the puck an arbitrary velocity at right angles to the string. Then, as wc saw in Chapter 3, the acceleration of the fixed object nitude purely toward the center of the circle and is We know 2 o' /r. is of mag- that the production of this "centripetal" acceleration necessitates the existencc of a corresponding force: F = ma = (7-5) In the hypothetical case that we have have to be supplied by the tension if the string magnitude, 7-7 the tethering string. And not strong enough to supply a force of the required it will break and the object, being of the motion), Basic dynamical situation for a parlicle traveling in a circle at constant speed. 198 T in would is (at least in the plane Fig. described, this force Using Newton's law now will fly off free of forces along a tangent. y \ \ \ S* > \ n* u «e 1 To center j of curve ~^j / / / / / v /r y F. (b) Rjf. 7-5 (a) View ofa curve in a road, as seenfrom Car on the banked curve, as seen vertically ocerhead. (b) front directly behind or m front. The motion ofa car on a banked curve problem of practical this type. presents an important Suppose that a road has a radius of curvature r (measured in a horizontal plane), as indicated in Fig. 7-8(a) A and is banked at an angle a as shown in Fig. 7-8(b). car, traveling into or out of the plane of the latter with speed v, has a centripetal acceleration v of banking is to make it 2 /r. diagram The purpose some possible for the car, traveling at reasonable speed, to be held in this curved path by a force exerted on it purely normal to the road surface tangential force, as there were horizontal. (And — i.e., would have to be thcre if would be no the road surface the inability of the road-tire contact to supply such a tangential force would rcsult in skidding.) Consider, then, the ideal case as shown in Fig. 7-8(b). Resolving the forces vertically and horizontally, and applying F = wa, we have Nsina = r (7-6) Ncosa Replacing tana Fa = F„ = by mg, and solving — for a, we find that (7-7) gr which defines the correct angle of banking for given values of v 199 Motion in a circlc and r. Altematively, given r and a, Eq. (7-7) defines the speed at which the curve should be taken. The situations that arise if greater or lesser values of v are used will require the introduction of a frictional force nitude ij.N) J perpendicular to N (and of limiting mag- acting inward or outward along the slope of the banked surface (see Problem 7-17). CURVILINEAR MOTION WITH CHANGING SPEED If a particle changes its speed as travels it along a circular path, has, in addition to the centripetal acceleration toward the it center of the curvature, a the path. component of acceleration tangent to This tangential acceleration component represents the rate of change of the magnitude of the vclocity vector. (This is in contrast to the centripetal acceleration component, which depends upon the rate of change of the direction of the velocity vector.) We derivcd the relevant results in Chapter 3 [Eq. (3-18)]. The situation is most components separately: radial easily handled by considering the two component of = acceleration (at right a, = angles to the path) (r is the radius - J of curvature) r 1 j transverse acceleration) } = a$ = (tangent to the path)| £o ,- lim a<-.o — — = do [note that this Al the change of di is « magnitude (only) of the velocity vector] (7-8) With only slight rcinterpretation, we may apply these results motion along any arbitrary curvilinear path. For every point along such a path, therc is a center of curvature and an associatcd radius of curvature (both of which change as one movcs along the path). Provided that we interpret the symbols to the case of r and u to mean the instantaneous values of the radius of curvature and the speed, the above expressions are perfectly general. They give the instantaneous acceleration components tangent to the — In our formal diseussion of coordinate systems in Chapter 2 we defined the outward radial direction as positive. We are introducing here the symbol a, to denote the acceleration component along this direction. The minus sign ' indicates that this acceleration as 200 we have diseussed. is in faet toward the center (i.e., Sce also the diseussion on pp. 106-108. Using Nevvton's law centripetal), Arbitrary path Fig. 7-9 Acceleration componenls at a poinl on an arbitrarily curved palh. path, and normal to the path — for the general curvilinear motion. In this case Fig. 7-9 indicates a more appropriate notation for these acceleration components. A on a phonoup provides a simple and familiar possessing both radial and tangential ac- particle of dust that rides, without slipping, graph turntable as example of a it particle starts components. celeration Its total acceleration a combination of a r and o» at right angles: a (Fig. 7-10) = (a r 2 + a» is the 2 1' ) 2 . The net horizontal force applied to the particle by the turntable must be in the direction of a and of magnitude ma. If the contact between the turntable and the particle cannot supply a force of this magnitude, such circular motion is not possible and the particle will slip relative to the surface. If we analyze the process of whirling an object at the end of a string, so as to bring circular motion (as, for it from rest up to some high speed of example, in the athletic event "throwing the hammer") we (1) supply the tangential force to increase the speed of the object, and Fig. (2) 7-10 vector ofa see that the string must perform two functions: supply the force of the appropriate magnitude Net acceleration particle attached to a disk ifthe angular velocity of the disk is changing. 201 Curvilinear motion with changing speed mv 2 /r Fig. 7-11 Pariide tracels in a circular path about O. To crease its speed, the slring in- PO' must provide a force component langenlial to the circle. toward the center of the supplied by the string supplied by it circle. To fiil must "lead" the this dual role, the puli object, so that the tension has a tangential component, as shown in Fig. 7-11. This can be maintained if the inner end, O', of the string moved around in a circular kind of thing we do more or less conis This tinually path, as indicated. the instinctively in practice. CIRCULAR PATHS OF CHARGED PARTICLES UNIFORM MAGNETIC FIELDS One of IN most important examples of the circular motion behavior of electrically charged particles in magnetic The separate section the magnetic force is is the fields. on pp. 205-206 summarizes the properties of and describes how this force is in general given by the following equation: W F mag = where q moving magnetic (7-9) the electric charge of the particle (positive or negative). is now imagine Let us q, X B in the fieid B a charged particle of (positive) charge plane of this page, in points perpendicularly a region down in which the into the paper. Then from Eq. (7-9) we have F = qvB This is (7-10) a pure deflccting particle's motion. change, but its although there ticle 202 Hence force, always at right angles to the the magnitude of the velocity v cannot Thus, direction changes uniformly with time. is no center of force in the usual sense, the par- describes a circular path of radius r with center Using Newton's law O [Fig. (b) (a) Fig. in 7-12 (a) Charged parlicle following a circular path a uniform magnelicfield. (b) Magnelicalty curved tracks ofelectrons The main fealure energy 30 and posilrons is the paih MeV going a cloud chamber. in ofan eleclron of inilial through a succession ofalmosl circular orbils ofdecreasing radius as U loses energy. (Courtesy oflhe Unicersily of California Lawrence Radiation Laboralory, Berkeley.) 7-1 2(a)]. The centripetal acceleration is v 2 /r, and so by Newton's law we have qvB = mu whence mo Thus 203 (7-11) the radius of the circle Charged particles in is a mcasure of the uniform magnetic momentum mv ftelds of a particle of given charge underlies the in a given nuclear physicist's magnetic method momenta of charged particles in a cloud [see Fig. 7-1 2(b)]. From the radius of the for This fact field. determining the or bubble chamber circular track which a charged particle generates in a cloud chamber placed between momentum can be readily found if the To get the period T (or the the poles of a magnet the charge of the particle known. is angular velocity), wc write qvB = mwv or o> = % = 1B which is independent of the particle's spced. (7-12) quency given by Eq. (7-12) and depends, for to mass of the led him fre- called the "cyclotron" frequency field, only on the ratio of charge was the recognition of this by E. O. magnetic a fixed particle. Lawrence that is The angular It 1929 to design the in first cyclotron, in which protons could be raiscd to high energies by the application of an alternating electric field The holdmeans of constant frequency. ing of charged particles in an orbit of a given radius by of a magnetic field is an essential feature of most high-energy nuclear accelerators. Having the velocity in course, a very special case. we can imagine ular to B a plane perpendicular to But if this v to be resolved into and a second component condition 7-13 parallel to B. Helical path of charged particle having a velocity vs component parallel t o a magnetic field. 204 (Jsing Newton's lavv = not is, of satisfied, one component perpendic- Eq. (7-9), has no magnetic forcc associated with Fig. is B const. The it; latter, by the former . in precisely the changes with time the resultant motion B perpendicular to CHARGED PARTICLE It is IN is a fixed circle, as described above. Thus shown in Fig. 7-13. A MAGNETIC FIELD a matter of experimental may, at a givcn point particle way a helix, whose projection on the plane is an fact that electrically in space, experience charged a force when moving which is abscnt if it is at the same point but staThe cxistence of such a force depends on the presence somewhere in the neighborhood (although perhaps quite far it is tionary. away) of magnets or reveal the following featurcs 1 The force is [cf. Fig. 7-14]: always exerted at right angles to the direction of the velocity v of the particle. direction of the velocity 2. The force is is 3. is force reversed is The force amount of charge, reverses if is proportional to the magnitude of For motion parallel to one certain direction tion in which a compass needle placed there would the direction of the magnetic _/?e W at that point. (o) Situa- of zero force and maximum force moving in a Field direction magnet ic field. (b) General veetor relationship ity, of veloc- magnetic field, (b) and magnetic force. 205 align itself. For any other direction of motion, the direction of the for a charged particle at a given This direction coincides with the direc- is zero. size v. Field direction tions q, the sign of the For a given value of q and a given direction of v, the It is called Fig. 7-14 the reversed. point, the force 5. if reversed. of the force 4. The proportional to the carried by the particle. charge Detailed observations electric currents. Charged particle in a magnetic field force is perpendicular to the plane formed by v and the field direction. 6. The magnitude of the force proportional, for given is values of q and v and for a fixed magnetic field arrangement, to and the the sine of the angle between v Ali the above results can be field direction. summarized in a very compaet We are going to intro'duce a quantitative mathematical statement. measure of the magnetic field veetor symbol B, in such a way strength and denote that Fig. 7-14 by the it shows the relation of the direetions of qv, B, and F for a charged partiele. that q may be of (Note normal to the plane Then the value of F, in both magnitude and either sign containing v and B.) direction, is and that F is given by the following equation, involving the cross produet of veetors (see p. 127): F = We «vXB can then use of the magnetic this equation to define the quantitative measure field strength, such that a field of unit strength, applied at right angles (sin 6 with a speed of 1 = 1) to a charge of m/sec, would produce a force of 1 1 C moving N. MASS SPECTROGRAPHS The charaeteristie curvature of a partiele of given magnetic field ing precise relative devices often make The and second whercas the on electric force in the charged partiele is not. difference V, the electric force fmag = first in the is is proportional to beam of charged partieles distance d apart, with voltage If a given by ^ But the magnetic force (7-13) is given by Eq. (7-10), $<-'£ These forees can be arrangcd to be 206 manner deseribed a travels between parallel plates a Fa = a velocity selection takes advantage of the faet that the magnetic force v, in all use of the magnetic force in two ways, as a velocity selector last seetion. q/m methods of obtainvalues of atomic and isotopic masses. Such provides the basis of nearly Using Newton's law in opposite direetions by Fig. 7-15 (a) spectrometer. Schematic design ofa simpleform ofmass (b) Exampfe ofisolopic separalion and mass analysisfor isolopes of xenon with a spectrometer like that in (a). [After A. O. Nier, Phys. Rev., 52, 933 (.1937).] 207 Mass spcctrographs applying the magnetic There is field at right angles to the electric field. then only one speed at which particles can travel through these "crossed fields" undeflected: V_ (7-14) Bd Figure 7-15(a) that uses a velocity a diagram of a simple mass spectrograph of this type, followed by a region in is filter which the chargcd particles (ions) travel in a semicircular under the influence of the magnetic alone. field path Figure 7— 15(b) shows an example of the separation of isotopes by such a device. THE FRACTURE OF RAPIDLY ROTATING OBJECTS The question of up in a rotating object, and the bccome excessive, provides another the stresses set possibility of fracture i they f good example of Newton's law applied to uniform circular motion. Whenever an object such portion of it as a wheel a corresponding accelcrative force is rotating, every Suppose, for required. example, that a thin wheel or hoop of radius r its is has an acceleration toward the axis of rotation, and is rotating about axis at n revolutions per second (rps) [see Fig. 7-1 6 (a)]. Then any small section of the hoop, such as the one shown shaded, must be supplied with a forcc cqual to by 2 its centripetal acceleration v /r. of the angular velocity to Fig. 7-16 = w 2irn (a) Rotat- ing hoop. (i) Forces aeting on a small elemen! of the hoop. 208 Using Newton's law is defined its mass, m, multiplied In this case the magnitude by the equation Thus the instantaneous specd = v If = ar (7-15) isolation diagram shown shaded portion of material on that the forces acting it on ' via one end had a component at its contact with These forces must, by symmetry, rim at each point. (For example, to the Am [Fig. 7-16(b)] for the small in Fig. 7-16(a), it is clear must be supplied adjacent material of the rim. exerted given by lirrn we make an be tangential is if the force radially outward, then by the equality of action and reaction the portion of material with which it was in contact would be subjected to a force But with a component radially inward. Am portions such as asymmetries of in are equivalent; there this kind.) Thus we can a uniform hoop, is no by all any picture the small portion of the rim being acted on by a force of magnitude If the length of arc represented basis for T at each this portion is As, it end. subtends an angle A0, equal to As/r, at the center O, and each force has a Thus to Tsin(A0/2) along the bisector of AS. component equal by Newton's law we have = Am 27"sin(A0/2) Putting sin(A0/2) « o 2 A0/2, this then gives us 2 7"A0 ^ Am — r i.e., T — ~ Am — r r or 7" A.? = Let us o2 Am now (7-16) express the mass Am in terms of the density, p, of the material and the volume of the piece of the rim. cross-sectional area o f the rim is A, If the we have Am = pAAs 'We shall assume that the spokcs of the wheel serve primarily to give just a geometrical connection between the rim and the axis and have almost gible strength. 209 The fraeture of rapidly rotating objects negli- — = Substituting this into Eq. (7-16) and substituting o Eq. (7-15), we CM7) T = 4t¥rV< Now, from lirrn obtain the following result: an cxperimental it is rod of a given fact that a bar or material will fracture under tension if the ratio of the applied force to the cross-sectional area exceeds a certain critical value Thus we can infer from Eq. (7-17) we have bcen discussing has a critical the ultimate tensile strength, S. that a hoop of the kind above which ratc of rotation, it will burst. We have, in fact, 1/2 = Wmox iw (7-18) Suppose, for cxample, that we have a (i.e., and about 0.3 m). The ultimate strength its is is N/m 2 about 10° hoop of radius 1 ft about 7600 kg/m 3 steel density of steel , . These values lead rpm— much to a value of n max of about 500 rps, or 30,000 than any such wheel would normally be driven. faster However, the rotors of ultracentrifuges arc, in fact, driven at speeds of this order, significant fraetion of the bursting speed for their up to a particular radius (sec p. 513). MOTION AGAINST RESISTIVE FORCES We shall an object consider an important class of problems in which now is subjected to a constant driving force, motion opposed by a resistive force, , but has an object is In general this resistive force is and so on. speed, so that the statement of Newton's law its a pulled along a solid surface, or the air resistance to falling raindrops, cars, in Typical of such direetion opposite to the instantaneous velocity. forees are the frictional resistance as F R, that always acts moving a funetion of must be written as follows: F - = m R(o) As we saw in fact nearly that Chaptcr -= Usili»; 5 5, independent of we can put R(c) 210 in (7-19) j ~ const. Newton's la w the resistive force of dry frietion u, is as indicated in Fig. 7-1 7(a), so Resistive force Fig. 7-17 (o) Dric- Driving force Driving force mg and resistive F„ RW / forces for an object resisted lion. Frictional force by dryfric- (b) Dricing and resistive forces for object resisted ff an by "m fluid friction. (b) (a) MHHHnfflBH MBBHHBHBimiHHi Equation (7-19) then reduces to the simplc case of acceleration undcr a constant net force, as dcscribed by the kinematics of The Eqs. (7-1). situation is very different in the case of fluid resistance, for which R(o) increases monotonically with indicated in Fig. 7-1 7(b) and as described [Eq. (5-4)] o, as by the relation In this case, F the force force is + = Ac R(c) (7-20) we consider an if , Bv 2 object starting out the initial acceleration FJm, is immediately rcduced to a value below as the object has any appreciable velocity from rest under but the net driving F it , is because as soon exposed, in its own frame of reference, to a wind or flow of fluid past it at the speed v. The statement of Newton's law as it applies to this problem must now be written m— dt = Fq — Av — (In this equation One must be F , Bu (7-21) A, B, and v are what careful to consider ment of Newton's law all taken to be positive. is the appropriate state- for this system if the direetion of v is reversed.) The solving of Eq. (7-21) is not at all such a simple matter as our familiar problems involving constant forces or forces perpendicular to the velocity. awkward solution to an to plunge into stead, this 211 is all are now faced with finding the We do not intend the formal mathematies of this problem. a suitable Motion against Wc differential equation. moment In- to point to the value of ap- resistive forces proximate numerical methods— in other words, the method of the digital computer, using finitely small intervals. the principle of the have a good some method in case for using Chapter it. First, We outlined 3 [Eq. (3-11)]; here however, let we us consider individual features of the solution: For some rangc of small values of v, the acceleration will be almost constant and o will start off as a linear funetion of t 1. F with slope 2. As c /m. increases, a decreases monotonically, giving a steadily decreasing slope in the graph of v versus 3. There is (Fig. 7-18). any given applied the speed at which the graph of R(v) versus v foree. It is seeted by a horizontal line at the ordinate Algebraically, 7-17(b)]. t a limiling speed, v m , under it is equal to F is inter- [see Fig. the positive root of the quadratic equation Ao - F = B v* (What the status of the negative root, is and why is it to be discarded?) Notice the contrast between the sharply defined value of v„, in the graph of R(v) versus o, and the gradual manner in which this velocity is approached (and in principle never quite reached) one considers u as a funetion of time (Fig. 7-18). It is well worth taking a moment to consider the dynamical It is a motion with zero acsituation represented by v = v,„. celeration under zero net foree, but it seems a far cry from the if moving under no foree at all; not an application of what we understand by But let us emphasize that, like static the principle of inertia. equilibrium, it is a case of ZF = 0. Every time we see a car unacceleratcd motion of objects and it is Fig. 7-18 ccrtainly lo terminal Asymplolic approach speed for object in a fluid resislice 212 medium. Using Newton's law hurtling along a straight road at a steady 80 racing through the air at a constant 600 mph, mph, or a jet plane we are seeing objects traveling under zero net force. This basic dynamical fact tends to be obscured because what matters in practical terms value of the driving force motion once it F is the large needed to maintain the steady has been established. DETAILED ANALYSIS OF RESISTED MOTION how Newton's law for resisted motion [Eq. practice we need to introduce one additional In order to see (7-21)] works feature in already describcd in Chapter 5 sistive force differ their dependence on the Fig. 7-19). — that the two terms in the rc- not only in their dependcncc on d but also in linear dimensions of the object (see Specifically, for a sphere of radius r we have A = C\r B = C 2 r'2 and thus = R(v) C\ro The two terms + C2'V in the resistance thus (7-22) become equal at a critical speed, v c defined by the formula , Vc = (7-23) c 2r We know picture if v Fig. 7-19 that the term proportional to v will is sufficiently small (since Bv 2 /Av (a) Linear and quadratic terms influid resistance for a small object, with viscous resistance predominant. (b) Similar graphfor a large object, for which the v 2 term predominates at a/l except low speeds. 213 Dciailcd analvsis of resisted motion ~ v), dominate the but we know equally that for speeds much quadratic term takes over. > in excess of vc (say v If the resistive medium 10uc ) the the is air, coeffkients c t and c 2 have the following approximate values: = ci = c2 Thus, if 0.87 r 10- 4 X 3.1 kg/m kgm- 1 sec _1 3 we have expressed in meters, is ... 3.6 = v c (m/sec) 10~ 4 X ,, ... (7-24) This means that for an object such as a small pebble, with r » cm, the value of vc 1 only a few centimeters per second; a is speed equal to 10 times this (say 0.5 m/sec) would be acquired under gravity within a time of about 0.05 sec and a in free fail distance of about 0.5 in. (Check thcsc numbers!) problems of practical (we interest shall consider Thus for most an exception in the next section) the motion under a constant applied force in a medium can be extremely resisting well described with the help of the following simplified version of Eq. (7-21): m « ^ di The F - B 2 (7-25) resistive term, Bv 2 , is quite important in the motion of ordinary objects falling through the air under gravity. becomes very apparent = putting dv/dt tional force, radius 1 in we if Eq. (7-25), with The = 2.5 times that « :s ^ pr F c 2r 4(2.5 is 2 ) X Bu, 2 a 10 ) X (10" 2 3 ) thus about 0.1 N. for an object of in the « 10" 2 kg The value of this size is about 10 the coef- -4 kg/m. equation = find Vi Under 214 by , B (= F - , equal to the gravita- :! Substituting these values we set cm. The density of stone or rock is about so we have i.e., about 2500 kg/m value of ficient F Take, for example, the case of our pebble of mg. of water, m This calculate the terminal speed, v t « 30 m/sec the assumptions of genuinely free Using Ncwton's law fail, this speed would : be attained about 150 We appreciably significant than these. Iess problems of meets can be sure, then, that the effects of the become quite sistance ized time of about 3 sec and a vertical distance of in a ft. free fail within This means that under re- times and distances many of the ideal- gravity, of the sort that everyone encounter with mechanics, do not correspond in his first very well with reality. now Let us to handle the consider the computer procedure for solving Given a computer, these problems. it is almost as equation [Eq. (7-21)] as full proximation represented by Eq. (7-25). it is little trouble to use the ap- But for the purposes of discussing the method, we shall use the simpler, approximate form. We and choose some convenient small increment of time, will measure time from t = in A/, terms of integral multiples of Af. In the simplest approach, we assume that the acceleration during each small interval remains constant at the value (n) calculated for the beginning of that interval. Thus, for an velocity equal to zero, the acceleration during the is The end At initial is set of this interval thus given by Using + = ci 2 At 02 calculate the acceleration d\ at / = At: 2 m Cl tfo this acceleration to the next interval, the velocity at given by is = We know this we B = Applying = aoA/ this velocity, ai t F Jm). (= equal to a velocity Vi at the first vi + fli At that the first step in the calculation, as performed in way, leads to an overestimate of Pj, but we see that in this is in some measure compensated by particular problem the error leading in turn to an underestimate of a%. The calculated values of a in successive time intervals are as indicated in Fig. 7-20(a). Applying an exactly similar treatment to the changes of position we would take our set of values of velocities, as given by the above calculation and represented graphically in Fig. 7-21 (a), and would use the following formula (jt„) x„ +i 215 = x„ + v„At Detailed analysis of resisted motion Fig. 7-20 (d) Basis ofsimple ealculation of acceleration versus time for objecl start- ingfrom resistive resi n a i medium. The acceleration is calculaied from the speed al the beginning ofeach time interval. {b)lmproved approach to the same problem, using acceleration calculated from speed at midpoint ofeach time Thus we should have interval. = = = x\ X2 X3 ro A/ = (clearly x\ + oi X2 + CiAt At Our graph of x versus = t an underestimate) vi At would be the result of summing the areas of the rectangles in Fig. 7-21 (a) up to successively greater values of n. A more sophisticated analysis takes account of the fact that a better average value of the acceleration or velocity during a given time interval is provided by the instantaneous value at the midpoint of that interval. Thus the acceleration between n At and (n + 1) At is set equal to the instantaneous value at (n This Ieads to the following formulas: Fig. 7-21 (a) Veloc- ity-time graph based on velocities at the beginnings of the successive time intervals. (b) Improved graph based on velocities evaluated at midpoints of the time intervals. 216 Using Newton's law + -J) At. v n +i = c„ + a n+ i /2 A/ *»+l = X„ + Vn+i /2 &l This looks fine, but (7-26) we now run into a slight snag when we try To find v u the velocity at the cnd we now need v (= 0) and a ll2 The lattcr, to get the calculation started. of the first interval, . however, by Eq. (7-25), depends on a knowledge of v U2, which we do not yet know. (Noticc that in the first, crude method, we were able to and a Fig. 7-22 (a) .) start We are out directly from the Com- parison ofidealized (resistanceless) and aclual dependence of speed on lime for a falling pebble of radius l cm. (b) Idealized and aclual dislances fallen I by such a pebble. 217 initial thus forced to compromise a Detailcd analysis of resistcd motion conditions v little, although we are with a better treatment than before. still Ieft to calculate an approximate value of » V112 vn 2 from What we do is the equation (7-27) «o -z and then we are under way. Figures 7-20(b) and 7-21 (b) show what this method means in terms of graphs of a and v against time. Figure 7-22 shows graphs of the actual calculated variation of speed and distance with time for our 1-cm-radius pebble falling in air, under gravity; the idcalizcd frce-fall curves are given for comparison. MOTION GOVERNED BY VISCOSITY If we are dealing with microscopic or near-microscopic objects, such as particlcs of dust, then, in contrast to the situations discussed above, the resistance is due almost term, Av, up to quite high values of of radius sider a tiny particle tells v. If, m (= 10 1 entirely to the viscous we con- for cxample, -6 m), then Eq. (7-24) us that the critical spccd, vc at which the contributions , and Bo 2 bccome equal, is of lower speeds for which the motion resistance alone, writtcn, without and Av 360 m/sec. This implies a wide range is controlled by viscous the statcmcnt of Newton's law can be any appreciable error, in the following form: *° m di mFo -Ao (7-28) with A = c\r Motion under these conditions played the central role in R. A. Millikan's cclebrated "oil-drop" experiment to determine the elcmentary electric charge. The basic idea was to measure the electric foree exerted on a small charged object by finding the terminal spced of the object in air. If the radius of the particle is completely determined, and the known, the resistive driving foree must be equal and opposite to foree is it at v = v t . In Millikan's original experiments the charged partieles were tiny droplets of oil in the mist of vapor from an "atomizer." Such droplets havc a high probability of carrying a net 218 Using Newton's law electric Fig. o d Basic paratlel-plate arrangement in Millikan 7-23 experiment. q charge of either sign when they are produced. In order to apply them, Millikan used the arrangement shown electric forces to Two schematically in Fig. 7-23. parallel metal plates, spaced by a small fraction of their diameter, are connected to the terminals of a battery. ' The force on a tween these plates q anywhere be- given by v _ where is particle of charge V (7-29) the voltage difference applied to the plates is their separation in meters. If q is and d is F is measured in coulombs, given in newtons by this equation. (See also our discussion of the Thus a dynamic balance cathode-ray oscilloscope, p. up between this force Av-, we should have electric driving force were set V q-j= a The Ad, = if and the resistive ciru, droplets of various sizes. 195.) randomly produced in a mist of oil vapor The ones that Millikan found most suitable are for his experiments were the smallest (partially because they had the lowest terminal speed under their lets own weight). But these drop- through a medium-power microscope tiny that even were so they appeared against a dark background merely as points of light; no direct measurement could be made of their size. Millikan, however, used the clever trick of exploiting the law of viscous resistance a second time by applying it to the fail of a droplet under the gravitational force alone, with no between the Fo = plates. mg = Under these conditions y pr g the density of oil (about 800 where p is speed of fail voltage we have under gravity is kg/m 3 ). The terminal then given by Millikan himself used plates about 20 cm across and 1 cm apart and several thousand volts. Most modern versions of the apparatus use smaller values of all three quantities. 219 Motion governed by viscosity a 4jt y pr g = (Gravitational) = u„ cirDB y— r 2 (7-30) Putting in the approximate numcrical values (Gravitational) Putting r f « « 10 lju -4 = ~ i?„ 10 m/sec -0 = 10 r 2 find (u in m/sec, ( /• m) in m, we have 0.1 mm/sec min to fail 1 cm in air under own weight, thus allowing prccision measurements of its speed. Such a droplet would take over its we It term Bo 2 is motions as incidentally, that for such (It is clear, sistive 1 is this the re- utterly negligible.) worth noting the dynamical stability of this system, and indeed of any situation involving a constant driving force and a resistive force that increases monotonically with speed. by chance the droplet should slow down force that will speed it up. Conversely, subjected to a net retarding force. motion of a a if it If little, air, being molecules, does not behave as a perfectly If a net is should speed up, it is one could observe the falling droplet in sufficient detail, this doubtless be found, because the there behavior would made up of individual homogeneous fluid. other words, the speed of the droplet would fluetuate about In some average value, although the fluetuations would be exceedingly small. In the Millikan experiment proper, the vertical the charged droplets is or opposing the gravitational force. Thus as positive and sign downward, the terminal will if we measure velocity in both V is velocities magnitude be defined by the equation mg + *r- cm, where motion of studied with the electric force either aiding (7-31) the voltage of the upper plate relative to the lower and q is the net charge on the drop (positive or negative). Although in principle the terminal velocity is approached but never quite reached (see Fig. 7-18), the small droplets under the conditions of the Millikan experimcnt do, in effect, reach this speed within a very short time — much less than 1 msec in most cases (see the next seetion). Millikan was able to follow the motion of a given droplet 220 Using Ncwton's lavv ' for many hours on end, using which to puli it up or down its electric at will. tracted observations the charge charge as a handle by In the course of such pro- on the drop would often change spontaneously, and several different values of v, would be ob- The tained. crucial observation was that any such experiment, in with a given value of the voltage V, the speed v was limited to a t set of sharp and comes itself distinct values, implying that the electric charge in discrete units. obtained the first But Millikan went further and precise value of the absolute magnitude of the elementary charge. You method might wonder why Millikan used such a roundabout to on a charged principle be possible to hang such measure the electric force exerted would in on a balance and measure the force in a However, in practice, when only a few static arrangement. elementary charges are involved, the forces are extremely weak and such a method is not feasible. For example, the force on a After particle. an all, it electrified particle particle with a net 1 cm charge of 10 elementary apart with 500 V between them, is units, between plates 13 N, only about 10~ equal to the gravitational force on only 10 ,upg! GROWTH AND DECAY OF RESISTED MOTION We have seen how the velocity under a constant force resistive fluid medium What happens value. We can rises asymptotically if the driving force is suddenly removed? guess that the velocity will decay away toward zero in a similar asymptotic way, as indicated in Fig. 7-24. speed is If the initial small enough, the whole decay depends on viscous sistance alone and Eq. (7-28) in which = m— dt is F governed by a is set special, simplified re- form of equal to zero: — Av (7-32) or do — = dt where a 'For his —at) = A/m. own full and interesting account (and much other good physics), The Electron (J. W. M. Du Mond, ed.), University of see R. A. Millikan, Chicago Press (Phoenix 221 in a toward the terminal Series), Growth and decay of Chicago, 1963. rcsisted motion Fig. 7-24 Growth and decay of the velocity ofa parlicle controlled by a viscous resistiue force proportional to v. You may equation of not, is recognizc Eq. all you may (7-32) as the basic differential how in effect just arithmetic, this equation can be solved by what using the approach of numerical We divide up the time analysis and digital-computer techniques. into a large Whether you do or forms of exponential decay. like to see number of equal intervals Al, and interpret Eq. (7-32) as telling us that the changc of v between (n + 1)A/ proportional to the is mean t = n At and t = velocity during that interval: Au = Solving t;„ v n+ i we have this, +i -*~-"\—2—) At 1 - 1 + a Al/2 aAt/2 _ where/is a constant ratio, less than unity. The velocities at equal intervals of time thus decrease in geometric proportion. initial velocity is u the velocity at time , Substituting the expression for and putting k = 1 - 1 + fO o(t) t/At, we / t (= k Al) is If the given by from the preceding equation thus get aAr/2\ a At/l) l/Ai we made shorter and shorter, and their number correspondingly greater. To simplify the discussion, we We shall now consider what happens to this expression as imagine the intervals At to be shall put aAt = z 222 Using Newton's law The quantity approach z is then a large number that oi/= Besides substituting u(l), we we shall allow to infinity. l/z in the above equation for exponent t/At by shall also replace the its equivalent Thus we have quantity, azt. - /l \/2z\ a" \TTtj2z) \i -r i/^z/ i.e., o (0 «» tfO/> where Let us look at the behavior of p(z) as z larger (Table 7-1). As z increases, the made is number p larger and clearly ap- is TABLE 7-1 Decimal value of p y(z) z 1 0/3)' 0.3333 2 (3/5) 2 0.3600 3 (5/7)3 0.3644 4 (7/9)* 0.3659 5 (9/1 l) 5 0.3667 (19/21)'» 0.3676 10 proaching a limiting value; reciprocal of the this value is famous number the base of natural . . 0.367879 . . . . and is the (= 2.71828 .), which forms or Napierian logarithms. Thus in Eq. (7-32) e . . we can put lim p(z) = 0.367879 . . . = -1 e Z—*X and hence we have the following expression, now value of 223 exact, for the v(i): v(t) = The reciprocal of v e~°' (7-34) a in Growth and decay of Eq. (7-34) resisteci is of the dimension of time molion and represents a characteristic time constant, ponential decay pjocess. moving through the _ We can m m_ A ar air, I r for t, the ex- n the particular case of small spheres is defined by the equation 0. express this in terminal velocity of much more fail vivid terms under gravity, vg . by introducing the For from Eq. (7-30) we have mg "'w It follows that we have r = 2» (7-35) • g Thus t equal to the time that a particle would take to reach is a velocity equal to the terminal Velocity with v„ » 0.1 mm/sec, For an oil drop of radius 1 -5 sec (= 10/tsec). time would be only about 10 free fail. this under conditions of /j., In a time equal to any substantial multiple of t, the value of v(t) as given by Eq. (7-34) For example, = v(t) and put v , voe-' = t we take if falls to a very small fraction of vo- the basic equation ir (7-36) . 10r, then the value of v becomes less than 10 -4 toward its limiting value, after a particle starts the action of a suddenly applied driving force, from rest under must be a kind of upside-down version of the decay curve (see Fig. 7-24). fact, if the terminal velocity is o«, conditions of viscous rcsistance, v(l) If = oAl — t v u(0 one can see are related. - the approach to is it, In under these described by - e-" T ) one chooses to write v, (7-37) this in the form — lir »( explicitly how closely the growth and decay curves Indeed, one could almost deduce Eq. (7-37) from Eq. (7-36) plus the recognition of this symmetry. 224 of which for many purposes can be taken to be effectively zero. It is more or less intuitively clear that the growth of velocity Using "New lon's law AIR RESISTANCE AND "INDEPENDENCE OF MOTIONS" When Galileo put forward the proposition that the motion of a projectile * could be analyzed into separate horizontal and vertical and a constant ac- parts, with a constant velocity horizontally celeration under gravity vertically, he to the conceptual basis of mechanics. made It a great contribution may be worth pointing down out, however, that this "independence of motions" breaks if one takes into account the resistive force exerted on objects of ordinary We have seen how for such objects the magnitude size. of the resistive force is proportional to the square of the speed. Consider an object moving in a vertical plane (Fig. 7-25). At a given instant let the horizontal, its velocity v be directed at as shown." The gravitational force, F„, and a object an angle 6 above then subjected to the is resistive force R(u), of magnitude Bv 2 , in a direction opposite to v. Newton's law applied to the x and y components of the motion at any instant, thus gives us d m -^= -R z = -Bo 2 cose at dCy = —mg — m-^ Since v cos 6 = vz , _ 2 Bu and v n sin 6 • sin 6 = vu , we can rewrite these equations as folio ws: m —= do* \ - la {BojOm m-— = -mg dt (flOPy Thus the equation governing each separate component of velocity involves a velocity knowledge of the magnitude of the and hence of what is other coordinate direction. the ( a) more important does ferent Fig. 7-25 partiele 225 happening at each instant along the The larger this cross the magnitude of v, connection between the components of the motion become. Thus we caleulate the vertical Resistive moving in the total motion of a falling really dif- cannot body without reference and gravitational force vectorsfor a a vertical plane. Air resistance and "independence of motions" to the horizontal component. salutary for those who Recognition of this fact may be are accustomed to taking for granted the In the folklore of physics idealized equations of falling objects. who would make money off his nonscientific acquaintances by a bet. He asked them to consider two identical objects. One is dropped from there is the story of the impecunious student rest at a certain height same horizontally at the The first? above the ground; the other instant. Which one victim (poor ignoramus!) vague idea that the must somehow keep is fired off reach the ground would often have some motion of the second object fast horizontal it in will the air longer. But of course that is it?— look at the analysis of motion under gravity in Chapter 3. And even a direct test would appear But the to prove the poin t if the velocities were kept small. demonstrably false, isn't complete equations, as written above, show that a large initial horizontal velocity would increase the time taken to descend a given vertical distance. Notice that Moral: Beware of if the resistance is facile idealizations. purely viscous, varying as the power of v, then the x and y components of the motion can be handled entirely separately, even though the problem is not that of idealized free fail. Thus, although one can always take a statement of Newton's law as the starting point of any problem first of a particle exposed to forces, the way of proceeding from there to the analysis of the complete motion may depend critically on the precise nature of the individual problem. SIMPLE HARMONIC MOTION One of the most important of of a mass to its distance from that point. confined to the F(x) all dynamical problems that is by a force proportional the motion is assumed to be attracted toward a given point x axis If we have = -kx (7 ~38) good approximation to this situation is provided by an object on a very smooth horizontal table (e.g., with air suspension) and a horizontal coiled spring, as shown in Fig. 7-26(a). The object A would normally rest at a position in compressed nor extended. slight 226 The force brought displacement of the mass described by Eq. (7-38). Using Newton's law which the spring is neither into play in either direction is The constant k is by a then well called the spring o (a) 7-26 Fig. (a) Mass-spring sysiem on a frictionless horizontal surface. (b) Mass hanging front a vertical Graph offorce versus displacement with a linear force law; the equilibrium silualions O and O' spring. (c) correspond to (a) and (b), respeclivety. and is measured in newtons per meter. This linear force law for a spring was discovered by Robert Hooke in 1676 and is An even simpler arrangement in practice, named after him. more complicated theoretically, is to suspend slightly although constant ' a mass at the bottom end of a vertical spring, as in Fig. 7-26(b). In this case the normal resting situation already involves an extension of the spring, sufficient to support the weight of the object. A further displacement up or down from this equilibrium position leads, however, to a net restoring force exactly of the form of Eq. (7-38). This is indicated in Fig. 7-26(c), which shows the magnitude of the force exerted by the spring as a funetion of its extension y. If one takes as a new origin of this graph the point O', then one has a net restoring force (F tional to the extra displacement (y The — — mg) propor- ya). great importance of this dynamical problem of a mass on a spring is that the behavior of very many physical systems under small displacements from equilibrium obeys the same basic equation as Eq. (7-38). We shall diseuss this in more detail in Chapter 10; for the present we shall just concern ourselves with solving the problem as such. announced it in a famous anagram—ceiiinosssttuv—which 2 years he revealed as a Latin sentence, "ut tensio, sic vis" ("as. the cxtension, so the force"). By this device (popular in his day) Hooke could claim prior publication for his discovery without actually telling his competitors what J He first later it 227 was! Simple harmonic motion This makes a good case in which to begin with the computer method of solution rather than with formal mathematics. The basic equation of motion, expressed in the form ma = F, is as follows: d\ Rewriting . (7-39) this as 2 we d x k dfi m k/m recognize that Denoting dfi We can is X this by w 2 , our is a constant, of dimension (time) basic equation thus 2 . becomes = —W X (7-40) read this as a direct statement of the way in which dx/dt changing with time and can proceed to calculate the approximate change of dx/dt i( di \<//^ i) in a small interval of time At 2 - - CO X and so (f)~- w! xM This is defined Fig. 7-27 the reverse of the process (cf. Chapter 3). Displace- ment versus time in simple harmonic motion. 228 Using Newton's law (7-41) by which d 2 x/dt 2 was originally : Suppose, to be x = x and v Then 7-27. (= at time = / *» xo + —« i>o » - w 2 A;oAr x rfx = we that specific, dx/di) both v = with shown in Fig. out at start positive, as f A/ we have f o Ar The displacement is a little bigger, the slope a little these new values, we take another step of A/, and so features can be read from Eq. (7-41): 1. As long as x we go from if dx/dt 2. to t + / negative is The graph has Using t as more change of dx/dt it gets smaller; proportional to x. is The — it becomes less and as x x becomes negative, dx/dt positive with each time increment A/. these considerations, a curve that positive is a straight line. As soon negative or \idx/dt becomes more negative. greatest curvature at the largest x, becomes almost 3. is, Several dx/dt decreases when positive, the slope A/. That it rate of its is Using less. on. we can construct the picture of always curving toward the line x is axis), necessarily = (i.e., the forming a repetitive wavy pattern. Now anyone who has ever drawn graphs of trigonometric functions will recognize that Fig. 7-27 looks remarkably like a sine or cosine curve. More specifically, it suggests a comparison with the following analytic expression for the distance x as a function of time x = A where A is sin(af the + ô) maximum value attained by a a constant with the dimension angle that allows us to Testing this fit (time) the value of trial function -1 x , at x during the motion, and <p an adjustable t = 0. against the original differential equation of motion [Eq. (7-40)] requires differentiating x twice with respect to v = /: — = aA cos(atf at + <ô) .2 a = -—2 = at We see a = w. 229 —oc A sin(a/ + <f>o) — —a that the solution does indeed x fit, provided that we put This then brings us to the following Simple harmonic motion final result: = A x(t ) Equation (7-42a) A what is + 1/2 (7-42a) J harmonic called a is of the sine function , I what is The oscillator. and v>o (Gk: phi) is The complete argument, the amplitude of the motion, is called the initial phase (at ip w = the characteristic equation of is equation of motion this at where motion (SHM), and any system that obeys called simple harmonic constant —„v -ay + *?o) sin(w/ motion at any given The instant. = / is 0). called just "the phase" of the result represented by Eq. (7-42a) could be equally well expressed by writing x as a cosine function, rather than a sine function, of x(l) = A + cos(at <p' /: (7-42b) ) <p' This form of the some purposes. with an appropriate value of the constant solution is found more convenient The harmonic motion is for . characterized by its period, T, which defines successive equal intervals of time at the end of which the state of the motion reproduces itself exactly in both displacement and velocity. The value of T is readily obtained Eqs. (7-42) by noting that each time the phase angle (at changes by 2w, both + tpi = 2vr = + ip ) v have passed through a complete Thus we can put cycle of variation. <pi x and from + o>(/i + uti </>o + T) vo Therefore, by subtraction, 2w = 10T or ' /' (r) !-(!)' The form of and if more result this knowledge that if the spring m is (7-43) corresponds to one's commonsense gets bigger the oscillation goes madc stiffcr (largcr more slowly, k) the oscillation becomes rapid. Example. a fixed point. A spring of negligible mass hangs vertically from When a mass of 2 kg of the spring, the spring is is hung from the bottom end stretched by 3 cm. What is the period of simple harmonic vibration of this system? First 230 we calculatc the foree constant k. Using Newton's law The gravitational force on the mass of extension 2 is N 9.8 = m; therefore, we then have 0.03 Using Eq. (7-43) . N. 19.6 k = This force causes an = 653 N/m. 19.6/0.03 1/2 "" '-*(&) (You might X like to =°- 35!sec note that we need not have cm when a specified the mass. mass Any spring that extends 3 will have a vibration period of 0.35 sec with certain is this hung on it same mass. Why?) MORE ABOUT SIMPLE HARMONIC MOTION Fitting the initial conditions We have come to recognize, both as a general principle and through various examples, that the complete solution of any problem in the use of Newton's law requires not only a knowledge of the force law but also the speciflcation of two independent quantities that correspond to the constants of integration intro2 2 duced as we go from a (= d x/dt ) to have talked of giving the velocity, v oscillator, . initial x. position, we also need initial conditions or their equivalent. We have already identified A <p initial x A and A and <p >n Eq- (7-42). as the amplitude of the motion and are given the values of x <po and v we as follows: Eq. (7-42a), = A sin(co/ + = we phase. If can readily solve for From , Here, in our analysis of the motion of a harmonic Actually they appear as the two constants as the Most commonly we and the initial x = wA ^ dt ifio) cos(o>/ + *>o) (7-44) Therefore, xo Vq = A sin v>o = o>A cos v>o It follows that 1/2 '-k+(?)T tanô = oixo oo 231 More about simple harmonic motion (7-45) A geometrical representation of There a is basic connection very motion and uniform motion a particularly simple Imagine that 7-28(a). peg will P between simple harmonic in a circular path. This fact leads to way of SHM. displaying and visualizing a horizontal disk, of radius A, rotates with constant u about a angular speed that a peg SHM vertical axis mounted on is Then, if the disk through its center. the rim of the disk as is Suppose shown in Fig. viewed edge on (horizontally) the seem to move back and forth along a horizontal straight Its motion along this line will correspond line [Fig. 7-28(b)]. exactly to Eq. (7-42a) if the angular position of the peg at is correctly chosen so that the angle to ut + <pq. SOP in Fig. In this representation the quantity actual angular velocity of P as it travels 7-28(c) w is around the is t = equal seen as the circle. The peg has a velocity v that changes direction but always has the magnitude u A. (7-44). 7-28 Fig. a uniformly disk. (a) Thus Resolving v parallel to the motion of the point Peg on rotating (b) Displace- ment of the peg as viewed in a plane conlaining the disk. (c) Detailed relation of circular motion to simple harmonic motion. 232 Using Newton's !aw B Ox at once gives Eq. in Fig. 7-28(c) corresponds in every respect to that of a particle along the x Dynamical performing SHM axis. relation between SHM and circular motion Although the analysis just described is a purely geometrical one, it suggests a close dynamical connection, also, between the actual linear motion of a harmonic oscillator and the projection of a The acceptance of F = ma means that same motion implies the same force causing it, whatever the uniform circular motion. the particular origin of the force may be. understood with the help of Fig. 7-29. kept moving in a circle of radius to a This equivalence can be A particle of A by means fixed support at the center of the circle, O. has a constant speed, v, mass m is of a string attached If the particle the tension, T, in the string must be given by mv T= ~A Here v 2/'A is the magnitude of the instantaneous a„ of m toward the center of the circle. Now the total force and the acceleration, total acceleration at any instant can be resolved into their x and y components in a rectangular coordinate system. (Normally we would not be interested in doing Fig. 7-29 this, because T and a, have well-defined constant mag- Dynamical relationship between and circular motion linear harmonic motion. 233 More about simple harmonic motion Taking the x components alone, we have nitudes.) mu 2 — cos e Fx = - T cos e = A 2 ax v = cos 8 r A Thus the x component of the complete vector equation, F = ma, is Fx = max with the values of Fx and ax stated above. In order to display the dynamical identity of this component , motion with SHM, we uA. We Fx and ax putting v = can take the expressions for separately, introducing the angular velocity « and then have Fx - —mco 2 Acosd = —mw 2x ax = —u 2 Acos8 = — u 2 x The first of these equations defines a restoring force proportional to displacement, exactly in accord with our initial statement of Hooke's law [Eq. (7-38)]. The second corresponds the equation [Eq. (7-40)] that was our exactly to starting point for the kinematic analysis of the problem: d x Thus we 2 see that the every respect. dynamical correspondence It tells us, is moreover, that we could, complete in if we wished, go the other way and treat a uniform circular motion as a superposition of two simple harmonic motions at right angles. This is, in fact, an extremely important and useful procedure in some contexts, although we shall not take time to follow it up here and now. PROBLEMS 7-1 Two identical gliders, each of mass m, are being towed through Initially speed and tow rope they are traveling at a constant the air in tandem, as shown. the tension in the A is 7"o. begins to accelerate with an acceleration a. A and B immediately 234 The tow plane then What are the tensions in aftcr this acceleration begins? Using Newton's law . Two 7-2 M on a horizontal A table. M as shown. to block M blocks, of masses and the There m = 2 kg, are F = 5 in contact is applied N m N be- but no frictional force between the M block first kg and 3 a constant frictional force of 2 is tween the table and the block table = constant horizontal force (a) Calculate the acceleration of the two blocks. (b) Calculate the force of contact between the blocks. 7-3 A mass sled of m is pulled by a force of magnitude to the horizontal (see the figure). surface of snow. It Draw an on P at angle 8 sled slides over a horizontal experiences a tangential resistive force equal to M times the perpcndicular force (a) The isolation W exerted on the sled by the snow. diagram showing the forces exerted all the sled. F = ma Write the equations corresponding to (b) horizontal and vertical the for components of the motion. Obtain an expression for the horizontal acceleration (c) terms of P, 6, m, n, and g. For a given magnitude of P, find what value of 6 gives (d) in the biggest acceleration. 7-4 it is A mi block of mass frictionless horizontal surface; on a rests connected by a massless string, passing through a frictionless eyelet, to a second block of mass mi that rests on a frictionless incline (see the figure). Draw (a) tion of isolation diagrams for the masses and write the equa- motion for each one separately. (b) Find the tension (c) Verify that, for 6 in the string = jt/2, and the acceleration of mi. your answers reducc to the expected results. 7-5 In the figure, F acts on (a) it P a pulley of negligible mass. is An external force as indicated. Find the relation between the tensions on the right-hand sides of the pulley. Find also the relation between F and left-hand and the tensions. (b) What relation among the motions of m. M, and P is provided by the prescncc of the string? (c) Use the above results and Newton's law as applied to each M, and P in terms of m. M, g, block to find the accelerations of m, and F. Check that the results make sense for various specialized or simplified situations. A man and the platform on which he stands 2 with a uniform acceleration of 5 m/sec by means of the rope-andpulley arrangement shown. The man has mass 100 kg and the platform is 50 kg. Assumc that the pulley and rope are massless and 7-6 move without 235 Problems is raising himsclf frietion, and neglect any tilting effects of the platform. Assume g = 10m/sec 2 (a) What are the tensions . (b) Draw draw a separate isolation What clearly indicate all the platform and the forces acting direction. its the force of contact exerted is and C? man and force diagram for each, showing on them. Label each force and (c) in the ropes A, B, diagrams for the on the man by the platform ? 7-7 In an equal-arm arrangement, a mass 5/no is balanced by the masses 3mo and 2mo, which are connected by a string over a pulley of negligible mass and prevented from moving by the string the figure). e.g., Analyze what happens if the string A is A (see suddenly severed, by means of a lighted match. - "S 5»i„ 3»i„ 2m 7-8 A prisoner in jail decides to escape by sliding to freedom He a rope provided by an accomplice. down attaches the top end of the rope to a hook outside his window; the bottom end of the rope hangs The rope has a mass of 10 kg, and the prisoner The hook can stand a puli of 600 N without prisoner's window is 15 m above the ground, what clear of the ground. has a mass of 70 kg. giving way. is If the the least velocity with which he can reach the ground, starting from rest at the top 7-9 (a) end of the rope? Suppose that a uniform rope of length L, resting on a tionless horizontal surface, length by means of a force is fric- accelerated along the direction of F pulling it at one end. Derive its an expression T in the rope as a function of position along its length. How is the expression for Tchanged if the rope is accelerated vertically for the tension in a constant gravitational field ? (b) A mass M is accelerated by the rope in part (a). Assuming the mass of the rope to be w, calculate the tension for the horizontal and vertical cases. 7-10 In 1931 F. Kirchner performed an experiment to determine the charge-to-mass ratio, e/m, for electrons. An electron gun (see the figure) produced a beam of electrons that passed through two "gates," each gate consisting of a pair of parallel plates with the upper plates Electrons could pass connected to an alternating voltage source. 236 Using Newton's law -07 -K o- / straight through a gate only momentarily Wilh zerc. the voltage if on the upper plate were the gates separated by a distance / X equal to 2.449 10 7 Hz (1 Hz = 1 / equal to at a frequency 50.35 cm, and with a gate voltage varying sinusoidally cycle/sec), Kirchner found that electrons could pass completely undeflected through both gates when Under these the initial accelerating voltage (K ) was set at 1735 V. conditions the flight-time between the gates corresponded to one half-cycle of the alternating voltage. (a) (b) (c) What was the electron speed, deduced directly from / and /? What value of eI m is implied by the data ? Were corrections due to special relativity significant? [For Kirchner's original paper, see Ann. Physik, 8, 975 (1931).] 7-11 A certain loaded car way between the front and is to have rear axles. It (at the rear) start slipping How known when is the car its center of gravity half- found that the drive wheels is driven up a 20° incline. back must the load (weighing a quarter the weight of the empty car) be shifted for the car to get up a 25° slope? (The distance far between the axles 7-12 A is child sleds 10 ft.) down a snowy hillside, starting from rest. level field at the foot. The The has a 15° slope, with a long stretch of ft up the slope and continues for 100 ft on the level field before coming to a complete stop. Find the coefficient of friction between the sled and the snow, assuming that it is constant throughout hill child starts 50 the ride. 7-13 A Neglect air resistance. beam of electrons from an electron gun passes between two parallel plates, 3 mm apart and 2 the electrons travel to farther on. 3 It is cm when 100 V if long. After leaving the plates desired to get the spot to deflect vertically through are applied to the deflector plates. the accelerating voltage general, that cm form a spot on a fluorescent screen 25 cm Vo on the electron gun? What must (Show first, be in the linear displacement caused by the deflector plates can be neglected, the required voltage is given by V o = V(lD/2Yd), where Y is the linear displacement of the spot on the screen. The notation 237 is that Problem s used on p. 197.) A 7-14 ball known It is of mass m is attached to one end of a string of length nine times the weight of the made table, is /. that the string will break if pulled with a force equal to ball. The ball, supported by a frictionless to travel a horizontal circular path, the other end of the What is the largest number make without breaking string being attached to a fixed point O. of revolutions per unit time that the mass can the string? A 7-75 cm mass of 100 g is attached to one end of a very light rigid rod The other end of the rod is attached to the shaft of a motor so that the rod and the mass are caused to rotate in a vertical 20 long. with a constant angular velocity of 7 rad/sec. circle Draw a (a) mass force diagram showing by the rod on the mass when'the rod points i.e., at0 = 90°? You 7-16 the forces acting all on the an arbitrary angle of the rod to the downward vertical. (b) What are the magnitude and the direction of the force exerted for in a horizontal direction, Camel at 60 mph and when suddenly you behind you fiying at 90 mph. are fiying along in your Sopwith 2000-ft altitude in the vicinity of Saint Michel Red Baron notice that the is just 300 ft Recalling from captured medical data that the Red Baron can withstand only 4 g's of acceleration before blacking out, whereas you can withstand 5 down you decide on the following plan. g's, You dive straight at full power, then level out by fiying in a circular are that out horizontally just above the ground. Assume comes that your speed is constant after you start to puli out and that the acceleration you experience in the are will follow is Since 5 g's. you know that the Red Baron you, you are assured he will black out and crash. Assuming that both planes dive with 2 g's ucceleration from the same initial point (but with initial speeds given above), to what altitude must deseend so that the Red Baron, are, is Assuming either crash or black out? must a poor shot down, in trying to follow will and must get within 100 ft that the your plan succeed ? After starting down you recall reading you excced 300 mph. if your plan sound on your plane? 7-17 A curve of 300 of 25 m/sec is (« 55 m mph) normal to the surface (a) What (b) The radius on a level Red Baron of your plane to shoot you that the wings of your plane will fail off in view of this limitation you your subsequent road is so that the force exerted Is banked for a speed on a car by the road at this speed. bank ? tires and road can provide a maximum tangential force equal to 0.4 of the force normal to the road surface. What is the highest speed at which the car can takc this curve without is the angle of frietion between skidding? 7-18 238 A large mass M hangs Using Ncwton's law (stationary) at the end of a string that smooth tube to a small mass m that whirls around path of radius / sin 6, where / is the length of the string passes through a a circular from m to the top end of the tube in (see the figure). Write down the m must dynamical equations that apply to each mass and show that v2 Consider whether time of 2ir(lm/gM) orbit in a complete one . there is any 7-19 A model rocket on the value of the angle 6 restriction rests joined by a string of length motion. frictionless horizontal surface on a / in this to a fixed point so that the rocket and is moves /. The string will break if its The rocket engine provides a thrust F of The constant magnitude along the rocket's direction of motion. in a horizontal circular path of radius tension exceeds a value T. rocket has a mass (a) Starting m that does not decrease appreciably with time. from rcst at / = 0, at what (b) What was (c) What / 1 is the rocket the magnitude of the rocket's instantaneous net 1/2? Obtain the answer in terms of F, T, and m. distance does the rocket travel between the time t\ acceleration at time when later time Ignore air resistance. traveling so fast that the string breaks ? 1 The rocket engine continues the string breaks and the time 2/i? to operate after the string breaks. has been suggested that the biggest nuclear accelerator we are likely to make will be an evacuated pipe running around the The strength of the earth's magnetic field at the earth's equator. 7-20 It equator is about 0.3 G or 3 X 10 -5 what speed would an atom of lead MKS (at. With units (N-sec/C-m). wt. 207), singly ionized (i.e., move around such an orbit carrying one elementary charge), have to so that the magnetic force provided the correct centripetal acceleration? (e = lead 7-21 1.6 X 10~ 10 C.) Through what voltage would a atom have A to be accelerated to give trick cyclist rides his bike form of a it around a "wali of death" vertical cylinder (see the figure). force parallel to the surface of the cylinder on the bike by the At what speed must the cyclist go the normal force exerted (a) 239 Problems singly ionized this correct orbital The maximum is speed ? in the frictional equal to a fraction n of wali. to avoid slipping down? (b) At what angle (c) If ~ m radius of the cylinder ride, to the horizontal (<p) must he be inclined? 0.6 (typical of rubber tires m, 5 is at on dry roads) and what minimum speed must the the cyclist and what angle does he make with the horizontal? 7-22 The following expression gives the resistive force exerted on a sphere of radius r moving at speed v through air. valid over a It is very wide range of speeds. = R(c) R -4r«J X 3.1 10 + 0.87/-V m, and v in m/sec. Consider water drops falling under their own weight and reaching a terminal speed. (a) For what range of values of small r is the terminal speed where N, r is in in 1% determined within by the first term alone in the expression for R(v)l For what range of values of Iarger r is the terminal speed 1% by the second term alone? (b) determined within Calculate the terminal speed of a raindrop of radius 2 (c) were no If there rest before 7-23 An paratus. oil experiment The mm When is the upper plate radii, mm is made 1100 V the radius of the droplet? is its net charge, 1.6 plates un- from one line to positive with respect measured as a number of elementary 10" 19 C.) What voltage would hold = fail and takes 22.0 sec to cover the same 4 is 3.1 X 10~ n; (MKS units). What What is With the apart. that the resistive force X deduccd precise value of the charge 7-24 The expcriment is done with The droplets are timed between Assume charges? (e (c) . observed to take 23.6 sec to to the lower, the droplet rises (b) mm. from apart. kg/m 3 lines that are 2.58 two horizontal (a) fail performed with the Millikan oil-drop ap- is plates are 8 charged, a droplet distance. it reaching this speed? droplets of density 896 the other. from what height would air resistance, the droplet stationary? [Use the in part (b)]. solid plastic spheres of the same material but of different Two R and 2R, are used equal charges q. The in a Millikan expcriment. The spheres carry Iarger sphere is observed to reach terminal speeds as follows: (1) plates uncharged: terminal speed ward), and (2) plates charged: terminal speed suming that the resistive force ciro, find, in terms of i>o and = on a sphere of radius vi, the = vo (down- lm (upward). r at As- speed o is corresponding terminal speeds for the smaller sphere. 7-25 Analyze in retrospect the legcndary Galilean experiment that Imagine such an experiment done with two iron spheres, of radii 2 and 10 cm, respectively, dropped simultaneously from a height of about 15 m. Make caleulations to took place at the leaning tower of Pisa. 240 Using Ncwion's lavv determine, approximately, the difference in the times at which they hit the Do you ground. think this could be detected without special « mcasuringdevices? (Density of iron 7-26 Estimate the terminal speed of balloon, with a diameter of 30 about inside) of cm 7500 kg/m 3 .) fail (in air) air-filled toy and a mass (not counting the air About how long would 0.5 g. of an it take for the balloon to come to within a few percent of this terminal speed? Try making some real observations of balloons inflated to different sizes. 7-27 A is What (b) The (c) / and the 2-kg What 0. cm 5 When is mass are placed on a smooth cm and table. then is the equation of the ensuing motion? is from rest, the mass were started m/sec in the direetion of increasing instead of being released with a speed of what would be 7-28 a mass of 2 kg the spring constant k ? spring = I f, x = is cm when pulled so as to extend the spring by 5 is released at x, extended by 10 (a) The mass off at and comhung from it. spring that obeys Hooke's law in both extension pression the 1 the equation of mass is doubled deseends an additional distance for the arrangement in motion? diagram in //. What is AH diagram (b)? (a), the end of the spring the frequency of oscillation shown individual springs are identical. 7-29 Any object, parlially or wholly submerged in a fluid, up by a foree equal to the weight of the displaced r rn: _L n 2m cylinder of density p length / is floating with its is A buoyed uniform axis vertical, in a of density po. What is the frequency of small-amplitude vertical oscillations of the cylinder? fluid 7-30 (a) and fluid. A (a) (b) at when 7-31 A D is of length SHM L and under is that the mass in the case where the mass is attached from one end instead of the midpoint. block rests on a tabletop that motion of amplitude A and (a) If the oscillation that will allow the is undergoing simple harmonic period T. vertical, is what is block to remain always (b) If the oscillatior. horizontal, is betwcen block and tabletop will attached to the midpoint of a string given a slight transverse displacement. Find the frequency a distance is The Find the frequency of the constant tension T. deseribes m small bead of mass string (itself of negligible mass). is p., what is the value of A in contaet with the table? and the the maximum coefficient of frietion maximum value of A that allow the block to remain on the surface without slipping? 7-32 The springs of a car of mass 1 200 kg give it a period when empty of 0.5 sec for small vertical oscillations. (a) How far passengers, each of 241 Problems does the car sink down when a driver and three mass 75 kg, get into the car? (b) The road whcn 2 in. it car with its passengers traveling along a horizontal is suddenly runs onto a piece of new road surface, raised above the old surface. Assume that this suddenly raises the wheels and the bottom ends of the springs through 2 in. before the body of the car begins to movc upward. In the ensuing rebound, are the passengers thrown clear of their seats? Consider the maximum acceleration of the resulting simple 242 Using Newton's law harmonic motion. If it universally appears, by experiments and astronomical obseruations, that all bodies about the earth gravitate towards the earth . . . in proportion to the auantity of matter that they severally contain; that the gravitates towards the earth . . . we must, in and all likewise . . . the planets one in like manner towards conseauence of this rule, universally towards another; and the comets the sun; moon allow that all bodies whatsoever are endowed with a principle of mutual gravitation. newton, Principia (1686) 8 Universal gravitation THE DISCOVERY OF UNIVERSAL GRAVITATION in chapters 6 and 7 we have built up the kind of foundation dynamics that Newton himself was the nutshell, it is first to establish. in In a the quantitative identification of force as the cause of acceleration, coupled with the purely kinematic problem of relating accelerations to velocities now consider, as a topic splendid example of i n its We and displacements. own shall and most right, the first how a law of force was deduced from the and historically study of motions. It is convcnient, not unrcasonable, to consider separately three aspects of this great discovery: 1. The analysis o f the data concerning the orbits of the planets around the sun, to the approximation that these orbits are circular with the sun at the center. Several people besides Newton were closely associated with this problem. 2. The proof that gravitation is universal, in the sense that the law of force that governs the motion of objects near the earth's surface is also the law that controls the motion of celestial bodies. It seems clcar that result, 3. Newton was The proof that the true planetary ellipses rather than circles, are law of the true discoverer of this through his analysis of the motion of the moon. force. orbits, which are explained by an inverse-square This achievement, certainly, was the product of Newton's genius alone. In the present chapter of these questions quitc we fully, shall be able to discuss the first using only our basic results in the 245 The second question kinematics and dynamics of particles. requires us to learn (as Newton himself had originally to) how to analyze the gravitating properties of a body, like the earth, which so obviously not a geometrical point is close to its surface. We shall present lem here and complcte the story Chapter in when viewed from one approach to the prob- feature of the gravitational problem is 11, where discussed. this special The third question, concerning the exact mathematical description of the orbits, is something that we shall not go into at all at this stage; such orbit problems will be the exclusive concern of Chapter 13. THE ORBITS OF THE PLANETS We have described in Chapter 2 how the knowledge of the classical planets — Mercury, Venus, Mars, Jupiter, — was already exceedingly well developed by the time motions of the and Saturn of the astronomer Ptolemy around 150 a.d. By this we mean that the angular positions of these planets as a function of time had been catalogucd with remarkable accuracy and over a long enough span for their periodic returns to the the sky to bc cxtremely well We known. same position in have pointed out previously, however, that the interpretation of such results on the model of the solar system that one uses. Let us carefully at the original observational data more clusions that can be The first and the con- drawn from them. thing to recognize is that, whether or not one accepts the earth as the real center of the universe, center as far as this to a all depends now look it primary observations are concerned. is the From vantage point, the motion of each planet can be described, first approximation, as a small circle (the epicycle) whose moves around a larger circle (the deferent). Now there are some facts about the motions of two particular planets— Mercury and Venus— that point the way to some far-reaching center conclusions. These are 1. That for these two planets, the time for the center C of the epicycle [Fig. 8-1 (a)] to travel once around the deferent i.e., the same time that it takes the sun is exactly 1 solar year — to complcte one circuit around the ecliptic. The planets Mercury and Venus never get far from the sun. They are always found within a limited angular range from the line joining the earth to the sun (about ±22J° for Mercury, 2. 246 Universal gravitation Fig. 8-1 (a) Motions of .'/.'( sun and Venus as seen from the Venus always earlh. lies within the range ±$m angular of the sun's direction. (6) Heliocentric picture of the same situation. ±46° for Venus). Both of these for if we go over to the heliocentric, Copernican system [Fig. We 8— l(b)]. facts are beautifully see that the larger circle of Fig. 8-l(a) corresponds in this case to the earth's and the smaller circle own orbit around the sun, of radius rs, — the epicycle— represents the other planet (Venus or Mercury, as the case interpretation, this accounted we can proceed to may make orbit of the be). Given quantitative inferences about the radii of the planetary orbits themselves. This is a crucial advance of the Copernican scheme over the Ptolemaic. Although Ptolemy had excellent data, they were for him just the source of purely geometric parameters, but with Copernicus arrive at the basis of a truly physical model. Thus we in Fig. 8-1 (b) angular deviation, dm , of the planet P from the earth-sun line ES defines the planet's orbit radius r by the the maximum equation — = sin (rB i The radius of > (8-la) r) the earth's orbit is clearly a natural unit for measuring other astronom ical distances, and has long been used for this purpose: 1 astronomical unit (AU) mean (1.496 In terms of this unit, 247 The we then have orbits of the planets distance from earth X 10" m) to sun For Mercury: For Venus: When it r « sin 225° r « sin 46° AU 0.72 AU 0.38 comes to the other planets (Mars, Jupiter, and These planets are not closely Saturn) the tables are turned. linked to the sun's position; they progress through the full 360° with respect to the earth-sun line. This can be readily explained if we interchange the roles of the two component circular motions, so that the large primary circle (the deferent) orbit of the planet, epicycle sun. is now is taken to be the larger than that of the earth, and the seen as the expression of the earth 's orbit around the In the case of Jupiter, for example, the Ptolemaic picture is represented by Fig. 8-2(a) and the Copernican picture by Fig. 8-2(b). now Thus the periodic angular swing, ±0m , of the epicycle is related to the ratio of orbital radii through the equation ^=sin- (rB < (8-1 b) r) r in which the (8-la). roles of r and r B are reversed with respect to Eq. Ptolemy's recorded values of 6 m for Mars, Jupiter, and Saturn were about 41°, 1 1°, and 6°, respectively. These would then lead to the following results: esc 41° r « « r « esc 6° For Mars: r For Jupiter: For Saturn: esc 11° « « ~ AU 5.2 AU 9.5 AU 1.5 Thus with the Copernican seheme (and this was its great triumph) construct it became possible to use the long-established data to Fig. 8-2 (a) Motions of the sun and Jupiter as seen the earth. from The angle 8 m here charaeterizes the magnitude of the retrograde (epicyclic) motion. (6) Helio- centric picture same of the situation. 248 Universal gravitation Fig. 8-3 Universe according lo Copernicus. (Reproduced from his hisloric work, De Revolutionibus.) of the planets in their orbits in order of their increasing from the sun. Figure 8-3 is a reproduction of the diagram by which Copernicus displayed the results in a picture distance historic his book {De Revolutionibus) in 1543. The data with which Copernicus worked (and Ptolemy, too, 1400 years before him) were actually far too good to permit a simple picture of the planets describing circular paths at constant speed around a common center. out a detailed analysis to find out orbit of each planet was offset Thus Copernicus how carried far the center of the from the sun. But even with this adjustment, the detailed change with time of the angular positions of the planets could not be orbit fitted unless the was made nonuniform. motion around the Copernicus, like Ptolemy before him, introduced auxiliary circular motions to deal with the not the answer and we For the moment we shall use the basic idealization of uniform circular orbits and set aside until later the rcfincmcnts that were first mastered by Kepler problem, but shall this, not discuss as its when he recognized we now know, was compIexities. the planctary paths as being ellipses. PLANETARY PERIODS The problem of determining the periodicities of the planets, like that of finding the shapes of their orbits, 249 Planctary pcriods must begin with what ' Fig. &-4 (a) Relative posilions the of the sun, earth, and Jupiter Et at the beginning fr (SEiJi) and end (SE2J2) ofone synodic period. (b) Comparable gram for earth, dia- the sun, the *£ E, y / \ // \ V/ 1 \ 2 i and Venus, allowing for the fact v 1 / > v, that Venus must be j E 1 x offset from the line between sun and earth (b) (a) ^^HBH BI ifit is to be visible. can be observed from the moving platform that is our earth. The recurring situation that can be most easily recognized one in some is the which the sun, the earth, and another planet return, after same positions relative to one characteristic time, to the The length of another. this recurrence time is synodic period of the planet centric in question. model of the solar system, true (sidereal) period this is known as the In terms of a helioeasily related to the of one complete orbit of the planet around the sun. Consider Jupiter. first the case of one of the outer planets, say Figure 8-4(a) shows a situation that can be observed from time to time. Jupiter lie in a The positions of the sun, the earth, straight line. established by finding the date the celestial and Observationally this could be on which Jupiter passes aeross it 180° away meridian at midnight, thus placing from the sun. Now if one such alignment is represented by the positions Ei and J t of the earth and Jupiter, the next one will oecur rather ycar later, when the earth has gained one whole more than revolution on Jupiter. This is shown by the positions E 2 and J2 1 . Jupiter has traveled through the angular distance 9 while the earth goes through 2k + 6. Both Ptolemy and Copernicus knew "The celestial meridian is the projection, on the celestial sphere, of a plane containing the earth's axis and the point on the earth's surface where the It is thus a great cirele on the celestial sphere, running is located. from north to south through the observer's zenith point, vertically above him. Noon is the instant at which the sun crosses this celestial meridian in its daily journey from cast to west. observer 250 Universal gravitation that the length of the synodic period separating these figurations by the symbol in general Then if the earth makes ns complete and Jupiter makes nj revolutions per r. revolutions per unit time we have unit time, = H£r + rtjr But n E and Te and Tj 1 are the reciprocals of the periods of revolution tij of the two planets. Thus and solving we have Tj we have this for T Tj = - (8-2a) - i. Te /t 1 Tb/t Putting two con- close to 399 days. Let us denote the synodic period is ~ 365/399 w 0.915, we thus find that The same type of observation and calculation can be applied Mars and Saturn and the other outer planets that we now know. When we come to Venus and Mercury, however, the situation, to as with the determination of orbital radii, First is the practical difficulty that we is a little different. cannot, at least with the naked eye, see these planets when they are in line with the sun, because it would rcquire looking directly toward the sun to do We so. can by considering any other 8-4(b)] in which the angle between the direcThis particular diagram shows is measured. easily get situation [see Fig. ES and EV Venus as a morning tions star, around this appearing above the horizon before the sun as the earth rotates value of the angle takes over l^yr case, however, earth. Thus = it is + Venus leading to the result 251 + Tb/t Planetary periods 1 hr or so The same one synodic period. that has gaincd This In this one revolution on the form of the equation that applies to we now have 1 east. 583 days, to be more precise. instead of the ngt 1 will recur after — about the outer planets, tivr a from west to TE /r « Putting 7V« 365/583 « 0.627, we find that ^»224 days a curious fact that Copernicus, in the introductory general account of his model of the solar system, quotes values of the planetary periods which are so rough that some of them It is These values are marked on his could even be called wrong. diagram (Fig. 8-3) and are repeated in his text: Saturn, 30 yr; Jupiter, 12yr; Mars, 2yr; Venus, 9 months; Mercury, 80 days. The worst (9 cases are Mars (2 yr instead of months instead of about 7$). about I i) and Venus This seems to havc led some peoplc to think that Copernicus had only a crude knowledge of the facts, which was Perhaps he was certainly not the case. careless about quoting the periods because his real interest was in the The geometrical details of the planetary orbits and distances. any event, truth of the matter, in is that his quantitative knowledge of both periods and radii, as spelled out in detail later in his book, was so good that the best modern values do not, most part, for the quoted. This ican and the is differ from the ones he significantly shown in Table 8-1, which lists both the Copern- modern data on the classical planets. (Incidentally, from Ptolemy's data are almost iden- the values to be extracted TABLE 8-1: DATA ON PLANETARY ORBITS Orbital radius, AU Synodic period, days Sidereal period Modern Planet Copernicus Modern Mercury Venus 0.376 0.3871 115.88 87.97 days 87.97 days 0.719 0.7233 538.92 224.70 days Earth 1.000 1.0000 — 224.70 days 365.26 days 365.26 days Mars 1.520 1.5237 779.04 1.882 yr 1.881 yr Jupiter 5.219 5.2028 398.96 11.87 yr 11.862 yr Saturn 9.174 9.5389 378.09 29.44 yr 29.457 yr tical Copernicus Copernicus with those of Copernicus, an astonishing tribute to those astronomers whose measurements, from about 750 the time of Ptolemy's own b.c. up to observations around 130 a.d., provided the basis of his analysis.) KEPLER'S THIRD LAW The data of Table 8-1 point clearly to a systematic relationship between the planetary periods and distances. 252 Universal gravitation This is displayed . Fig. 8-5 Smooth curue relating the periods and the orbital radii of the planets. graphically in Fig. 8-5. The first the following year in his In was precise form of the relationship discovered by Johann Kepler in 1618 and published by him book The Harmonies of he triumphantly wrote: "I it first believed I the World. was dreaming . . and exact that the ratio which exists between the periodic times of any two planets is precisely the ." Table 8-2 ratio of the ^th powers of the mean distances But it is absolutely certain . TABLE . . KEPLER'S THIRD LAW 8-2: Radius r of of planet, orbit Period T, r s /T2 , Planet AU Mercury Venus 0.389 87.77 7.64 0.724 224.70 7.52 Earth 1.000 365.25 7.50 Mars 1.524 686.98 7.50 Jupiter 5.200 4,332.62 7.49 Saturn 9.510 10,759.20 7.43 days (AU)'i /(dayy2 X 10" shows the data used by Kepler and a test of the near constaney 2 of the ratio r 3 /T Figure 8-6 is a different presentation of the . planetary data (actually in this case the data of Copernicus from Table 8-1) plotted in modern fashion on log-log graph paper so as to show T~ This is this relationship: r3 ' 2 known (8-3) as Kepler's third law, having been preceded, 10 years earlier, by the statement of his two great discoveries (quoted 253 Kepler's third law !|HBHnaHMMngHBmHHn|HmaHgnBMjfl Fig. 8-6 Log-log plot of planetary period T versus radius r, orbit using data guoted by Copernicus. The graph shows that T is proporlional to r3 '" (Kepler's third law). in the Prologue) concerning the elliptical paths of the individual planets. The dynamical expIanation of Kepler's third law had to await Newton's discussion of such problems in the Principia. A very simple analysis of it is possible if we again use the simplified picture of the planetary orbits as circles with the sun at the center. It then becomes apparent that Eq. (8-3) implies that an inverse-square law of force of radius r we have Expressing v in terms of the v 254 = 2itr Universal gravitation is at work. known For in a circular orbit quantities, r and T, we have Z 4 ==- = ar From Newton's =ma From we law, then, infer that the force on a mass r =- *&£ we have the relation Y=K (8-6) 2 K might applies to all a (8-5) Kepler's third law, however, where in must be given by circular orbit Fr (8-4) (toward the center) be called Kepler's constant — the same value of From Eq. the planets traveling around the sun. we thus have l/T 2 = K/r z and , it (8-6) substituting this in Eq. (8-5) gives us = - Fr The ^ (8-7) implication of Kepler's third in planet proportional to is its portional to the square of inertial its is mass therefore, that the force m Huygens when on a and inversely pro- distance from the sun. contemporaries Halley, Hooke, and arrived at law, terms of Newton's dynamics, analyzed all Newton's appear to have some kind of formulation of an inverse-square law problem, although Newton's, the planetary in in terms of his concept of forces acting on masses, seems to have been most clear-cut. The general idea of an influence falling off 2 as l/r was probably not a great novelty, for it is the most deflnite the natural-seeming of all conceivable effects — something spreading out and having to cover spheres of larger and larger area, in 2 proportion to r so that the intensity (as with light from a source) , gets weaker according The required an inverse-square relationship. to proportionality of the force to the attracted mass, as by Eq. appreciated the (8-7), was a feature of which only Newton With his grasp of the concept full significance. of intcractions exerted mutually between pairs of objects, saw that the mean that the force object just as Newton reciprocity in the gravitational interaction it is is proportional to the mass to the must of the attracting Each object is concerned. Hence mass of the attracted. the attracting agent as far as the other one is the magnitudo of the force exerted on either one of a mutually 255 Kepler's third law : gravitating pair of particles must be expressed famous the in mathematical statement of universal gravitation Gmxm2 F_ _ where G (8 _ g) a constant to be found by experiment, and is are the inertial masses of the particles. G matter of determining the famous problem that in led We m, and m 2 shall return to the practice, but first we shall discuss Newton toward some of his greatest discoveries concerning gravitation. THE MOON AND THE APPLE but It is an old as a young man of story, still an enthralling one, of how Newton, came to think about the motion of the nobody had ever done before. The path of 23, moon in a way that the moon through space, as referred to the "fixed" stars, is a line of varying curvature (always, however, bcnding toward the sun), which crosses and recrosses the earth's orbit. But of course there is a much more striking way of looking earth-centered view, which shows the proximatcly circular orbit around the earth. quitc like the planetary-orbit problem that But Newton, with discussing. structed an intellectual at moon it — the familiar describing an ap- To this extent it is we have just been his extraordinary insight, con- bridge between this motion behavior of falling objects— the and the being such a commonplace latter needed a genius to rccognize its relevance. He saw the moon as being just an object falling toward the earth like any other— as, for example, an apple dropping off a tree in phenomenon his garden. A that it very special case, to be sure, because the moon was away than any other falling object in our experience. But perhaps t was all part of the same pattern. As Newton himself described it, he began in 1665 or 1666 to think of the earth's gravity as extending out to the moon's orbit, with an inverse-square relationship already suggested by so much farther i ' We Kepler's third law. tripetal acceleration could of course just restate the cen- formula and apply it to the cussing the problem. at any point A in its In effect hc said orbit (Fig. 8-7). 'See the Prologue of this book. 256 Universal gravitation this: moon, but it is own way of dis- Imagine the moon illuminating to trace the course of Newton's If freed of all forces, it Fig. 8-7 portion Geometry ofa small ofa circular orbit, showing the deviation y from the tangential straiglu-line displacement AB ( = x) lltat would be followed in the absence of gracity. would the AB, tangent travel along a straight line Instead, it moon follows the arc AP. O is its far the moon radial distance r falls, in this the distance of about 16 ft is AB as x, and bit of analytic the distance 1 sec, BP toward O, even Let us caleulate how and compare with it that an object projected horizontally near the earth's surface would a unehanged. sense, in to the orbit at A. the center of the earth, has in effect "fallen" the distance though First, If fail in that geometry. BP as y, it If will same time. we denote the distance be an exceedingly good approximation to put y (8-9) ~2r One way of obtaining this ONP, in which we have ON = r - NP y result = x is to consider the right triangle OP = r Hence, by Pythagoras' theorem, (r - y) 2 + x2 = x r2 = 2ry - y 2 « r for any small value of the angle d, Eq. (8-9) follows y good approximation. Furthermore, since (again for small d) Since as a 2 the arc length AP (= s) is almost equal to the distance can equally well put 257 The moon and the apple A B, we y *- « (8-10) In order to put numbers into this formula we need to know both the radius and the period of the moon's orbital motion. The distance to the moon, as known to Newton, depended on the two-step process devised by the ancient astronomers— finding the earth's radius and finding the moon's distance as a multiple A of the earth's radius. rcminder of these classic measurements discussion is given in Figs. 8-8 and 8-9 and the accompanying everyone, is that the (pp. 259-261). The final result, familiar to moon's orbit radius r is about 240,000 miles ~ 3.8 Its period T is 27.3 days « 2.4 X 10 sec. Therefore, travels a distance along its orbit given by fi r (m , 1 During denote sec)s this y>i s = 2ttX 3.8 2A x same time to identify it, it falls X 0* - in 10 1 8 m. sec it _ 1000 |WM m « - 1Q| a vertical distance, which we will given [via Eq. (8-10)] by >-i«7: -^lo« (ini sec) 1 X =1 3X,0 ~ 3m - 6 In other words, in a distance of mm, 1 indeed or about 5V in.; slight. sec is On projected surface, 1 sec, while traveling "horizontally" 1 km, the moon 1 its falls vertically through through just over deviation from a straight-line path is the other hand, for an object near the earth's horizontally, the vertical displacement in given by y2 = yr- = 4.9 m Thus 'i - «x 'O"'' Newton knew - m that the radius of the moon's orbit was about as the ancient Greeks 60 times the radius of the earth itself, had first shown. And with an inverse-square law, if it applied equally well at all radial distances from the earth's center, we would expect yjy* to be about 1/3600. It must bc right! And yet, what an astounding result. Even granted an inverse- square law of attraction between objccts separated by many times their diamctcrs, one still has the task of proving that an 258 Universal gravitation object a few feet above the earth's surface is attracted as though the whole mass of the earth were concentrated at a point 4000 miles below the ground. Newton 1685, nearly 20 years after his He first did not prove this result until great insight into the problem. published nothing, either, until complete, in the Principia in 1687. follows on p. earth's One way came out, perfect and of solving the problem 262 (after the special section below). MOON FIND1NG THE D1STANCE TO THE The it all radius About 225 B. c. Eratosthenes, who lived and worked at Alexandria near the mouth of the Nile, reported on measurements made on the shadows cast by the sun at noon on midsummer day. At Alexandria (marked A in Fig. 8-8) the sun's rays made an angle of 7.2° to the local vertical, whereas corresponding measurements made 500 miles farther south at Syene (now the site of the Aswan Dam) showed the sun to be exactly overhead at noon. (In other words, Cancer.) It Syene lay almost exactly on the Tropic of follows at once from these of length 500 miles, subtends an center of the earth. angle of 7.2° or Hcnce _ Re ~ 8 Re ~ 4000 miles 500 figures that the arc 1 or Fig. 8-8 Basis of the method useci by Eratosthenes to find the earth's radius. When the midday sun was exactly overhead at Syene (S) its rays fell at 7.2° to the vertical at Alexandria (A). 259 Finding the distance lo the moon g rad AS, at the The moon's distance measured in Hipparchus, a Grcek astronomer who made observations of Rhodes, in earth radii lived mostly about 130 B.c. on the island from which he obtained a remarkably accurate estimate of the moon's distance. His method was one suggested by another great astronomer, Aristarchus, about 150 years earlier. The method involves a clear understanding of the positional and moon. relationships of sun, earth, moon Hipparchus measured this angle to be 0.553° (« a. that sun and at the earth. 1/103.5 rad); he — knew what Aristarchus before him had found that the far more distant than the moon. Hipparchus used this also sun We know subtend almost exactly the same angle is knowledge in (Fig. 8-9). The shaded plete an analysis of an eclipse of the shadow; its moon by region indicates the area that boundary lines with one another, because this is the earth comis in PA and QB make an angle a the angle between rays coming from the extreme edges of the sun. The moon passes through the shadowed region, and from the measured time that this passage took, Hipparchus deduced that the angle subtended at the earth by the arc BA was 2.5 times that subtended by the moon Thus Z AOB « 2.5a. Let us now do some moon earth's center to the geometry. is If the distance D, the length of the arc nearly equal to the earth's diameter PQ AB « 2RB - aD 8-9 Basis of the method used by Hipparchus tofind moon's distance. The method depended on obseroing Fig. the duration {and hence the augular widtli) total eclipse in the the arc 260 shadow of the AB. Universal gravitation of the moon's earth, as represented from the BA is very diminished by the amount aD: the itself. by But we also have AB ~ _ 2.5a AB « 2.5aZ> D or Substituting this in the first equation WC have 3.5aDi» 2R F , or D _ 2 3.5o Since a Z) 1/103.5 rad, this gives «s _ 5- Combining D» this with the value of .Re itself, we have 236,000 miles Modern methods Refined triangulation techniques give a mean value of 3,422.6", or 0.951°, for the angle subtended at the radius. Using the modern value of the (Re = 6378 km = moon by the earth's earth's radius 3986 miles) one obtains almost exactly 240,000 miles for the moon's mean distance. Such traditional methods, however, are far surpassed by the technique of making a precision measurement of the time for a radar echo or laser reflection to return to earth. The fiight time of such signals (only about 2.5 sec for the roundtrip) can be measured to a fraction of a microsecond, giving range determinations that are not only of unprecedcntcd accuracy but are also effectively instantaneous. THE GRAVITATIONAL ATTRACTION OF A LARGE SPHERE It has long been suggested that Newton's failure to publicize his discovery 261 The about an invcrse-square law of the earth's gravity gravitational attraction of a large sphere extcnding to the moon was due discrepancy, resulting earth's radius. from an erroneous value for the This would, via Eq. (8-10), falsify the value the moon's distance of orbit) primarily tO an actual numerical his use of since fail, calculated, according to the was of moon's method discovered by (the radius of the /• When Newton Hipparchus, in tcrms of the earth's radius. first was home in the countryside, out of reach of reference books, and it is reliably recorded that he calculated the earth's radius by assuming that 1° of latitude is 60 miles, instead of the correct figurc of nearly 70 miles. Be this as it may, it remains almost certain that Newton, with his outstandingly thorough and critical approach to problems, would never have did the calculation he regarded the theory as complete until he had solved the problem of gravitation by large objects. analyzing this problem. more in a (In Let us Chapter 1 1 now we consider a shall tackle way of it again sophisticated way.) Suppose we have a large solid sphere, of radius R as shown in Fig. 8-10(a), and wish to calculate the force with which it attracts a small object of mass m at an arbitrary point P. We shall , assume that the density of the material of the sphere may vary with distance from the ccnter (as is the case for the earth, to a very marked degree) but that the density equidistant from the center. We is the same at all points can then consider the solid sphere to be built up of a very large number of thin uniform Fig. 8-10 (a) A solid sphere can be regarded as built up ofa set ofthin concentric splierieal shells. gravitational effect of an individual shell can by treating 262 it (b) as an assembtage ofeireular zones. Universal gravitation The befound — The spherical shells, like thc successive layers of an onion. total gravitational effect of the sphere can be calculated as the superposition of the effects of is the by a the force exerted assuming that the funda- thin spherical shell of arbitrary radius, mental law of force Thus these individual shells. all becomes that of calculating basic problem that of the inverse square between point masses. we show In Fig. 8-10(b) a shell of mass negligible thickness, with a particle of from the center of the shell, AP. shell. we If near point A, the force that mass m M, radius R, of at a distance r consider a small piece of the it exerts on m is along the line from the symmetry of the system, howcver, that It is clear the resultant force due to the whole shell must be along the line OP; any component of A near will from material near we need near A, force transverse to OP due to material be canceled by an equal and opposite contribution A'. Thus if only consider we have an element of mass dM contribution to the force along its OP, i. e., the radial direction from the center of the Hence we have shell to m. GmdM s2 now Let us shaded is consider a complete belt or zone of the shell, in the diagram. shown represents the portion of the shell that It and contained between the directions 6 + dd to the axis OP, and the same mean values of s and <p apply to every part of Thus, if we calculate its mass, we can substitute this value dM in the belt 2irR 2 R dd is and sin 6 dd. mass of the _. = dM belt 2TrR OFr 263 The task now circumference is is .. . sin Q Now the width of 2irR sin 9; thus the whole shell —sm6d6 M = —M is 4irR 2 ; its area hence the „ d6 gives us GMm j = its The area of is given by 2 Thus Eq. (8-11) Our as Eq. (8-11) to obtain the contribution of the whole belt to the resultant gravitational force along OP. is it. to co s f sin 9 dd .. sum the contributions such as ,. (o-lZ) Ti dFr gravitational attraction of a largc sphere over the whole of the and e. over thc whole range of values of shell, i.e., <p, .?, This looks like a formidable task, but with the help of a calculus (another of Newton's inventions!) the solution little turns out to be surprisingly straightforward. From the geometry of the situation [Fig. 8-10(b)], and possible to exprcss both of the angles <p in and R, and the variable distance separate applications of the cosine rule we have fixed distances, r ,»+ *»cose = - From the • „ sin 6 first n dd = , l -uar r s. By two R2 ~^T of these, by differentiation, we have — sds -rR Hence, substituting the values of cos we + s2 - 2 =~ "* v is it terms of two <p and sin 6 dd in Eq. (8-12), obtain dFr The 2 = " GMm ~4^R total force minimum is + s2 - (r R 2 )ds ^ found by integrating value of s (= r — R) to its this expression maximum from the value (r + R). Thus we have d±4^'f "--îr - The integral is just the sum of two elementary forms; r Inserting the limits, r+B f /T-R we 2 2 "*"% _ » 2 J. S* — 2 -R 2 then find that ds m [(r + R)- (r = 2R - (r- R) = 4R 264 Universal gravitation - 2 R)] - R _ r - R2\ \r+ R " r- R J _ (r 2 2 + (r + R) we have Substituting this value of the definite integral in Eq. (8-13) we have ^ Fr = - (8-14) What a wonderful the radius R result! It is of extraordinary simplicity, and of the shell does not appear at all. It is uniquely a consequence of an inverse-square law of force between particles; effect of would force law no other yield such a simple result for the net an extended spherical object. Once we have Eq. (8-14), the total effect of a solid sphere Regardless of the particular follows at once. way which the in density varies between the center and the surface (provided that it depends only on R) the complete sphere does indeed act as It does its total mass were concentrated at its center. though not matter how close the attracted particle of the sphere, as long as it is in fact outside. consider what a truly remarkable result this it obvious that an object a few feet P is to the surface Take a moment to is. Ask yourself: Is above the apparently fiat ground should be attracted as though the whole mass of the earth (all 6,000,000,000,000,000,000,000 tons of it!) were condown? It is from obvious as could be, and there can be little centrated at a point (the earth's center) 4000 miles about as far doubt that Newton had to convince himself of own hc could establish, to his between satisfaction, the this result before grand connection gravity and the motion of the terrestrial moon and other celestial objects. OTHER SATELLITES OF THE EARTH Newton's thinking quite least theoretically 8—1 1 is an — of having other embraced the satellites possibility — at of the earth. Figure from Newton's book, The System of the incorporated in the Principia); it shows the illustration World (which transition explicitly is from the effectively parabolic trajectories of short- range projectiles (although the apparent parabolas are really small parts of ellipses) to a perfectly cireular orbit and then to other elliptic orbits of arbitrary dimensions. Let us derive the formulas for the required velocity v and the period T of a satellite orbit at a distance r 265 Other satellites from launehed horizontally in a cireular the center of the earth. of the earth The necessary Fig. 8-11 Newlon's diagram showing the transit ion front normal parabolic trajectories to complete orbits encircling the earth. (From The System of the World.) force to maintain circular motion is provided by gravitational attraction: = G Mnm mv r* r Mg where and G is the mass of the earth, m the mass of the satellite, the universal gravitational constant. Solving for v, (8-15) .-(«T often convenient to express this result in terms of It is familiar quantities. object of on it, by Eq. F. But mass - m We at the earth 's surface, the gravitational force (8-8), is GMatn Re 2 this is the force that can be set equal to question. Hence we have GMgm mg = 2 Rs or GME = gRE Substituting this in Eq. (8-15), 266 more can do this by noticing that, for an Universal gravitation we get mg for the mass in -W 1/2 v The period, T, of the satellite is then given 2xr _ 2^/2 (8-16) g ll2 R E v g = Putting by 9.8 m/sec 2 /?g , = 6.4 X 10° m, we have a numerical formula for the period of any satellite in a circular orbit of radius r around the earth: Tw (Earth satellites) where T is in seconds For example, a and 5.3 The first X 10 ;i sec w man-made an orbit as shown 10~ 7 r 3/2 X (8-17) r in meters. satellite (about 200 km, say), has r T« 3.14 « minimum at 6.6 X practicable altitude 10° m, and hence 90 min satellite, Sputnik in Fig. 8-1 2(a). Its I (October 1957) had maximum and minimum distances from the earth's surface were initially 228 and 947 respectively, giving a Fig. 8-12 satellite (a) Orbit mean value of Sputnik I, man-made the first (October 1957). (A) Synchronous satellite com- munication System. Orbital diameter in relation to earth's diameter is approximately to scale. 267 Othcr satellites of the earth km, of r equal to about 6950 km. With Eq. (8-17) gives an orbital period of about this value of r, 96 min, which agrees closely with the observed Particular interest have an orbital period equal on that satellites to the period of the earth's rotation such If placed in orbit in the earth's equatorial plane, its axis. remain above the same spot on the earth's surface, satellites will and a figure. attaches to synchronous set of three of them, ideally in a regular triangular array as shown in Fig. 8-12(b), can provide the basis of a worldwide Communications system with no blind one finds in Eq. (8-17), r « Putting spots. km 42,000 T= 1 day Thus or 26,000 miles. must be about 22,000 miles above the earth's surabout 5£ earth radii overhead. The first such satellite such satellites face, i. e., to be successfully launched was Syncom II in July 1963. Equation (8-16), on which the above calculations are based, has a very noteworthy feature. A satellite traveling in a circular orbit of a given radius has a period independent of the mass of Thus a massive spaceship of many tons will, for the same value of r, have precisely the same orbital period as a flimsy object such as one of the Echo balloons, with a mass of only the satellite. about 100 kg — debris with a direct G, a small piece of interplanetary This result is a consequence of the fact that the gravitational force on any object THE VALUE OF or, for that matter, mass of only a few kilograms. is strictly proportional to its own mass. AND THE MASS OF THE EARTH Although the result expressed by Eq. (8-14) was obtained by considering a large sphere and a small particle, one can quickly convince oneself that it is also the correct statement of the force between any two spherical objects whose centers are a distance r apart. For suppose that we have two such spheres, as shown in Fig. 8-1 3(a). The calculation that we have carried out shows on the left) attracts every particle of the other as if the left-hand mass were a point [Fig. 8-1 3(b)]. This therefore reduces the problem to the mutual gravitational that one sphere (say the one mass m) mass M. But now we can apply the result a second time. Thus we arrive at Fig. 8-1 3(c), attraction between a sphere (the right-hand sphere, of and a point of the particle of Iast section with two point masses separated by a distance r, as a rigorously correct basis for calculating the force of attraction between the two extended masses shown The above result is ment, already described 268 in Fig. 8-1 3(a). important in Universal gravitation in the analysis Chapter 5, for of the experi- finding the universal Fig. 8-13 lion. (a) Two treating it gracilaiing spheres at small separa- ofone (b) Effect spliere as a poinl mass. (M) (c) can be calculated by The argumen! can be repeated, so thal the attraction between the spheres can be calculated as lhougli bolh were poinl masses. from the measured force between two gravitation constant. G, spheres of effect known In order to get the biggest possible masses. with an interaction that is so extremely weak, is it usual to arrange things so that the centers of the spheres are separated by only a little more than the sum of the radii. It is then a great convenience to be able to calculate the force, even under these Notice, however, that conditions, on the basis of Eq. (8-14). Some the result holds only for spheres. determine G have made them greater ease of machining it of the measurements to use of cylindrical masses, because of the In such cases to high precision. becomes necessary to calculate the net force by an explicit integration over the spatial distribution of material. from laboratory measurements of the force exerted between two known masses, is (as already quoted in Chapter 5): The G = prescntly accepted value of G, as obtained 6.670 X 10-" m 7kg-sec 2 (8-18) : know the value of G, although he made a celebrated guess a t the mean density of the earth, from which he could have obtained a conjectural figure. In Book III of the Newton himself did not Principia, he common remarks one point as follows at : "Since ... the matter of our earth on the surface thereof twice as heavy as water, and a little lower, in mines, three, or four, or even five times heavier, about it that the quantity of the whole matter of the earth or six times greater than If we denote the if it mean consisted at the earth's surface is given The value of G', of water density of the earth as p as R, the gravitational force exerted 269 all on a . is is about is found probable may be . . five ." and its radius mass m just partiele of by and ihc mass of ihc earth 9Mn F= (8-19) R2 where Hence 4* F = -- (GpR)m Since, however, this acceleration g is just the force that gives the particle an in free fail, we also have F = mg It follows, then, that ^GpR g = (8-20) If in this cquation we put g « 9.8 and (using Newton's estimate) p m/sec 2 ~ , 5000 R ~ to 6.37 X 10° m, 6000 kg/m 3 , we find that G« (6.7 ± 0.6) X 10-" m :, /kg-sec 2 Thus Newton's estimate was almost exactly on target. In practice, of course, the calculation is done the other way around. Given the directly G [Eq. (8-18)] we suband (8-20) to find the mass and the mean The result of these substitutions (with determined value of stitute in Eqs. (8-19) density of the earth. R = 6.37 X M 5.97 p = = 10° m) X X 10 5.52 10 is 24 3 kg kg/m 3 LOCAL VARIATIONS OF g If we take the idealization of a perfectly spherical, symmetrical earth, then the gravitational force distance h above the surface F = 270 GMm + A)2 (R Universal gravitaiion is on an object of mass given by m at a If we identify times the value of g at the point in ques- m we have tion, » For F with m /i - « R, this (8-2,) would imply an almost exactly linear decrease Using the binomial theorem, we can rewrite of g with height. Eq. (8-21) as follows: GM(.,h\— ,,, m Hence, for small * 2~ * \ h, 7 we have gCA)-w(l-f) = GM/R 2 where g the earth's surface. , (8-22) the value of g at points extremely close to [Alternatively, we can use a calculus method we want to consider the that can be extremely useful whenever fractional variation of a quantity. It is based on the fact that the differentiation of the natural logarithm of a quantity leads at once to the fractional variation. In the present case we have .. «W GM = -^T Therefore, ln g = const. — 2 ln r = R, g Differentiating, As k _ 2 Ar r g Hence, putting Ag r = go, and Ar = h, we have ^ - 2go - which leads us back to Eq. (8-22). Notice how this method frees us of the need to concern ourselves with the values of any multiplicative constants that appear in the original equation 271 GM in the equation for g. A recognition of this e.g., the value of fact can enable one to avoid a Local variations ofg lot of unnecessary arithmetic in the computation of small changes of one quantity that depends upon another according some to well-defined functional relationship.] He height. Hooke, Robert contcmporary, Newton's made several of the gravitational attraction with efforts to detect a variation did this by looking for any changes in the measured weights of objects at the tops of church towers and the bottoms Not surprisingly, he was unable to find any difBy Eq. (8-22) one would have to ascend to a point about 1000 ft above ground (e. g., the top of the Empire State of deep wells. ference. g was even as great as I part in see in a moment, however, such variations greatest of ease by modern techniques. Building) before the decrease of As we 10,000. shall are detected with the Superimposed on the systematic variations of the gravitaforce with height are the variations produced by in- tional homogeneities in the material of the earth's crust. if one is For example, standing above a subterranean deposit of salt or sand, much lower in density value of to be reduced. g than ordinary rocks, one would expect the Such changes, although extremely measured with remarkable accuracy by modern small, can be become a very valuable instruments and have tool in geophysical prospccting. Almost all which a mass gravity and an other words, in g modern is in gravity meters are static instruments, in equilibrium under the combined action of elastic restoring force supplied it is by a spring. A just a very sensitive spring balance. as the instrument is moved from one change point to another leads to a minute change in the equilibrium position, and this tected by sensitive optical methods or In electrically is de- by, for example, making the suspended mass part of a tuned circuit whose and hencc frequency, capacitance, displaccment. To is changed by the slight be useful, such instruments must be capable of detecting fractional changes of g of 10 -7 or Figure less. 8-14(a) shows the basic construetion of one such device. With it one can trace out contours of constant g over a region of interest. Figure 8-14(b) shows the results of such a survey, after allowance has been made for effects surface features, and so on. 1 272 in these gravity gal = 1 cm/sec 2 surveys « is \Q~* g Universal gra vital ion altitude, in- The primary unit of mcasureknown as the gal (after Galileo): dications of ore concentrations. ment due to varying Such contours can give good (a) Fig. 8-14 ia) Sketch of basic features ofa sensitiue grammeter, made offused quarlz. The arm marked acts as the carries a pointer P. control spring The restoring force S\ and a gravity surcey ouer made by Sweden, and reproduced be oblained . Example g indicating an ore deposit. the Boliden Mining Co., in D. S. Parasnis, Mining Geophysics, Elsevier, Amsterdam, 1960.) 273 (.b) an area about 400 by 500 m, with contours of conslant (After a survey 52 provided by a is null reading can wilh the help ofthe ca/ibrajed spring ofa W U is picoied al A and B and main weight. Local varialions of g This too large for convenience, so most surveys, like that far is of Fig. 8-14(b), show contours labeled in terms of milligals -3 cmsec 2 « 10~"g). Under the best conditions, (1 mgal = 10 relativc mcasurements accurate to 0.01 mgal may be achievable. One can appreciate how change of g by 0.01 mgal in elevation of only impressive this is by noting that a 8 (1 part in 10 ) corresponds to a change about 3 cm at the earth's surface! THE MASS OF THE SUN Let us return to the simple picture of the solar system in which each planet deseribes a cireular orbit about a fixed central sun We (Fig. 8-15). how law, have seen, the use of in the diseussion of Kepler's third Newton's Iaw of motion implies that the on the planet is given, in terms of its mass, orbital radius, and period, by the following equation [Eq. (8-5)]: foree 2 . F = - 4tt r mr /- According to the basic law of the foree, however, as exprcssed by Newton's law of universal gravitation [Eq. of Fr is (8-8)], the value given by the equation Fr = - GMm r* where M is expressions, mass of the sun. From the equality of we obtain the following result: the these 4*y (8-23) GM We may Fig. 8-15 maled by a again note the feature, already Planelary orbit approxicirele wiih the sun at the cemer. 274 l two Iniversal gravitation commented on in connection with earth satellites, that the period mass of the orbiting object, What does other planet. turn Eq. (8-23) around, value of this mass, M, in is independent of the in this case the earth itself or matter is the mass of the sun, and we have an equation that tells some if we us the terms of observable quantities: *-££ <H4> 3 2 Kepler's third law expresses the fact that the value of r /T The statement of indeed the same for a U the planets. does not, however, rcquire the use of absolute values of r for that matter, of of r and T for T either. It is sufficient to know is this result — or, the values the various planets as multiples or fractions of the earth's orbital radius and period. In order to deduce the mass of the sun from Eq. (8-24), however, the use of absolute values We essential. have seen, earlier of the earth's year has been days of antiquity. to the sun is, A known with how is the length great accuracy since the knowledge of the distance from the earth however, rather recent. knowledge makes an interesting the special seetion following. mean in this chapter, The The development of this is summarized in final result, expressed as a story, which distance in meters, can be substituted as the value of r in Eq. (8-24), along with the other necessary quantities as folio ws: r*** 1.50 TK ~ G = We 3.17 6.67 XlO H m X 10 7 sec X 10-" m 3 /kg-sec 2 then find that M„ m « 2.0 X 10 30 kg FINDING THE DISTANCE TO THE SUN The by first attempt to estimate the distance of the sun was the great Greek astronomer, Aristarchus, in the made third century B.c, and he arrived at a result which, although quite erroneous, held the field for but made many centuries. His method, sound ineffectual in principle by the great remoteness of the sun, dicated in Fig. 8-16(a). He knew is in- moon was moon were that one half of the illuminated by the sun and that the phases of the the result of viewing this illuminated hemisphere from the earth. 275 Findina the distance to ihc sun Sun Earth Fig. 8-16 (a) Melhod allempied by Arislarchus Ihe sun's dislance moon. by measuring ihe angle (b) Triangutation to find SEM at half- melhod of cslablishing the of ihe solar sysiem by finding ihe dislance o/ Mars, iising Ihe earth' s radius as a base line. (c) Direel scale delerminalion of ihe sun's dislance by obserc'mg Ihe transit of Venus front When the moon differenl poinls is on exactly half Ihe earth. full, the angle SME is 90°. If, in an exact measurement can bc made of the angle d, the difference in directions of the sun and moon as seen from a point on the earth, we can deduce the angle a (= 90° - 6) this situation, subtcnded at the sun by the earth-moon distance r m- Aristarchus a ~ 3° ~ 55 rad and measured angle is 9 and judged to be about 87°, which gives hcnce « /\s 20/\u. not a, the error Our Since, however, the in the final result present knowledge tells may be (and us that the value of represented by Fig. 8-16(a) is is) very great. in the situation actually about 89.8° instead of 87°; this relatively small change in 6 raises the ratio rs /r M to several A hundred instead of 20. complctcly different attack on the problem was initiated by Kcpler, although its full exploitation was not possible until later. Even so it at once becamc clear that the sun is more distant than Aristachus had concluded. The basis of the method is indicated in Fig. 8-16(b). It involves observations on much 276 Universal gravitation the planet Mars. both planets line joining distance between Now radii. earth, When Mars if them Mars more to the vastly to the sun. the difference between their orbital is viewed from two different points on the is should appear it with respect in slightly different directions The A and B distant background of "fixed" stars. particular angular difference, is the earth, it lies on a Under these conditions the is closest to called the parallax; it is for observers placed at S, the angle subtended To measure radius at the position of Mars. by the earth's this angle one does not need to have observers at two different points on the earth; would carry an observer from A. to B in about 6 hr during a given night. Now Kepler was able to deduce from the very careful observations of his master, Tycho Brahe, that the value of S must be less than 3 minutes of the rotation of the earth itself arc, which distance to is about 1/1200 rad; he could conclude that the Mars must be greater than 1200 in this configuration Then, using the known earth radii or about 5 million miles. from the Copernican seheme values of orbital radii relatioe (Table 8-1), it follows that the distance of the sun from the earth more than 10 million miles. John Flamsteed, a contemporary of Newton to whose observations Newton owed a great deal (he was Astronomer Royal from 1675 to 1720), reduced the upper limit on the parallax of Mars to about 25 seconds of arc, and concluded that the sun's is more than 2400 earth distance was at least radii, i.e., 80 million An miles. Italian astronomer, Cassini, arrived at a specific value of about 87 million miles at about the same time, using observations made by himself in Europe and by a French astronomer, Richer, at Cayenne in South America. Another contemporary of Newton's Edmund Halley proposed a method that finally led, 100 years later, to the first precision measurements of the sun's distance. The method — 1 — involved what is known Venus aeross the sun's as a transit of Venus, disk, as seen 8-16(c) illustrates the basis of the method. As the sun, Venus looks like a small black dot. and also the times at which the the position of the observer Edmund Halley, best known on i.e., from the Its a passage of earth. it apparent path, transit begins or ends, earth. for the Figure passes aeross depend on Since the motion of Venus comet named after him, succeeded this, however, he had been very active in physics and astronomy. He became a devoted friend and admirer of Newton, and it was largely through his help and persuasion that Flamsteed in 1720 as Astronomer Royal. Long the Principia was published, 277 Finding the distance to the sun before is known, the timing of the accurately yield accurate measures can be used to transit of the differences in angular positions of Venus as seen by observers at different positions. From such observations the parallax of Venus can be inferred, after which one can use an analysis just like that for Mars. These transits arc fairly rare, because the orbits of the earth and Venus are not in the same plane, but Halley pointed out that a pair of would occur then in them and 1769, and again in 1874 and 1882, and From the first two of these (both occurring own death) the solar parallax was found to be in 1761 2004 and 2012. long after Halley 's between about 8.5 and 9.0 seconds of arc, corresponding to a distance of between about 92 and 97 million miles. Thus the currently accepted result was approached. (The best meadcfinitely surements of its elosest this type have been made on the asteroid Eros at approach to the earth.) Further refinements came with the observations made in the late nineteenth century. One of the most notable of these was the use of an accurately known value of the speed of light from the accumulated to deduce the diameter of the earth's orbit time lag, over a period of 6 months, in the observed eclipses of the moons of Jupiter. The situation is indicated in Fig. 8-17. While the earth moves from J, to J\. E x to E2 , Jupiter moves only from This introduces an extra transit time of about 16 min for the light that tells us that one of the speed of light is moons has, for Knowing that space, we can empty Jupiter's example, just appeared from behind the planet. 186,000 miles/sec in deduce that the earth's orbital radius is equal to this speed times about 480 sec, or about 90 million miles. (The caleulation was originally donc just the other way around, by the Danish astronomer Roemer in 1675. Using an approximate value of the distance from earth to sun, he made the first quantitative estimate of the speed of Fig. 8-17 light.) Measure- ment of the diameter of the earth's orbit by observing the eclipses of Jupiter's moons and the apparent delays due to the travel time of light through space. 278 Universal gravitation Although the modern measurements of the sun's distance are of great accuraey, we must reckon with the still this distance varies diiring the course of a year. however, we can relatively small variation, If fact that we ignore make this use of the average value, already quoted near the beginning of this chapter: = rE 1 AU = X 1.496 10 n m MASS AND WEIGHT Perhaps the most profound contribution that Newton made to was the fundamental connection that he recognized beinertial mass of an object and the earth's gravitational force on it a force roughly equal to the measured weight of the object. (Remember, we have defined weight as the magnitude science tween the — of the force, as measured for example on a spring balance, that holds the object at rest relative to the earth's surface.) had been known since Galileo's time that It the earth it was with about the same acceleration, fail just a kinematic fact. took on a much deeper this acceleration, there have by F = ma, F, = It But If objects near Until g. Newton Newton's law an object is it observed to must be a force F„ acting on given it i.e., mg (8-25) then becomes a vitally significant dynamical fact that, since the acceleration g it is strictly remarkable same the is for all objects, the force F, causing proportional to the inertial mass. this result is, static experiment. a springlike device and has, as qg . appreciate — a torsion the force fiber. One it finds a quantity, with which electrical interaetions) a gravita- This "charge" inertial purely in a by balancing is charaeteristie of any object far as these experiments are concerned, do with the how serateh to in- between two objects One measures might be called (by analogy with tional charge, To imagine starting from vestigate the force of attraction to in terms of significance. all mass, which is nothing at acceleration (under the aetion of forees produced, for example, stretehed springs). One all defined solely in terms of cxperiments with objects of materials, in different states of aggregation, all sorts and so on. It by of then turns out that, in each and every instance, the gravitational charge is strictly proportional to an independently established quantity, the inertial mass. 279 Mass and weieht Is this just a remarkable coincidence, or does this point to something very fundamental? For a long time it apparent coincidence was regarded as one of the unexplained mysteries of nature. It took the sagacity of an Einstein to suspect may, that gravitation be equioalenI to acceleration. in a sense, Einstcin's "postulate of equivalence," charge q and the inertial mass m quantity, provided the basis of his embodied that to this in Chapter 12, gravitational are measures of the own theory in the general theory of relativity. when we the same of gravitation as We shall come back discuss noninertial frames of reference. We of F„ to are quite accustomcd to exploiting the proportionality m in our use of the equal-arm balance [Fig. 8-1 8(a)]. two forces, but are doing is balancing the torques of What we what we are actually interested in of material. We make the valuc of g is the is the equality of the amounts use of the fact that, to very high accuracy, same at the positions of both masses, and we do not need to bother about what its particular value is. Thanks to the proportionality of gravitational force to mass, we could, with an equal-arm balance and a set of Standard weights, measure out a requircd quantity of any substance equally well on the earth, the moon, or Mars. The spring balance [Fig. 8-1 8(b)], on the other hand, has a calibration that depends on a particular value of g. Its readings are in effect readings of force, even though we use them as a basis for measuring out required amounts of mass. We might find it convenient to use a spring balance for this purpose on the moon, directly but if its mask this scale were marked out and Standard masses. Fig. 8-18 (a) Weigh- ing with an egual-arm — balance in effect a comparison of masses, valid whatever direct the value (b) of g. Weighing with a spring balance —a measurement of the gravitational force, directly dependent the value on of g. 280 in kilograms, we should have to attach a fresh calibration with the help of Universal gravitation Fig. 8-19 Forces acling on the bob ofa simple pendulum. Newton himself recognized that the strict proportionality of the gravitational force to the inertial mass, as evidenced by the identical local acceleration of falling objects, own in his was a key feature statement of universal gravitation as expressed in He Eq. (8-8). therefore made a series of very careful pendulum experiments to test whether a pendulum of a given length, but with a variety of objects used as the bob, always had the same To period. mass see how m hung on a this works, consider an object of inertial The two string (Fig. 8-19). ing air resistancc) are the tension T and forces on it (ignor- the gravitational force F„. The tension T is at every instant pcrpendicular to the path of the pendulum bob and so has no effect on the tangential acceleration a$. The tangential acceleration is due to the tangential component of Fe From F„. = ma» = Fig. 8-19, —Fa smd from which = — (Fg/m)sin 9 a« At every angle (8-26) depends on the 6 the acceleration ag — for given ratio F„/m. — the vehcity of the bob So also the period for one completc round trip will depend upon the ratio (Fs /m). Newton observed the periods of pcndulums with difTherefore initial at every angle 6 will be conditions determined by From bobs but with equal lengths. ferent the periods of all More 1 his observation that such pendulums were equal within his experimental error, Newton concluded that to better than this ratio. F was proportional to m part in 1000. recent experiments (beginning with Baron Eiitvos in Budapest in the nineteenth century) have made use of a very clever idca that permits a static measurement. It depends on recognizing that an object hanging at rest relatioe to the earth i n an acceleration toward the fact has virtue of the earth's rotation, radius r = R cos \, whcrc X it is is the latitudc (see Fig. 8-20). means that a net force of magnitude where m answer is is the inertial mass. that, when How mu 2 r is this Mass and weight This must be acting on force provided? it, The a body hangs on a string ncar the earth's surface, the string, exerting a force T, 281 by earth's axis because, traveling in a circular path of is not in quite the same : Basis of the Eotvos Fig.8-20 methodfor comparing mass and mi object that the earth the inerlial the gravitational is mass of at rest relatice to and hence is being aceeler- ated toward the earth' s axis. And direction as the gravitational force F„. proportional to m, the angle between To for different objects. sensitive torsion balance T if and F„ F„ is will not strictly be different search for any such variations, a very is used, carrying dissimilar objects at two ends of a horizontal bar [Fig. 8-21 (a)]. If the directions of the tensions T, and T a are different [Fig. 8-21 (b)], there will be small horizontal components [Fig. 8-21(c)] that act in opthe posite directions with respect to the horizontal bar but that givc Fig. 8-21 Principle torques in the same sense. oftlte Eoluos lorsion- balance measurement of T! and T2 Two approximately quite the same, there (o) To fiber. from a the whole apparatus torsion bar. f the objects the other hand, if the directions is no net torque tending to twist the torsion any such torque, Eotvos placed test for the existence of egual masses hang (b) I On are identical, even if their magnitudes are not i n a case that could be rotated. The do not have identical ratios T, sin B of inerlial to gracitational mass, the lensions in the suspending strings must be in slightly different directions. In eguilib- jyi rium, the direction of the main supporting fiber must be intermediate between the directions T2 . ofTi and r, sin (c) This implies the possibility of a net '6 torque that twists the torsion bar abotit a (a) vertical axis. 282 Universal eravitation (b) (c) i* hori- Fig. 8-22 To see whether a net torque Eotvos exisls in the experiment, the whole apparatus turned is through 180". This would recerse the sense ofthe torque. zontal beam carrying the two masses was aligned in an east-west and a reading was taken of its position with respect to the case. The whole system was then rotated through 180°, as inFig. 8-22(b). If you analyze both situations direction [Fig. 8-22(a)] on the basis of Fig. 8-21, you will find that, center line of the case, the angle of twist this operation; would be if any net torque existed, its existence revealed. More some elegant modernized experiments of recently, have been performed by R. H. Dicke and his collabBy such experiments it has been shown that the strict this type orators. hence, with respect to the would be reversed by ' proportionality of FB to m holds to 1 part in 10 10 or better. The description of the above experiments points to a closely phenomenon a systematic variation with latitude of the measured weight of an object. If we take the idealization of a — related perfectly spherical earth (Fig. 8-23), the equilibrium of the object is maintained by applying a force of magnitude W at an angle a to the radius such that the following conditions are satisfied: Wsina = — Jfcosa = F„ where r moi 2 rsin\ mui 2 rcos\ = R cos X. Since a justifiable to put cos « « 1 is certainly a very small angle, in the the result W« 'See R. 283 Fg - H. Dicke, mo> 2 R cos 2 X Sci. Am., 205, 84 (Dec. 1961). Mass and weight it is second equation, thus giving Fig. 8-23 The force needed to balance the weiglil of an object different 'm bolh direction is and magnitude from t he force of gravity. It follows that « Wo + mu 2 R sin 2 X W(\) W where = iy is the measured weight on the equator. Putting mgo, we can also obtain a corresponding expression for the latitude dependence of g: l?(X) 80 + <* 2 R sin 2 X we this expression If in R = = 6.4 X « with g g(\) « 10° m, 9.8 9.8(1 This formula we 2 m/sec is + find w 2 /? « X 3.4 -1 and 10~ 2 m/sec 2 , which 27r/86,400sec gives us 0.0035 sin 2 X) more w = substitute m/sec 2 successful than it deserves to be, for we have no right to ignore the significant flattening of the earth, due again to the rotation, which makes the equatorial radius of the earth about 1 part in 300 greater than the polar radius. This eliipticity has two consequences: It puts a point on the equator farther away from the earth's center than would bc, but it otherwise also in effect adds an extra belt of gravitating material around the equatorial region. The resultant value of at sea level, taking these effects into account, scribed it is g quite well de- by the following formula: g(\) = 9.7805(1 Thus our simplc + 0.00529 sin 2 X) calculation has the correct form, but (8-27) its value for the numerical coefRcient of the latitude-dependent correction is 284 only about two thirds of the true figure. Universal gravitation WE1GHTLESSNESS It appropriate, after thc detailed discussion of the relations is among mass, gravitational force, and weight, to devote a few The very we have drawn between the gravitational force on an object and its measured weight makes use of what is called an operational definition of the latter quantity. The weight, as we have defined t, is the magnitude of the force that will words to the propcrty that is called weightlessness. explicit distinction that i hold an object at rest relative to the earth. Our definition of An weightlessness derives very naturally from this: weight less whenever it is in object a state of completely free fail. each part of the object undergocs the same acceleration, state of whatcvcr value corresponds to the strcngth of gravity at location. (In saying this we assume g that its does not change appreciably over the linear extent of the object.) is is In this An object that prevented from falling, by being restrained or supported, and deformations inevitably has internal stresses librium state. This may become in its equi- when a drop very obvious, as of mercury flattens somewhat when it rests on a horizontal surand deformations are removed in the The mercury drop, for cxample, is fail. Ali such stresses face. weightless state of free free to take on a perfectly spherical shape. The above any The definition of weightlessness can be applied in and gravitational environment, this is the phenomena of bizarre dynamical life way should be. it in a space capsule do not depend on getting into regions far from the earth, where the much gravitational forces are the capsule, acceleration, ample, if and everything which it, is is in orbit same For ex- falling freely with the consequence goes undetected. in a spacecraft reduced, but simply on the fact that in around the earth, 200 km above the gravitational force on the spacecraft, and on everything iri it, is still about 95% of what one would measure at sea level, but the phenomena associated with what we call weightlessness are just as pronounced under these conthe earth's surface, ditions as they are in another spacecraft 200,000 miles where the earth's gravitational attraction is down from earth, to j^Vo of that an object released inside The same would be true in a spacecraft that was simply falling radially toward the earth's center rather than pursuing a cireular or elliptical at the earth's surface. In both situations the spacecraft will remain poised in midair. orbit 285 around the earth. Weightlessness When viewcd in these terms the phe- — nomena of weightlessness are not in the least mysterious, though they are still al- startling because they conflict so strongly with our normal experience. LEARNING ABOUT OTHER PLANETS The recognition of the universality of gravitation gives us a powerful tool for obtaining Information about planets other than the earth, and indeed about particular, In celestial objects in general. a planet has satellites of if its own, we can find its mass by an analysis exactly similar to the one we used for deducing the mass of the sun from the motions of the planets This provides the simplest way of finding the mass themselves. of any planet that has Such satellites. more than one of them, satellites, if a planet has also provide a further test of Kepler's third law, taking the planet itself as the central gravitating body. Newton himself applied an analysis of this kind to Jupiter, four most prominent satellites. These were "moons") that made history when Galileo discovered them with his new astronomical telescope in 1610 (see pp. 287-290). Figure 8-24(a) shows their changing positions as seen through a modern telescope, and Fig. 8-24(b) reproduces a using data for its the satellites (or few of the sketches that Galileo himself made, night after night, over a period of many months. Figure 8-24(c) is a graph constructed from Galileo's quantitative records, using the readings unambiguously associated with the outermost of the that can be A four satellites. had no period of about 16 days can be inferred. Galileo hesitation in interpreting his observations in terms of the four satellites following circular orbits that were seen edgewise giving, as motion we would describe at right angles to it, our the appearance of simple harmonic line of sight. measurements Galileo arrived On the basis of further at rather accurate values of the orbital periods of all four satellites and at moderately good values of the orbital radii expressed as multiples of the radius of Jupiter itself. Newton, in the Principia, cision, obtained (p. 290) lists rithmically by his used similar data of greater pre- contemporary John Flamsteed. Table 8-3 these data, and in Fig. 8-25 they are plotted loga- (cf. Fig. 8-6) in such a way as to show how they give another demonstration of the correctness of Kepler's third law. The slope is accurately f. The use of Jupiter's radius orbital radii 286 as a unit for measuring the was not merely a convenience. As we have already Universal gravitation noted in our discussion of the mass of the of the solar system was not Newton's day. known suri, the absolute scale with any great certainty in interesting, however, that using the data as It is presented in Table 8-3, without absolute values of the radii, one can deduce the mean density, pj, of Jupiter. By analogy and in particular with our analysis of Eq. (8-15)], we have 266, (GMj\ 1/2 2xr whence earth satellites [p. — Mj = Putting Mj = we ^ Pj Rj 3 get i _ 3t n '> n*/T 2 ~ 7.5 3 m /kg-sec 2 wefind Substituting 10- n of 10 -9 sec -2 , m mass may be inferable — called disturbing effects one another's and Venus. orbits. from a X 6.67 its own, the magnitude detailed analysis of the tiny perturbations — that planets exert on This technique has been used for Mercury The unraveling of complicated and case G - 1050 kg/m 3 about the same density as water. If a planet does not have satellites of its and , pj i.e., X difficult these mutual interactions matter, however, and in at least is a one posed a problem that was not adequately answered for a it long time. This was the interaction between the two most massive planets, Jupiter and Saturn, which caused irregularities It was even considered possible that the basic law of gravitation would need to of a very puzzling kind in the orbits of both. be modified slightly ship. The away from a precise inverse-square relation- solution to the mystery was finally achieved, almost a century after the publication of the Principia, by the Frcnch mathematician Laplace, building on work by his great fellow 287 Learning about othcr planels The moons of In 1608 Jupiter Hans Lippershey, successful lelescope. own design. in Holland, patented what Galileo learned of this, may hace been the and soon made lelescopes first of his His Ihird instrumen!, of more than 30 diamelers' magnification, led him lo a dramalic discouery, as recounted by him in his book, The Slarry Messenger: "On the secenth day of January in this prcsenl year 1610, al Ihe first hour of Ihe night, when I was ciewing ihe heavenly bodies with a teleseope, Jupiter presented itself lo me, and I perceiced that beside Ihe planet . . . there were three small starlets, small indeed, bui cery bright. belieced them lo be among Ihe hosl offixed stars, they aroused Though I my curiosily somewhat by appearing lo lie in an exacl straight line parallel to the ecliplic I paid no attenlion to the distances between them and Jupiter, for at Ihe oulsel l thoughl them lo be fixed stars, as l hace said. Bui returning lo the same incestigalion on January eighth ledby what, l do not know I found ." a uery different arrangemenl. . . . — . Afew more — . nights ofobseruation were enough lo concince Galileo what he was seeing: "I hadnow [by January 11] decided beyond all question thal there exisled in Ihe heavens three stars wandering aboul Jupiter as do Venus and Mercury about the sun. . . . Nor were there just three such stars; four wanderers complele measured the distances between them Moreocer I recorded the times of the observaby means of the teleseope. tions ..for the revolutions ofthese planets are so speedily completed that it is their revolutions aboul Jupiter. ... Also I . . . . usually possible to lake eien their hourly varialions." Fig. 8~24(a) and its Jupiter four mosi prominent salellites as seen through a modern teleseope. The first and second photographs illustrale Galileo's observation that noticeable changes oecur within a single night. ( Yerkes Obsercatory photographs) 288 Universal gravitation 4-J«- d -7- 4-S-A»? '•™y i M r. f-.tv-r- *^_^>D^ #r>_ -7» <a.6?M/ (8& T' .=: * r V Kr. W <-* r- tf- + =*-*=^V* jf * J?Tb 4- «' o v*—* «"'*•' 0***-* + Fig. 8-24(b) Facsimile of a page of Golileo's own handwritten records ofhis observations diiring the later months (July-October) of 1610. Fig. 8-24(c) A graph conslrucied from Galileo's own records, showing the periodic mol ion ofCallisto, the outermost of the four satellites visible to Mm. The period of about 16 3/4 days is clearly exhibited. 289 M " The raoons of Jupiter *" '° " " * countryman Lagrange. It turned out that a curious kind of resonance effect was at work, resulting from the fact that the periods of Jupiter and Saturn are almost in a simple arithmetic TABLE 8-3: n = r/Rj Satellile Fig. 8-25 A DATA ON SATELLITES OF JUPITERn 3 /T 2 sec~ 2 Period (T) Io 5.578 1.7699 days Europa 8.876 3.5541 days Ganymede 14.159 7.1650 days Callisto 24.903 16.7536 days « « « « , 1.53 X 10 5 sec 7.4 X 10-° 3.07 X X X 10 5 sec 7.5 X 10~ 9 6.19 1.45 10 5 sec 10 6 sec 7.5X10-° 7.4 IO" 9 X log-log grapli displaying the applicabilily of Kepler's thitd law IO the Galilean satellites of Jupiter. It may be seen that Calileo's results are little different from those obtained by John Flamsteed nearly 100 years later. "Thcse same data have been presented in a striking way in Eric Rogers, Physicsfor the Inauiring Mind, Princeton University Press, Princeton, N.J., 1960: Satellile T2 (milesy (hours) 2 X X 10 3 7.264 29.473 lfl 29.484 X 10 3 160.440 10 16 160.430 X 10 3 1.803 Europa 7.261 Callisto How X would you convince your friends cidencc is , X 10" X 10 10 X 10 Io Ganymede 290 r\ 10 3 that this close numerical coin- not evidence of a new fundamental law? Universal gravitation 1.803 relationship (5Tj term in (~900 » 2TS )- This made large an otherwise negligible long the perturbation, with a repetition period so yr) that the mystery it was When seemed to be increasing without limit. finally resolved the belief in Newton's theory was, of course, strengthened further. still THE DISCOVERY OF NEPTUNE Probably the most vivid illustration of the power of the gravitational theory has been the prediction and discovery of planets whose very existence had not previously been suspected. It is noteworthy that' the number of known planets remained unchanged from the days of antiquity until long after Newton. Then, in 1781, William Herschel noticed the object that we now He was engaged in a systematic survey of the only clue to start with was that the object seemed know as Uranus. and stars, his slightly less pointlike than the neighboring Then, having stars. a telescope with various degrees of magnification, he confirmed that the size of the image increased with magnification, which is — they remain below the not true for the stars of even the biggest telescopes, distinguishable from Once returned to in- those due to ideal point sources. object, Herschel had been drawn to the his attention it limit of resolution and always produce images night after night and confirmed that it was moving with respect to the other stars. Also, as has happened in various other cases, it was found that the existence of the object had in fact been recorded i maps n earlier star (first by John Flamsteed These old data suddenly became extremely valuable, because they were a ready-made record of the object's movein 1690). ments dating back through nearly a century. new measurements with showed that the object member of our solar with a mean radius of a carried out over (finally to When combined many months, they be called Uranus) was indeed system, following an almost circular orbit AU and a period of 84 years. This is where our main story begins. Once it was discovered, Uranus and its motions became the subject of a continuing study, and cvidence began to accumulate that there were some ex19.2 tremely small irregularities in its motion that could not be aseribed to perturbing effects from any known source. Figure 8-26(a), a tribute to the wonderful precision of astronomical observation, shows the The 291 suspicion began to The anomaly grow as a funetion of time since 1690. that perhaps there discovery of Neptune was yet another Fig. 8-26 (a) Unexplained residual devialions in the observed positions of Uranus between 1690 and 1840. (b) Basis ofascribing the devialions lo the influence an extra planet. The arrows indicate the nitude of the perturbing force at planet beyond Uranus, unknown Two men — J. Francc 292 C. Adams relalive of mag- dijferent times. in in mass, period, or distance. England and U. — indepcndcntly workcd on the problem. Universal gravitation J. LeVerrier in Both men used as a starting point the assumption that the orbital radius of the unknown was almost exactly twice that of Uranus. The was a curious empirical relation, known as Bode's law (actually discovered by J. D. Titius in 1772, but publicized planet basis of this by J. Bode), which expresses the fact that the orbital radii of the known planets can be roughly fitted by the following formula: R„ (AU) 2, 0.4 + is = 4, 5, and 6 one and Uranus. asteroid belt.) = — cc, is hard to defend). Using good values for Jupiter, Saturn, (The missing integer, n = 3, corresponds to the Figure 8-27 shows this relation of orbital radii (Mercury requires n n (0.3X2") an appropriate integer for each planet. Putting n = 0, we get the approximate radii for Venus, earth, and Mars where n 1, = which gets quite with the help of a semilog plot; relation (linear on this graph) accepts Bode's law, then n = it is 7 gives r what Adams and LeVerrier used. Fig. 8-27 Graph for predicting the orbital radius of the new planet with the help of Bode's law. 293 clear that a simple exponential works almost as The discovery of Neplune = 38.8 well, but AU, and if one this is Given the definite picture, as the new is automatically defined by becomes possible to construct a the period radius, Kepler's third law, and then shown in it Fig. 8-26(b), of the way in its orbital which in planet could alternately accelerate and retard Uranus motion, depending on their relative positions. With the help of laborious analysis, onc can then deduce where in new planet should be on a orbit the its Adams particular date. supplied such information in October 1845 to the British Astron- omer Royal, G. B. who acknowledged Adams' Airy, raised a question of detail, but otherwise did nothing. did not complete his the astronomer to own whom made an immediate calculations until he wrote and search (Neptune) on his very first (J. next night it had August 1846, but G. Galle, identified Germany) new planet It was only 8-28). The in the night of observation. about a degree from the predicted position letter, LeVerrier (see Fig. visibly shifted, thereby confirming its planetary status. Although the discovery of Neptune great success story, and of human received it is frailty. Adams was no support from bachelor's degree some respects a good and bad, in is also a story of luck, both really first in the field, (he his seniors when he bcgan was fresh -. v'-- 8-28 Star map showing the discovery of Neptune, September 23, 1846. O (From . Herbert Hal/ Turner, Astronomical Discovery, Edward « m *. m f -, - . a » • ' Arnold, London, i 1904.) • 294 Universal gravitation (•••«« • • 4 l t r • his Airy missed his calculations). • Fig. but he from • ' man Adams the credit, which he might readily have won, of being the But the locations that both who first identified Neptune. and LeVerrier predicted might well have been hopclessly misleading, for in their reliance on Bode's law they used an orbital radius (and hence a period) that was far from correct. value is about 30 AU The true instead of nearly 40 as they had assumed, which means that they overestimated the period by nearly 50%. It was therefore near to its largely a lucky accident that the planet was so predicted position on the particular date that Galle sought and found it. But let this not be taken as disparagement. A great discovery was made, with the help of the laws of motion and the gravitational foree law, and it remains as the most triumphant confirmation of the dynamical model of the universe Newton invented. 2 The discovery of Pluto by C. Tombaugh 1930, on the basis of a detailed record of the irregularities of that in Neptune's own motion, provided an echo of the original achievement. GRAVITATION OUTSIDE THE SOLAR SYSTEM When Newton known about wrote his System of the World, nothing was the distances or possible motions of the stars. They simply provided a seemingly fixed background against which the dynamics of the solar system proceeded. A ceptions. There were ex- few prominent stars— e.g., Sirius— known since antiquity by naked-eye astronomy, were found to have shifted But the serious and systematic was begun by William Herschel. motions investigation of stellar His observations, continued and refined by his son John Her- position within historic time. schel The and by other astronomers, revealed two first classes of results. was the continuing apparent displacement of individual stars in a way that suggested that the solar system volvcd in a general movement of is itself in- the stars in our neighborhood, at a speed of the order of 10 miles/sec (comparable to the earth's own orbital speed an empirical faet. around the sun). This, as it stood, was just But the second type of result pointed directly This also means that they overestimated the mass necessary to produce the observed perturbations of Uranus. LeVerrier gave a figure of about 35 times the mass of the earth; the currently accepted value is about half this. For a detailed account of the whole matler, see H. H. Turner, Astronomku! Edward Arnold, London, 1904. A shorter but more readily accessible account may be found in an essay entitled "John Couch Adams and the Discovery of Neptune," by Sir H. Spencer Jones, in The World of Mathemaiks (J. R. Newman, ed.), Simon and Schuster, New York, 1956. 2 Discovery, 295 Gravitation outside the solar system Scale (secondsof arc) 8-29 Fig. willi '1868 /*" 1864 lime oflhe relalive position vector the 1880 Varialion 1821 of members ofa double-star syslem. 1885 (After Arthur Berry, S:1827 A Short History of Astronomy, 1898; reprinted by Dover Publications, New 1781 • 1897 1890 1830 1895 * •l84o" York, 1961.) For Newton's dynamics. to the operation of the Herschels discovered numerous pairs of stars that were evidently orbiting Figure 8-29 shows one around one another as binary systems. of the best documented early examplcs, and the first to be sub- jccted to a detailed analysis in terms of Kepler's laws. £-Ursae, in one of the hind paws of the constellation (It is known as the Grcat Bear.) period of a binary star depends on the total mass of the The system, not on the individual masses. This is easily proved in the case in which the orbits are assumed to be circles around the common stars are the center, C. stars, If [see Fig. 8-30(a)]. we mi»t /W2WI1 (r. + write the statement of ' The individual line passing through F = ma for common to both stars. one of the we have say mi, , mass center of always at opposite ends of a straight = miu r\ r2 )2 where w (= 2tt/T) is the angular velocity Hence 2 CO = Gnt2 ri(ri For a 296 full + r2 )2 discussion of the concept of center of mass, sce Ch. 9, p. 337. Universal gravitation (a) Fig. 8-30 (a) Motion ofthe members of a binary star system with respect to the center of mass. C, for the case ofcircular (b) Direct orbits. visual euidence of the motion of a binary system Krueger 60, — photographed by E. E. Barnard. (Yerkes Obseroatory photograph.) (b) definition of the center of mass, However, by the = 'i mi + we have rri2 where r It = n + ri follows at once that 2 03 Thus if = G(m\ + m2 ) r, the distance r between the stars can be obtained by from a knowledge the masses is at once sum of of their angular scparation), the determined. Finding the individual masses entails the somewhat direct astronomical observation (e.g., starting harder job of measuring the motion of each star in absolute terms against the background ofthe "fixed" stars. Figure 8-30(b) shows convincing direct evidence of the orbital actual binary system. 297 Gravitation outside the solar system motion of an Fig. 8-31 Rotating galaxy (spiral galaxy NGC 5194 in the constellation Venatici). Canes (Photo- graph from the Hale Obsercatories.) With the development of modern astronomy, the systematic its neighbors came to be seen as part motions of our sun and of a greater scheme of movements controlled by gravity. AU around us throughout the universe were the immense systems— the galaxies— most of them vividly suggesting a state stellar of general rotation, as i difficult structurc to elucidate are embedded, clear, however, that Fig. 8-31, and that was the onc the Milky i.e., its in it The most which we ourselves It finally became n Fig. 8-31, for example. Way basic structure our sun is in galaxy. is much very describing some 20 like that of kind of orbit m (= 30,000 around the center, with a radius of about 3 X 10 light-years) and an estimated period of about 250 million years ,5 (=» 8 X 10 sec). Using these figures we can infer the ap298 Universal gravitation orbit, that proximate gravitating mass, inside the motion. From Eq. (8-24) we have would define this .23 With G= 7 X 10~" m 40 m~ v 3 3 2 /kg-sec , we X1 ° G1 « 3 find that x 41 Since the mass of the sun (a typical star) we see that a core really a figure that of about 10" ke 10 X about 2 is 10 This stars is implied. can be independently checked. It is 30 kg, is not a kind of ultimate tribute to our belief in the universality of the gravitational law that it is confidently used to draw conclusions like those above concerning masses of galactic systems. EINSTEIN'S THEORY OF GRAVITATION We how Newton have described carlier proportionality of weight to inertial significance; it mass played a central role clusion that his law of gravitation nature. For Newton this was a is rccognized that the a fact of fundamental in leading must be strictly him to the con- a general dynamical law of result, expressing the basic properties of the force law. But Albert Einstein, in 1915, looked at the situation through new eyes. For him the fact that all objects fail toward the earth with the same acceleration g, whatcver their size or physical state or composition, implied that this must be in some truly profound way a kinematic or geometrical result, not a dynamical one. being on a par with Galileo's law of inertia, He regarded it as which expressed the tendeney of objects to persist in straight-line motion. Building on these ideas, Einstein developed the theory that a planet (for sun because example) follows in so — that geodesic line doing is it is its charaeteristie path traveling along to say, the what around the is most economical way called a of getting His proposition was that although massive objects the geodesic path is a straight line in the Euclidean sense, the presence of an extremely massive object such as the sun modifies the geometry locally so that the from one point to another. in the absence of geodesies 299 become curved lines. The state of affairs in the vicinity Einsteirrs theorv o! gravitation of a massive object of a gravitational space" —a facile not in this view, to be interpreted is, field of force but in terms in terms of a "curvature of phrase that covers an abstract and mathematically complex description of non-Euclidean geometries. For the most part the Einstein theory of gravitation gives results indistinguishable from Ncwton's; the grounds for premight seem to be conceptual rather than practical. ferring it But one celebrated instance of planetary motions there in real discrepancy that favors Einstein's theory. This nomenon is liptical in shapc, very gradually rotates or precesses in planc, so that the major axis is is a the The phe- so-called "precession of the perihelion" of Mercury. that the orbit of Mercury, which is in is distinctly elits own along a slightly different direction at the end of each complete revolution. Most of this precession (amounting to about 10 minutes of arc per century) can be understood in terms of the disturbing effects of the other planets according to Newton's law of gravitation. tiny, obstinate residual rotation century. But there remains a ' equal to 43 seconds of arc per The attempts to explain it on Newtonian theory — for example by postulating an unobserved planet inside Mercury's own orbit — all came to grief by conflicting with other observation concerning the solar system. facts Einstein's theory, of on the other hand, without the use of any adjustable parameters, led to a calculated precession rate that agreed exactly with observation. It corresponded, in effect, to the existence of a very small force with a different dependence on distance than the dominant l/r 2 force of Newton's theory. The way in which a disturbing effect of this kind causes the orbit to precess cussed in Chapter basic law of gravitation square law — had been is dis- Other cmpirical modifications of the 13. — small departures from the inverse- tried before Einstein developed his theory, but apart from their arbitrary character they also led to false predictions for the other planets. it emerged automatically that the In Einstein's theory, however, size of the disturbing term was proportional to the square of the angular velocity of the planet and hence was much more important for Mercury, with its short period, than for any of the other planets. The about apparent amount of precession as viewed from the earth is aclually 1.5° per century, but most of this is due to the continuous change in the direction of the earth's Chapter 300 own 14). Universal gravitation axis (the precession of the equinoxes — see PROBLEMS Given a knowledge of Kepler's third law as it applies to the solar system, together with the knowledge that the disk of the sun subtends an angle of about £° at the earth, deduce the period of a 8-1 hypothetical planet in a circular orbit that skims the surface of the sun. 8-2 well It is that the gap between the four inner planets known the five outer planets is the asteroid belt instead of occupied by and by a This asteroid belt extends over a range of orbital radii from about 2.5 to 3.0 AU. Calculate the corresponding range of tenth planet. periods, expressed as multiples of the earth's year. 8-3 proposed to put up an earth It is satellite in a circular orbit with a period of 2 hr. How high (a) above the were (b) If its orbit the earth's surface would direction as the earth's rotation, for same it have to be? and in the plane of the earth's equator how long would it in be continuously visible from a given place on the equator at sea level? be placed in synchronous circular orbit around the planet Jupiter to study the famous "red spot" in Jupiter's lower atmosphere. How high above the surface of Jupiter will the satellite 8-4 A be? The satellite is to rotation period of Jupiter is 9.9 hr, Rj its mass is Mj about is about 11 times that 320 times the earth's mass, and its of the earth. You may find it convenient to calculate radius first the gravitational acceleration gj at Jupiter's surface as a multiple of g, using the and Jt and then use a relationship analogous to above values of that developed in the text for earth satellites [Eq. (8-16) or (8-17)]. R Mj A 8-5 satellite is to surface of the moon. be placed in a circular orbit 10 km above the What must be its orbital speed and what is the period of revolution ? A 8-6 The satellite is to satellite's power supply is in synchronous circular earth orbit. If the maxi- expected to last 10 years. acceptable eastward or westward drift in the longitude of the mum satellite during radius of 8-7 be placed The its lifetime is 10°, its what is the margin of error in the orbit? springs found in retractable ballpoint pens have a relaxed cm and a spring constant of perhaps 0.05 N/mm. Imagine that two lead spheres, each of 10,000 kg, are placed on a length of about 3 frictionless surface compressed (a) state, so that onc of thesc springs just between How much would about 301 1 1,000 Problcms kg/m . in its un- the spring be compressed by the gravitational attraction of the two spheres? 3 fits, their nearcst points. The mutual density of lead is (b) Let the system be rotatcd in the horizontal plane. At what frequency of rotation would the presencc of the spring become ir- relevant to the separation of the masses? 8-8 During the cighteenth century, an ingenious attempt to mass of He Maskelyne. was made by find the Astronomer Royal, Nevil observed the extent to which a plumbline was pulled the earth the British The figure The change of direction of the out of true by the gravitational attraction of a mountain. of the method. illustrates the principle plumbline was measured between the two sides of the mountain. made (This was donc by sighting on stars.) After allowance had been for the change in direction of the local vertical because of the curvature of the earth, the residual angular difference a was given by 2Fm/Fb, where ±Fm is the horizontal force on the plumb-bob due to the mountain, and Fe = E m/RE^. (m is the mass of the plumb-bob.) The value of a is about 10 seconds of arc for measurements on GM opposite sides of the base of a mountain about 2000 m high. Suppose mountain can be approximated by a cone of rock (of density 2.5 times that of water) whose radius at the base is equal to the height and whose mass can be considered to be concentrated at the center that the Deduce an approximate value of of the base. (The true answer these figures. a to gravitational deflection is about 6 X the earth's mass from 24 kg.) Compare the 10 the change of direction associated with the earth's curvature in this experiment. 8-9 Imagine that a certain region of the ocean floor there in roughly cone-shaped mound of granite 250 m high and 1 km is a diam- in The surrounding floor is relatively flat for tens of kilometers in The ocean depth in the region is 5 km and the density of the granite is 3000 kg/m 3 Could the mound's prcscnce be deeter. all directions. . tected by a surface detect a change in g of Assume (Hint: equipped with a gravity meter that can vessel 0.1 mgal? that the field produced by the mound at the surface can be approximated by the field of a mass point of the total mass locatcd at the level of the surrounding floor. Note same that in g you must keep in mind that the mound own volume of water. The density of water, even at calculating the change in has displaccd its such depths, can be taken as about equal to 1000 302 kg/m 3 . Universal gravitation its surface value of about 8-10 Show that the period of a particle that moves in a circular orbit and the mean close to the surface of a sphere depends only on G Deduce what this period would be for any density of the sphere. sphere having a mean density equal to the density of water. (Jupiter almost corresponds to this case.) 8-11 Calculate the density of the sun, given a knowledge of G, mean sun's diameter the length of the earth's year, and the fact that the subtends an angle of about 0.55° at the earth. 8-12 An astronaut who can lift km (roughly spherical) of 10 50 kg on earth is exploring a planetoid diameter and density 3500 kg/m 3 (a) assuming that he finds a well-placed handle? (b) The astronaut observes a rock falling from a rock's radius is 1 is m 1 and as One would not problem. rocks, even It is if it approaches the surface it? (This is velocity its obviously a fanciful expect a planetoid to have an astronaut were The cliff. cliffs or loose to get there in the first place.) pointed out in the text that a person can properly be termed when he "weightless" satellite, only Should he try to catch m/sec. 8-13 . How large a rock can he pick up from the planetoid's surface, yet it is is in noted in our normal weight there. a satellite circling the earth. The moon many discussions that we would weigh Is is | a of there a contradiction here? dedicated scientist performs the following experiment. After elevator shaft, installing a huge spring at the bottom of a 20-story-high 8-14 A bathroom scale he takes the elevator to the top, positions himself on a and pencil to inside the airtight car with a stopwatch and with pad record the scale reading, and directs an assistant to cut the car's Presuming that the scientist survives the support cable at I = 0. encounter with the spring, sketch a graph of his measured weight up to the beginning of the second bounce. versus time from / = first (Note: Twenty stories is ample distance for the elevator to acquire terminal velocity.) 8-15 A planet of mass M and a single satellite of mass M/10 revolve of mass, being held in circular orbits about their stationary center between their distance together by their gravitational attraction. The centers is (a) (b) D. What What is the period of this orbital fraction of the total motion? kinetic energy rcsides in the satellites? (Ignore any spin of planet and satellite about their own axes.) have considered the problem of the moon's orbit around which the earth as if the earth's center represented a fixed point about the moon earth and however, the fact, the motion takes place. In 8-16 We revolve about their 303 Problems common center of mass. Calculate the position of the center of mass, given that the (a) mass earth's is is How much (b) moon and limes that of the 81 tween their centers 60 earth longer would the month be mass compared to the earth negligible 8-17 The sun appears to be moving if at a speed of 10 (One light-year^ 10 What do m.) The earth takes M. (Note G value of 1 year to X 10" m around these facts imply about the total for keeping the sun in sun's mass of about 250km/sec describe an almost circular orbit of radius about 1.5 the sun. moon were the ? of radius about 25,000 light-years around the center in a circular orbit of our galaxy. that the distance be- radii. its mass responsible orbit? Obtain this mass as a multiple of the that you do not need to introduce the numerical to obtain the answer.) 8—18 (A good problem for discussion.) In 1747 Georges Louis Lesage explained the inverse-square law of gravitation by postulating that numbers of vast invisible particles directions at high speeds. particles, leading to a were flying through space in all Objects like the sun and planets block these shadowing effect that result as a gravitational attraction. has the same quantitative Consider the arguments for and against this theory. opaque objects (Suggestion: First consider a theory in which block the particles completely. This proposal is easy to refute. fairly Next consider a theory in which the attenuation of the objecls incomplete or even vcry small. is This theory is particles by much harder to dismiss.) 8-19 The continuous output of energy by the sun corresponds (through Einstein's relation the rate of about crease we have T~ A/ " 2 M away from the sun. of-magnitude estimate of the by assuming that side, for a must take into account the decreases the orbital radius gradually spiral low mass M, at progressive in- [cf Eq. (8-23)]. precise analysis of the effect that as its the orbital periods of the planets, because for an orbit of a in given radius A E = Mc 2 ) to a steady decrease in 4 X 10° tons/sec. This implies a size r itself increases However, one can of the — the get an order- effect, albeit a little bit remains constant. fact planets on the (See Problem 13-21 more rigorous treatment.) Using the simplifying assumption of constant approximate increasc decrease in in r, the length of the year resulting estimate the from the sun's mass over the timc span of accurate astronomical observations— about 2500 years. 8-20 It is mentioned at the end of the chapter how Einstein's theory of gravitation leads to a small correetion term on top of the basic Newtonian speed o 304 foree of gravitation. in a circular orbit For a planet of mass m, traveling at r, the gravitational foree becomes of radius Universal gravitation in effect the following: GMm _ r2 where c is (•«a the speed of light. (Correction terms of the order of v 2 /c 2 are typical of relativistic effects.) (a) Show GMm/r 2 is that, if denoted by the period under a pure Newtonian force 7"o, the modified period T is given approximately by -»(«-3») (Treat the relativistic correction as representing, in effect, a small fractional increase in the value of G, sponding to the Newtonian and use the value of v corre- orbit.) Hence show that, in each revolution, a planet in a circular would travel through an angle greater by about 24T 3 r 2 /c 2 To 2 than under the pure Newtonian force, and that this is also expressible as 6irGM/c 2 r, where is the mass of the sun. (c) Apply these results to the planet Mercury and verify that the accumulated advance in angle amounts to about 43 seconds of arc per century. This corresponds to what is called the precession (b) orbit M of 305 its orbit. Problcms God and . . . . . . created matter with motion and rest now conserves operations, as in the universe, in its parts, by His ordinary much of motion and of rest as He originally created. rene descartes, Principia Philosophiae (1644) — 9 and Collisions conservation laws some concepts and results that are at the very heart of mechanics. They are rooted in the fact that it takes at least two particles to make a dynamical Until now we have rather glossed over that fact, by system. in this chapter we shall be discussing talking in terms of individual particles subjected to forces of various kinds. Thus, for example, the motion of a planet around the sun, or of an electron between the deflection plates of a cathode-ray tube, was discussed as the problem of a single par- exposed to the force supplied by some completely immovable body or structure. But this is a very special way of looking at ticle things and and gives it is The sun not in general justified. an acceleration, to be doesn't play favorites — it sure, pulls on a planet but the law of gravitation has a completely symmetrical form back on the sun with an equal and opposite and force. So the sun must accelerate, too, and will follow some kind of path under the combined action of all the planets. It happens the planet pulls that the sun is hundreds of times more massive than the rest of first approximation the solar system put together, so that to a we can ignore disparity. The its wanderings. But this is only basic dynamical system is an accident of made up acting particles, the motions of both of which of two inter- must be con- The experimental study of such systems, via collision was in fact the true starting point of dynamics. The study of collisions has lost none of its importance to physics in sidered. processes, 307 ; the 300 years since let it first us then consider became a subject of exact investigation with care. it THE LAWS OF IMPACT In 1668 the Royal Society of London 1 put out a request for the of collision phenomena. experimental clarification Contribu- tions were submitted shortly afterward by John Wallis (mathematician), Sir Christopher Wren (architect), and Christian Huygens (physicist, and Newton's great Dutch contemporary). The results embody what we are familiar with today as the rules governing exchanges of momentum and energy in the collision Most importantly, they involve at one between two objects. stroke the concept of inertial mass and the principle of con- momentum. servation of linear These experiments revealed that the decisive quantity in a collision is what Newton called "the quantity of motion," defined as the product of the velocity of a body and the quantity of matter in it— this latter being what we have called the inertial mass. We have already Chapter 6) discussed Newton's straighthow he was quite ready to assume that (in forward concept of mass: mass of two objects together is just the sum of their separate masses, and how he took it for granted that the masses of difierently sized portions of a homogeneous material are proYou will recall from our earlier portional to their volumes. the total diseussion that these assumptions, however reasonable they may seem, are not always strictly justified. On the other hand, their use does not lead to detectable inconsistencies in the dynamics of ordinary objects. found was that, if And what Newton and these commonsense his contemporaries ideas were accepted as a quantitative basis for relating masses, then a very simple deseription of all collisions could be matically m\u\ made, which expressed mathe- as follows: is + m-iu-i = mwi + (9-1) /M2t"2 and v2 are the velocities after impaet (a one-dimensional motion is wherc M] and u t are the assumed). This is a velocities before impaet, and v x very powerful generalization, because it 'The Society was formally chartered in 1662. (Newton was its President from 1703 until his death in 1727.) Several other great European seientific academies came into existence at about the same period. 308 Collisions and conservalion Uiws applies to quite different sorts of collisions— those with almost two glass spheres, down no rebound at all, as between two lumps of putty. Qualitatively, collisions can be described in terms of their degree A quantitative but purely emof elasticity— i.e., bounciness. perfect rebound, such as occur between to thosc with measure of pirical of v 2 — v y (the this, used by called completely inelastic. called elastic (somctimes, hardened steel others, is ball is the ratio — u> (the If this ratio is zero, the collision relative velocity before impact). is Newton and relative velocity after impact) to Ui If the ratio is unity, the collision is more vividly, "perfectly elastic"). much more elastic A than a rubber ball in In the early investigations Wallis confined his studies this sense. Wren and Huygens to completely inelastic collisions, perfectly elastic objects, and Newton, some time experiments on objects of intermediate to almost later, added elasticity. THE CONSERVATION OF LINEAR MOMENTUM The physicist conserued is (i.e., always on the lookout for quantities that remain unehanged) in physical processes. Once he has discovered such quantities, they become powerful tools in his analysis of phenomena. codifying pasi experience. to to more and more new make They start out by being just an aid in But as they are found to be applicable instances, their value grows and one begins confident predielions with their help. about conservation of a particular quantity is The statement promoted to the some new instance the condown, one's faith in the law may be so great that one hunts around for the missing piece. Should status of a conservation law. If in servation law appears to break it be found, the conservation law is strengthened Thus, for example, the law of conservation of mass reaetions is accepted as a guide to all the masses of the reaeting materials. was first still in further. chemical possible measurements When on the chemical balance applied to the study of chemical reaetions, it appeared some processes mass was gained, in some it was Iost, and But when Lavoisier proved, in some it remained unehanged. through numerous examples, that mass was merely transferred, and that in a elosed system it was conserved, the whole seheme became clear. Chemists could then exp]oit the conservation law. For example, they could confidently infer the mass of a gaseous produet (eseaping from an unclosed system) from easily made measurements on solid and Iiquid reactants. that in 309 The conservation of linear momentum Some of the most powerful aspects of our physical description of the world are embodied in statements about conserved quantities. In mechanics the law of conservation of linear mentum such a statement is — one might even claim that the most important single principle in dynamics. directly on the marized in If, word momentum a compact statement: of the system of two colliding particles remains unchanged by the collision, i. e., the total linear momentum a conserved quantity. is Underlying —the concerning two-particle col- this generalization lisions is the tacit assumption that the system effectively isolated is each other but not with anything particles interact with In the experiments of else. on impact as sumwe introduce the product mv, then we have for a given particle, to describe the momentum total it is based results of the experiments Eq. (9-1). single Tbe It is mo- Newton and others this was achieved by hanging the colliding objects on long strings, so that they swung as pendulums and collided at the lowest point of their In the brief duration of the impact, therefore, the objects swing. were by essentially free their of all horizontal forces except those provided mutual interaction. Since the momentum by an object carried is given, in Newton's words, by "the velocity and the quantity of matter conjointly," values of — m i.e., by the value of the product and v separately 1 — single symbol, p, to represent the along a given Pi then + P'i line = describes with velocity it is mv and not by the convenient to introduce a momentum of a mass m moving v. The relation (9-2) const. the conservation collisions of the type studied of momentum in two-body by Newton and his contemporaries. MOMENTUM AS A VECTOR QUANTITY We have based our discussion so far on one-dimensional colas is lisions. It is important to appreciate, however, that — apparent from its definition — the momentum of a particle is a vector quantity having the same direction as the velocity of the So that, for example, a same momentum 310 as m body of mass \m traveling with traveling with velocity 2v has the v. Collisions and conservation laws mo- particle. Thus our statement of mentum in a collision between two bodies should really be the conservation of linear written as follows: + Pli P2i = pi/ where the subscripts values, respectively + P2/ i and (9-3) / are used to denote precollision (i.e., and and initial final postcollision). This single vector equation defines the magnitude and the direction of any one of the are known. momentum vectors if the other three very often be convenient to separate Eq. (9-3) It will into three equations in terms of the resolved parts of the vectors along three mutually orthogonal axes example, if two bodies, of masses final velocities U|, «mi + u 2 and Vi, v 2 + m2U2 = mivi m and i Each of these (x, y, z). component equations must then be separately satisfied. m2 , have Thus, for initial and Eq. (9-3) becomes , /H2V2 (9-4) which contains the following three independen! statements: m\u\ x miui,, ntiui, + nt2U2r = + nt2U = + JW2«2. = fltiPia miviy 2y mivi, + m2D 2z + m 2 02 V + /W202. (9-5) In carrying out actual numerical calculations this resolution of the vectors will often be necessary. and unspecified masses But manipulations involving velocities are best made terms of the in unresolved equations (9-3) or (9-4), without reference to any particular coordinate system. This is, indeed, one of the main strengths (and economies) of using vector notation. Example. An object of mass 5 kg, traveling horizontally on a frictionless surface at 16 m/sec, strikes a stationary object of mass 3 kg. After the collision, the 5-kg object have a velocity of magnitude direction of motion, as of the 3-kg object? all shown We the velocity vectors. 1 in Fig. 9-1. What momenta Let the Pi/ + P2/ P2/ = Pl. - Pl/ : +x Momentum original the velocity axis lie along the original Then we have (sincep 2 = its 1). Let us denote of the 5-kg object as p u, and the as pi/ and p 2 /. = Pii 311 momentum is observed to can choose an xy plane that contains direction of motion of the 5-kg object (particle the initial is 2 m/sec at 30° to Pli , = 0) + ( — Pl/) as a vector quantity final Fig. 9-1 Conserva- ofthe total vector momentum 'm a simt ion | | ple collision. Figure 9-1 shows the vector construction by which p 2/ can be found. The length of p h represcnts, on some appropriate scale, momentum the initial 60 kg-m/sec, and its is as shown. direction of p 2 / can be found directly and v 2 / The of 80 kg-m/sec. direction component form. m\ = u\ z = W2 = 5 kg 16 m/sec v\ x = = 6\/3 i;i„ = 6 m/sec «1„ — Pi/ First, let 3 triangle, Along list momentum the known conserva- quantities: = K2„ = m/sec 80 x: us kg «2x Thus we have [using Eq. = (9-5)] 30\/3 = 30 Along^: + + 3t> 2 , 3l>2» Hence U2l l- 2„ f2 The - 30\/3)/3 « 9.3 m/sec — = 10.0 m/sec = [(9.3) 2 + (10.0) 2 ]" 2 « 13.6 m/sec = (80 direction of v 2 „ tan 5 312 is -SU = = V2 «2x 9 is length and . Alternatively, tion in The from the vector found as p 2 /Aw 2 we can write down the is length of « at an angle d to the x axis such that — 10.0 :- 9.3 -47° Collisions and conservation laws Notice that has been obtained through this result conservation alone; requires it momentum no knowledge of the detailed momentum changes interaction between the objects. Relating the of the individual objects to the forces acting on them during the however, be our next concern. collision will, ACTION, REACTION, AND IMPULSE We an analysis along the shall begin this discussion with Newton we lines draw attention to a somewhat different approach, which can be more readily adapted to relativistic dynamics, although it is in harmony with Ncwton's that Later himself used. shall analysis within the confines of classical mechanics. Newton interpreted the collision experiments from the standpoint that the concept of inertial mass of an individual object is we a summary already established by experiments and arguments like those made use of in Chapter In that case, Eq. (9-4) 6. is of the actual experimental observations; the *'quantity of motion" is One can conserved. then, by using F about the forces acting during the = ma, draw conclusions collision. A collision is a process involving two objects, each of which exerts a force on the other. Object F2 a force i 1 exerts a force Fi on object 1 2 (Fig. 9-2). on object We make about the relationship between the two act for equal times. This last no assumptions forces, except that they assumption because we recognize that the forces 2; object 2 exerts certainly reasonable, is come into being as a result of the collision, and surely the duration of the collision must be the same for both objects. F = ma as it We can then take the statement of applies to each object separately: = miai F21 ' F12 = »1282 Isolation l.e., boundary ai No. = F21 &2 = F12 (9-6) 1 Fig. 9-2 Inferring the equalily of action and reaction forces from the fact of momentum conservation. Yet this is anolher of those intuitively "obvious" conditions that is not binding and indeed has to bc qualified in some of the collision problems that require the use of special relativity. 313 Action, reaction, and impulse Suppose, to simplify this present argument, that each force remains constant throughout a collision that lasts for a time At. Then we have Tl = + ui — A/ - u2 v2 mi + — At (9-7) rri2 where Ui and a 2 are the initial velocities of the two objects and From these equations we ?i and v 2 are their final velocities. therefore have mivi = /miui W2V2 = W2U2 + + F2jA/ F\2At Adding these two, we thus + /KlVl WJ2V2 = /MlUl Experimentally, however, get + /M2U2 + (F21 we have Eq. + Fi2>A/ (9-4). We (9-8) deduce, therefore, that F21 + F21 = -F12 F J2 = i.e., (9-9) This is, of course, the famous statement known usually as Newton's third law, that "action and reaction are equal and opposite." We have already made extensive use of this result, tacitly or explicitly, earlier in this book, and in Chapter 4 we indicated the kind of experimental support that one can supply for it in Newton, in developing the result static equilibrium situations. for dynamic situations, with forces changing from instant to it quite clear that he regarded Eq. (9-9) as an inIn the Principia he describes an from observattons. experiment in which he floated a magnet and a piece of iron on water, and released them from rest. He noted that there was no instant, made ference motion of the combined mass after the magnetic attraction had pulled them together, and took this as a demonstration of what he himself called the third law of motion. His description of his pendulum collision cxperiments, also in the Principia, is couched same terms. Kaving introduced these forces of interaction, we can now relate them to the changes of momentum of the individual in the 314 Collisions and conservation laws objects in a collision. and F 2 i Thus, At if is the duration of the collision a constant force exerted by is the change of momentum Api particle 2 of particle on particle 1, in the collision is 1 given by F21A/ = Api More generally, if some short time generates any constant force F interval A/, the on a acts change of particle for momentum that it given by is FA/ = Ap As we noted in Chapter 6, Newton's own approach to dynamics was rooted in this equation rather than in F = ma. We also introduced the terminology by which the product the impulse of the force in A/. If F is F At is called varying in magnitude and/or direction during the time span At, one can proceed to the limit of vanishingly small time and so obtain the following equation: F = (Newton's law reformulated) — (9-10) at This eguation, written on the assumption that F is the net force acting on a particle, then becomes the basic statement of Newtonian It is, in a sense, broader in scope than F = ma, or more emcient statement of it. For example, a given force applied in turn to a number of different masses causes the same rate of change of momentum in each but not the same acceleration. In short, we come to recognize that momentum is a valuable single quantity to be accepted in its own right and most importantly for its property of being conserued in an isolated dynamics. at least a system of interacting particles. In our earlier discussion of the collision problem, the forces of interaction as being constant in time. that would be quite unrealistic. It is In we took most cases easy to see, however, that through the integration of Eq. (9-10) we obtain the net mentum change caused by a force has the some arbitrary variation between momentum change Ap = / mo- a varying force. Thus, for example, that it generates is / = and / = if At, given by Fdt (9-11) Jo In a two-body collision, the duration of which 315 Action, rcaction. and impulse is Ar, all that we is actually observe is equal to the total that the total momentum momentum after the collision before the collision. In terms of the impulses Api and Ap 2 given to the separate objects + Api From this Ap 2 = we F2 / this by the condition result is expressed i infer that dt = - / Pu dt In principle, Fi a and F 2 could have quite unrelated values at any particular instant, as long as the above integrals are equal. i However, failing any evidence to the contrary we assume that they are equal and opposite at each and every instant. a one-dimensional collision the Thus in graphs of these forces as a function of time, whatever their exact form, are taken to be mirror images of each other as shown in Fig. 9-3. realize, true! however, that this There is no is a postulate. difficulty as far as It is And important to it is not always "contact" collisions between ordinary objects are concerned. But in situations in which objects influence one another at a distance, as for example through the long-range forces of electricity or gravitation, Newton's law may cease to apply. transmitted instantaneously, and if For no interaction is the propagation time cannot be ignored in comparison with the time instantaneous action and scale of the motion, the concept of reaction can no longer be used. simple mechanical model of such a delayed interaction is Fn O O Fig. 9-3 action and Corresponding variatioiis of F„ reaction forces during the course of a collision. 316 Collisions third and conservation laws A/ A/ A sug- Fig. 9-4 Interaction wilh a time delay, mediated by particles trauersing the gap between two separated objects. gested in Fig. 9-4. A At a of bullets with speed V. cart, B, carrying a A, carries a gun that cart, block a stream is a second bullets are caught. Suppose distance which the in fires off there L away momentum, thanks that the bullets carry plenty of to a large value of V, but are so Iight that they represent a negligible transfer of mass from the they are invisible Suppose further that to an observer standing some distance away. cart to the second. first (We might make them extremely black and bullets is fired off from A, we time L/V is, later does in effect, a and reaction B them a brief burst of if see this cart begin to recoil, ap- B remains parently spontaneously, whilc There small, or perhaps paint up a dark background.) Then set at rest. Not until a begin to recoil in the opposite direction. breakdown of the equality between action in such a case. the time the whole interaction By we has been completed, with all the bullets reabsorbed in B, recognize that one momentum has ultimately been conserved, but restricts attention to the carts it if does not appear to be conserved, instant by instant, during the interaction. The above example may seem we artificial because, after could save the action-reaction principle by looking closely and observing the striking B. Nevertheless, bullets at the instant of leaving it offers all, more A or an interesting parallel to what — those are probably the most important delayed interactions of electromagnetism. We know that the interaction between separated charges takcs place via the electromagnetic field, two and the propagation of such a field takes place at a speed which, although extremely large, transfer of let us say, is still finite— the speed of light. momentum from one The charge to another (resulting, from a sudden movement of the first charge) involves a time lag equal to the distance between them divided by c. If we looked at the charged particles alone, we would see a sudden change 317 in the momentum of the first Action, reaction. and impulse charge without an equal and opposite change in the momentum There would thus appear by tion, instant to bc a failure of is we unless instant, with the electromagnetic that of the other at the same time. field momentum conservasome momentum the interaction. And associate that carries more vivid when we introduce tion of the electromagnetic field momentum and an have associated a There is we the quantiza- and recognize that radiation carried in the form of photons, or light quanta. individually, The what electromagnetic theory suggests. precisely picture becomes even l By the time is we energy with each photon are remarkably close to our mechanical model. even a well-developed theoretical description of the between static interaction electric charges in terms of a continual exchange of so-called "virtual photons." In this case, however, since the forces are constant in time, the equality of action reaction holds good at every instant, and the existence of a and finite time for propagating the interaction ceases to be apparent. EXTENDING THE PRINCIPLE OF MOMENTUM CONSERVATION 2 perhaps worth amplifying the remarks of the It is We have little. tion of There momentum is, last section a seen how, in Newton's view, the law of conservais closely tied in with the action-reaction idea. however, an alternative approach to simple collisions which loosens connection and eases the transition to non- this Newtonian mechanics. This approach can be defined in terms of a question do we actually observe in a collision experiment? : What The answer is — measurements that our observations are purely kinematic ones of the velocities of the two objects before and after impact. Suppose that two u2 , respectively, collide velocities Vj and A objects, v2 - In and B, with initial velocities n,, and with one another and afterward have any individual collision of this type, it is always possible to find a set of four scalar multiples (a) that permit one to write an equation of the following form: <*iui + a2U2 = 'In circumstances same time" — «:»vi where such i.e., + itself — comes into question. section recommended approached 318 in may concept of "the It is just here that ceases to be adequate and the revised formulation of dynamics according to special 2This (9-12) elfects are important, the very simultaneity Newtonian mechanics a 4 \2 relativity becomes essential. be omitled without destroying the continuity, but it is you want to see how the bases of classical mechanics can be more than one way. if Collisions and conservation laws This as stands it ments for is But experi- a quite uninteresting statement. of values of u x and u 2 reveal the remarkable two given objects, we can all sorts result that in every such collision, for (a scalar <*! = a 3 = aA corresponding scalar prop(a obtain a vector identity by putting property of A) and a 2 = on = aB In other words, the purely kinematic observations erty of B). on a collision process permit us to introduce a unique dynamical property of each object. to hold if any of Notice that this simple situation ceases the velocities involved In that case that of light. it is still become comparable to possible to construct a vector balance equation in the form of Eq. (9-12), but only made parameters a are if the explicit functions of speed. In fact, we arrive at the relativistic formula for the variation of speed [Eq. (6-3)]. Let us return at low velocities. mass with • now to the results of experiments The basic statement of these results two given the impact of any objects, the velocity on collisions is that, in change of one always bears a constant, negative ratio to the velocity change of the other: V2 — = — const(vi — m) u2 It is precisely come out (9-13) because this ratio of velocity changes same to the duration of the interaction between the objects, that that it provides a measure of We objects themselves. One can 1, 2, 3, up an set is found to value, whatever the type, strength, or some we can infer property of the intrinsic define this property as the inertial mass. inertial mass scale for a number of objects ... by finding the velocity changes pairwise in such interaction processes of objects 1 and mm |Av[i «II |Av|2 and defining 2, |Av'|i m\ |Av'|.3 and so on. If m t mass ratio, let us say by Similarly for objects rm _ the inertial 1 is and 3, the Standard kilogram, we then have an operation to determine the inertial masses of any other objects. 'For further discussion, see the volume Special Relaliciry 319 Exlending ihe principle of momentum in this series. conservation We must however, do more. If we let and 3 objects 2 interact, then the ratio m3 _ |AT"| a m2 |Av"| 3 must be consistent with the two measurements. In fact same it is, from the ratio obtained and consistency then allows us to use the values m m 2, first experimental this internal z, mea- ... as sures of the inertial masses of the respective objects. Having set up a consistent measure of inertial mass in the form in this way, we can then rewrite Eq. (9-13) vi ~ »i — El V2 U2 mi (9-14) which when rearranged gives us mim + wi2U2 - Thus an equation mm + (9-15) »I2V2 identical in appearance with the usual mo- how mentum-conservation relation emerges, but notice trasts with the Newtonian analysis. We this con- have used the collision processes themselves to define mass ratios through Eq. (9-13). Once this has been done, the terms in Eq. (9-15) automatically add up to the same total before and after the collision. What we have done here, in effect, is to give primacy of place to momentum conservation. The question as to whether or not action equals reaction does not For when one valuable. interactions (i.e., is arise. And this can be very confronted with non-Newtonian those for which action and reaction are not equal opposite forces at each instant) one faces the problem of how to incorporate them into physics— whether to abandon the law of momentum conservation in its limited form or to extend the idea of momentum and retain the conservation law. The momentum-conservation principle has proved so extremely powerful that the latter course has been chosen, and conservation of linear momentum is a central feature of relativistic dynamics. This way of analyzing cesses exposes the the basic results of collision pro- very intimate relation that exists between kinematics and dynamics. If we changc our description of space, and motion, then we can expect that our dynamics must be changed also. This is, in fact, precisely the situation as we make time, the transition from the kinematics of Galileo and Newton kinematics of Einstein's special theory of relativity. 320 Collisions and conservation laws to the THE FORCE EXERTED BY A STREAM OF PARTICLES The impact of a stream of provides an the conservation of linear momentum. of particles, each having mass m and solid surface and Suppose that a stream speed M [Fig. 9-5(a)] and that the particles mass on a particles or fluid instructive application of the laws of collision Let the rate (i.e., number through an imaginary v, strikes all per second) at a block of lodge in the block. which particles pass fixed plane at right angles to the stream be denoted by R. We know that momentum must be conserved, although if the block were extremely massive one might, in watching the process, receive the impression that the particles was simply destroyed, because the unnoticeably small. when the first particle time At the number of particles momentum brought in by the velocity acquired is Suppose, for example, that the block stationary carrying momentum hits it. At that has arrived mv. If the stream were cut is restrained in cerned). any way as By conservation RmvAt = (M far as its horizontal of linear motion + AM)u AM = Rm At (o) Stream of individual particles striking a massive object. (b) Initial phase ofbuilding up to a constant average force produced by the particle stream. 321 The is is momentum and mass we where 9-5 move with velocity u (we are assuming that the block have Fig. RAt, each off at this instant, the block and the particles together would continue to some constant is the end of a short forec cxerted by a stream of particles not conthen Letting At approach zero, we two equations that describe arrive at mass of the block: the acceleration and the rate of increase of d - Rm -§ at (9-17) M where we introduce the of transport of mass symbol n to denote the mean rate single the in beam. M is sufficiently rest; its displacement Clearly, if appear to remain at large, the block will and its increase of velocity can remain negligible for But in any event, force F if on exerted the block it is stationary at some some time. instant, the at that instant must be equal to M du/dt. Hence from Eq. (9-16) we have F =* M~rat Rme - Thus a strcam of exerts an average fo'rce (9-18) po particles striking a stationary surface on it stream as truly continuous, calculating F. But this is Our we At let — * in a purely mathematical step that does reality. the impact of individual discrete is calculation treats the in the sense that not correspond to the physical short time scale The word given by Eq. (9-18). "average" should be emphasized. we might be The particles, force is produced by and on a able to detect this. sufficiently Figure 9-5(b) an attempt to portray the hypothetical results of such observations on the basis of the following very simple model. that each individual particle, a constant this time upon deceleration that brings it Suppose striking the surface, undergoes it to rest in a time T. During must be subjected to a force,/, given by S- T If the force exerted on the block as a result of the arrival of one would then be a particle were plotted as a function of time, it rectangle of height/and width T, as indicated by the small shaded area in Fig. 9-5(b). The force would suddenly come into existence a certain instant and, a time T later, it would suddenly fail to zero. Suppose now that we consider what happens as a function of time after the beam of particles is first turned on (e.g., by at 'The correctness of Eqs. (9-16) and (9-17) depends, in fact, on this condition. which the particles strike it is reduced from R to R{\ — u/6), and the values of du/dt and dM/dt are reduced If the block has a speed u, the rate at accordingly 322 — see Problem 9-13. Collisions and conservation laws opening a shutter that was previously preventing them from reaching the block). Then as successive particles arrive, each adds / to contribution its the total force, which thus rises in an However, at a time irregular stepwise fashion. of the the particle, first latter's T after arrival contribution to the total force would vanish. Thereafter there would be no net increase in the total force, because on the average the earlier particles drop aut of the picture as fast as the new ones appear. thus levels off at tions about that constant some mean total force value F but value. The force will appear to be almost the effects of the individual particles overlap con- if shown siderably in time, as making The will exhibit statistical fluctua- in the figure. This corresponds to the average time interval between successive particles very short compared to the average time that it takes for an individual particle to be brought to rest. We can use the above microscopic picture to recalculate the total force F. It is equal to the force per particle,/, multiplied number of particles that are effective at any one instant. This number is equal to the total number of particles that arrive by the within one deceleration period, T. Since the rate of arrival this clear.] [Figure 9-5(b) should is R, this number is make just RT. Thus we can put F„ = {RT)f RT™ = i.e., F = Rmu as before. It is noteworthy that the average force the beam exerts against the plate (and hence also the average force the plate exerts against the beam of atoms) is quite independent of the actual magnitude of the deceleration that each The average mentum of force the is atom undergoes. mo- simply equal to the rate of change of beam. This force exerted by a stream of particles has been exploited of a by W. Paul and G. Wessel to measure the average speed silver atoms. A beam of atoms, evaporated from beam of ' downward onto the pan of a very delicate an "oven," was directed balance. The force exerted on the balance pan was thus made up of two parts: a steady force, as given by Eq. (9-18), and a 'W. Paul and G. Wessel, Z. Phys. 124, 691 (1948). 323 The force exerted by a stream of particles Force on balance pan Fig. 9-6 Time dependence of a total force due t o the momentum tveight and transfer -Time O the of a stream ofparan object. Beamturned on ticles striking at this instant force increasing linearly with time, due to the increasing mass as given by Eq. (9-17) (see Fig. 9-6): ^total + M' = V» The forces involved were equivalent to the weight of a small number of micrograms. The experimental values for an oven temperature of 1363°K (= 1090°C) were approximately as follows: tiv = M = 3.4 5.6 N X 10- 7 X 10- ,0 kg/sec leading to a value for the average speed, v, of silver atoms at 1363°K: v «= 600 m/sec REACTION FROM A FLUID JET on a surface produces a push, so the production of such a stream in the first place must cause a force of reaction on the system that gives the stream its Just as the impact of a stream of particles momentum. In a of the stream is normal jet of liquid or gas, the basic granularity too fine to be noticed, and any device that sends out such a jet will experience a steady force of reaction, as given by Eq. (9-18). Thus if we imagine a bench test of a rocket engine, for example, with the engine clamped to a rigid structure, then the burnt fuel in Fig. 9-7(a), thrown out backward with speed vq, as shown and the forward push, P, exerted on the rocket is is given by 324 Collisions and conservation laws Sehemat ic diagrams o/a lest-bed arrange- Fig. 9-7 ment of(a) a roeket englne and (b) a jet engine with air intake at the front. P= One l*VO sees this same work on a smaller principle at scale in garden sprinklers, fire hoses, and so on. A jet engine of an aireraft presents a somewhat more complicated application of these dynamical results. air that enters at the front of the engine, exhaust gases process of that is an important role at the rear, plays momentum in the over-all The main funetion of transfer. carried with the plane In this case the and leaves as part of the the fuel to give the ejected gases a high is speed with respect to the plane, and most of the moving mass supplied by the It is very air. is convenient to analyze the dynamics of this system from the standpoint of a reference frame in which the engine is instantaneously at forward at a velocity v, the air rest. If the plane the front with an equal and opposite velocity, as velocity v a in at the rate total /x fucI , Thus, if air (kg/sec) and this frame. through the engine at the shown rear, all the ejected material has a Fig. 9-7(b). Then, at the ward traveling is seen as entering the engine at is rate /x ftir the total rate of change of is in back- being carried fuel is being momentum burnt defines a forward foree on the engine according to the following equation: P= ~ ") (9-19) A jet aireraft is traveling at a speed of 250 m/sec. Wuel»0 Example. + H\t{O0 Each of its engines takes in 100 m 3 of air per second, corresponding to a mass of 50 kg of air per second at the plane's flying altitude. the gases The air is used to burn 3 kg of fuel per second, and all coming from the combustion chamber are ejected with a speed of 500 m/sec relative to the aireraft. of each engine? 325 What Substituting directly in Eq. (9-19) Reaction from a fluid jet is the thrust wc have P = = 3 X 1.4 500 X 10 + 4 50(500 - 250) N Four engincs of this type would thus give a total driving force of about 12,000 lb, a more or less realistic figure. The most spectacular manifestation of these reaction forces, man-made systems, is of course the initial thrust from rocket engincs in a launching at Cape Kennedy (Fig. 9-8). at least in the Fig. 9-8 Launching of a Saturn V rocket. (N.A.S.A. photograph.) 326 Collisions and conscrvalion laws The following approximate on the first Total mass 2.5 = = = lb from thesc 2 quarter of the process initial as itself, total 10 10 kg/sec 7 N 6 kg initial acceleration is puli down is one must on the rocket!) fuel have to only about a analysis of the accelerative continually is X X about 2250 tons of mass The value. mass 3.4 4 that in a vcrtical liftoff the end of the first-stagc burn, been consumed, and the 10 figures that the speed of the ejected (Remember . X 2.8 overcome the downward gravitational first 1.4 =150sec min about 2500 m/sec and that the is about 2 m/sec By 10 6 3100 tons at liftoff Burntime gases X 7.6 infer system: 15tons/sec Total thrust One can V stage of the Saturn Rate of burning on published data 1 figurcs are based lost, the subject of the is next section. ROCKET PROPULSION This has bccome such a very large subject lying in recent years that do no morc than touch upon the underdynamical principles. For anything like a substantial dis- clear that it is we shall cussion you should look elsewhere. 2 The fact remains, however, any rocket does depend on the basic laws of conservation of momentum, as applied to a system made up of the rocket and its ejected fuel. For simplicity, let us consider the that the operation of motion of a rocket out n a region of space where the i gravity are sufficiently small to be ignored in the tion. Under of the rocket this is l shown plus its NASA 2 + At, 1960; S. remaining Facts: Space L. Am a mass The of fuel fuel at / Launch M. Barrere + m At, is Between time t burnt and becomes sep- the total and o and after ejection mass of the rocket is its velocity at time t. Vehicles. et al., Rocket Propuhion, Elsevier, Amsterdam, Bragg, Rocket Engines, Geo. Newnes, London, 1962; G. P. Sulton, Rocket Propuhion Elements, Wiley, 327 is situations before in Fig. 9-9, where Scc, for cxamplc, of the thrust from the ejected fuel. Suppose that the arated from the rocket. are effects approxima- assumption, the only force acting on the body burnt fuel has a speed v n relative to the rocket. and time first Rocket propulsion New York, 1963. MM|mnBMH|BnnH N/ 'Vi'» time i (b) f/g. 9-9 tion of an element Situalions jusi before The (w + Am)u = m(v is a kind of inelastic collision in the masses By conservation of velocity. after the ejec- rocket. ejection of the fuel since initially reverse, and jusi Am ofmass by a m and Am have the same momentum we have linear + Av) + Am(v — vo) Therefore, mo + u Am = mo + mAv + o Am — voAm Therefore, mAv = voAm or Ao = —m v This equation (9-20) not quite exact. is (Why?) But as we let At approach zero, the error approaches zero. As long as Au/t?o is much less than unity, Eq. (9-20) is an excellent approximation. Given the final initial total mass w/ of mass w, of the rocket plus rocket can be evaluated. If the initial the rocket are i\ and v f Eq. (9-20) , V — Vi » fuel, and the the rocket at burnout, the velocity gained »o tells and by the final velocities of us that 2 ~ Am numerically, by drawing a graph l/m against m, and finding the area (shaded) between the limits m/ and m,. This is a pure number, which The answer could be obtained (Fig. 9-10) of when if m multiplied by v is fuel) at 328 gives the increase of velocity. Analytically, used to denote the mass of the rocket (plus any instant, it is more its satisfactory to interpret Collisions and conservation lavvs remaining dm as the Craph of l/m versus m. The relatlve measure of Ihe gain of velocity resulling from a given mass ehange. Fig. 9-10 shaded area giues a change of mass of the rocket dm in a time di. actually a negative quantity, which is in this way, integrated from Defined when the beginning to the end of the burning process gives the value of m; - In these terms the change of velocity can be rrii(< 0). written as the following integral in closed form: "/ You — Vi = — 00 /f J mi —m = dm — (m.\ . Poin i (9-21) \mf J should satisfy yourself that this is i indeed equivalent to the of the numerical-graphical method described above. Notice that the time does not enter into the calculation at final result all, although of course it would do so if we wanted to consider the rate of increase of v or the magnitude of the thrust. Notice also that we would be entitled, at each and every instant, to look at the situation the rocket always just It is The first —v from the frame of reference of In this frame the velocity of the ejected fuel itself. , worth examining some of the implications of Eq. (9-21). thing to notice is that the gain of velocity Thus proportional to the speed of the ejected gases. make v through is and Eq. (9-20) follows immediately. as great as possible. chemical 5000 m/sec, and other losses, it is in burning The highest values of v processes are of the is directly it pays to attainable order of practice, because of incomplete burning and hard to do better than about 50% of the ideal a given fuel (cf. the figure of 2500 m/sec for These the LOX-kerosene mixture of Saturn V, first stage). they are in ordinary terms, but of course, very high velocities are, theoretical value for small compared to the velocities that can be given to charged 329 Rocket propulsion " particles by electrical acceleration. Hence the interest in developing ion-gun engines, or even using the highest available speed by making an exhaust jet of radiation. The trouble (that of light) with both of these, however, is the very small rate of ejection of mass, which makes the attainable thrust very small. The other main feature of Eq. (9-21) is the way in the increase of velocity varies logarithmically with the ratio. fuel which mass This places rapidly increasing demands on the amount of needed to confer larger and larger final velocities on a given Suppose, for example, that we wanted to attain a payload. the exhaust velocity v velocity equal to , starting from rest. Then, by Eq. (9-21) we have v/ — = V( vo = foln © Therefore, »-. -2.718... m, But to attain twice this velocity, we need to have 2 -fe) i.c, — = m, Uli - . «7.4 2 e Table 9-1 presents the results of such calculations in more convenient form. The last column represents the extra mass needed, TABLE 9-1 v, - vt mi/m, (w< - m,)/mf vo 2o 2.7 1.7 7.4 6.4 3uo 20.1 19.1 4»o 54.5 53.5 as a multiple of the payload. The practical problems of producing very large mass ratios are prohibitive, but the use of multistage rockets (which also have other advantages) avoids this difficulty (see 330 Problem 9-12, p. 359). Collisions and conservation laws — 1 A is may seem result that surprising at sight first is that there nothing, in principle, to prevent us from giving a rocket a forward velocity that of the exhaust gases. is considerably greater than the speed v Thus at a late stage in the motion one see both the rocket and the ejected material moving forward with respect to the frame in which the rocket started out from rest. No violation of dynamical principles would be involved, and if one made a detailed accounting of the motion of all the material that was in the rocket initially, one would find that the total momentum of the system had remained at zero would (as long as the effects of external forces, including gravity, could should of course be emphasized that our whole analysis would hold good, as it stands, only in the absence of be ignored). gravity It and of resistive forces due to the air. COLLISIONS AND FRAMES OF REFERENCE The seventeenth-century which momentum investigations of collision processes, conservation was established, were chiefly experimental. But one of the men involved — Christian Huygens applied the spirit of twentieth-century physics in a analysis bound. by brilliant of the particular case of two objects with perfect reHe based his argument on symmetry and on the equivalence of frames related by a constant velocity. The analysis is simple but of great intrinsic interest. Figure 9-1 ground to the 1 provides a suitably seventeenth-century back- situation. But though the picture may appear quaint and archaic, the thinking is sharp and fresh. Huygens imagines two equal elastic masses, colliding with equal and opposite speeds ±u. velocity is He assumes exactly reversed; this that they is rebound so that each a symmetry argument. Huygens imagines a precisely similar collision boat that is itself moving with speed v relative Next, to take place on a to the shore. This viewed by a man on terra firma, appears as a stationary mass being struck by a mass with velocity 2v. After the impact, collision, From Ernst Mach, The Science of Mechanics, trans. T. J. McCormack, Open Court Publishing Co., La Salle, 111., 1960. The original figure comes from Huygens' Ireatise on impact (ca. 1700). Mach's book (which is extremely readable and uses a minimum of mathematics) is itself a landmark in the discussion of the fundamental principles of physics. His speculations on the origin of inertia were a significant part of the background to Einstein's thinking about general relativity and cosmology (see Chapter 12). 331 Collisions and frames of rcfcrcnce — m* m i u n i L m O Fig. 9-11 O Huygens' visualization of a n elaslic collision between two equal masses, as shown 'm his book, De Motu Corporum ex Per- cussione (1703). (Reprinled in Vol. 16 ofC. Huygens, Oeuvres Completes, Marlinus Nijhoff, The Hague, Netherlands, 1940.) the is first mass has acquired the velocity 2v, and the second mass stationary. More generally, if the boat has a speed from u, different v, the velocities exchange as follows: Body Body 2 1 Before impact: u + v u — d Afler impact: u — o u + v Thus Huygens on predicts, theoretical grounds, the results of all possible one-dimensional experiments collision of two identical masses. on the The perfectly elastic common feature to all of them is that the magnitude of the relative velocity has the same value after the collision as it had before the collision. Huygens went even further. Again using a kind of symmetry argument, he deduced a general property of perfectly elastic collisions between unegual masses. If a moving object A strikes an unequal stationary object B, then after the collision they be moving with Imagine that velocities v this is and w as indicated will in Fig. 9-12(a). viewed from a boat moving to the right with B moving Huygens argues that the exact reversal of the motion of B, as seen from this second frame, must also imply the exact reversal of the motion of A in this perfectly elastic process. Thus the final velocity of A must be — (m — w/2) n this frame. But this velocity is also equal to v — w/2, because the final velocity of A as seen from velocity w/2. to the left, Then before with velocity collision the object i 332 Collisions is seen — w/2, as shown in Fig. 9-12(b). and conservation laws — After Before Fig.9-12 Elastic collision between two Seen from (a) unequal objects as laboratory frame, and A B A B O- Seen from seen (d) from the O^ O the shore (b) the boat B A O*- -*o ""2 {b) from a frame in B w w 2 2 which both celocities or: v are simply reversed. the shore Hence we have is v. - (u - - -x = = w/2) w — v v - w/2 (9-22) u that in the elastic collision of Thus Huygens concludes any two objects whatsoever, the magnitude of the relatioe velocity remains Notice, however, that something beyond the con- unaltered. momentum servation of linear thing involved here. is what we have learned to is call The extra some- kinetic energy, and conservation defines a very special class of collisions. its Let us consider this further. KINETIC ENERGY IN COLLISIONS Suppose that a one-dimensional in Fig. 9-1 3(a). If this system we have conservation of fluences m\U\. + W2«2 = WlJ-'l + collision takes place as is shown effectively free of external in- linear momentum: »12 2 1-" Suppose further that the collision is perfectly elastic, in the sense defined by Eq. (9-22). In the present problem this implies the following relationship: Ml — M2 t>2 — Di We can solve these two equations for the v2 The . "i 1$ results are = »11 — »12 — r— "i + »11 + »»2 2»ii = + «I IW2 Kinetic energy - 2»I2 r - — «2 , »11 333 = »11 + »12 mi — mi »n + U-2 »12 in collisions final velocities, v y and Using these values of Vi and v 2 , it is a straightforward piece of algebra to arrive at the following result: + mivi 2 You W2^2 2 = miui 2 + /W2W2 2 (9-23) will recognize that this equation, apart from the absence of factors of \ throughout, corresponds to a statement that the energy after the collision total kinetic energy We initially. have arrived equal to the kinetic is at this result symmetry considerations and without any We of forces or work. basis of mention shall bring these latter concepts into the picture in Chapter 10, but for the Some of on the explicit moment we do not need them. the early workers in mcchanics recognized the conservation property represented by Eq. (9-23) for perfectly elastic and collisions, referred to the quantity (Latin, vis vioa) Having, of a moving in mv 2 as the "living force" object. introduced kinetic energy into the de- effect, we shall now develop an imporgood whether or not the total kinetic energy scription of collision processes, tant result that holds is Suppose that Fig. 9-13(a) represents an arbitrary conserved. one-dimensional collision as observed in a reference frame S. Let us consider this same collision from the standpoint of another frame S' that has the velocity v with respect to 5 and words, We then have u\ v\ We (In other S. S' are related by the Galilean transformations.) = = shall initial «i — V U2 Pl — B PJ now show and = = U2 — v V2 — o that the change of kinetic energy between final states is an invariant — i. e., it has the same value in both frames. In S: înitial #fii.al = hm u 2 + = i'Wlfl 2 + > AK =K,- 4"»2«2 2 l Ki £'M2U2 2 = (imim 2 + im 2 (^wiwi 2 + (.-2 2 ) 2 Jm2«2 ) In S': ^'initial ^'final AK' = i«l(Ml - V) 2 + \tm(M2 •- V) 2 = i«l(d - V) 2 + im 2 (V2 - v) 2 = (%mivi 2 + fffttoa 2) — (2"»i«i 2 - 334 VftmiVl + W2f2> — (miUi Collisions and conservation laws + + i"J2«2 7M2«2)] 2 ) Fig. 9-13 Arbilrary (a) InS: (b) ms'-. one-dimensional collision as seen in two different frames related by fW a velocity v. By momentum 0*r o*r conservation, for a«;> kind of collision, the bination of terms in squarc brackets in is if is com- zero. It follows that AK AK' = Thus, AK' the change of kinetic energy for a given collision process one frame, specified in it has the same value in quantity, and the fact of Given a knowledge of this conservation, we can predict the values of u, and v> dimensional collision process for which the all frames. momentum in any one- initial velocities wi and u> are specified. The value of AK may be positive, negative, or zero. The first of these corresponds to an explosive process, in which extra kinetic energy as a result of the interaction. is given to the separating particles The second and third possibilities correspond to inelastic and elastic collisions, respectively, sense in which we have already been using in the these terms. THE ZERO-MOMENTUM FRAME The momentum of a particle or of a system of particles is not depends on the frame of reference in which one observes the motion. If, however, one compares the descriptions of the motion in two different inertial frames, related by a conan invariant; it stant velocity, the difference between the momentum frame is is always a constant vector of magnitude m is the mass of the particle and v is the velocity of one m\, where frame of a particle measured values of the One can always find momentum of any particle relative to the other. in which the total evidently a frame in which the particle of time one determines its is zero, and this it at rest at the instant momentum. One can reference frame in which the total particles is a reference vanishes; momentum likewise find a of any system of zero-momentum frame of reference is of great importance, not only as a convenience for looking at collisions 335 and interaetions in general The zero-momentum frame but also, as we shall show Fig. 9-14 Basis of defining the zero- momenlum {center-ofmass) reference frame. shortly, for its To dynamical implications. see how we identify zero-momentum framc, let us start with a simple example; two particles /n, and m 2 move along the x axis with constant velocities t>i and v% in a frame of reference S (Fig. 9-14). Our this task is to find the velocity D of a reference frame S' relative to such that momentum m\v\ in S' the total + m-iv'i ' S s ecl ua ' to zero. Let O' be the origin of S' moving with velocity v relative to O. From relative to v\ v'2 = = 9-14 we see that the Fig. O are given c\ — v 02 — u and + if this is (/Ml + /M2C2 = miL-i to be zero, m2)v = /Hl»l Equation (9-24) relative to trary. We S m2 — rrtiU particles in S' is + /W202 — mzv we must have + (9-24) OT2l>2 fixes the velocity of the reference frame S' but leaves the choice of the position (x) of O' arbi- use this freedom to choose the location of O' relative to the positions of m, and Eq. (9-24) of tn^ and by Hence the momentum of the two miv'i velocities in the m 2 as simply as possible. If werewrite equivalent form or better as — [(mi + m 2 )x] = dt evidently the simplest 336 — [mixi + m 2 X2] di way of satisfying this is to equate the Collisions and conscrvalion laws two Fig. 9-15 Basis ofdefining the cenler of granity, not necessarily identical "llg sets of m *t center ofmass. tvilli tJte The square brackets. difference between them must be a constant and we choose the constant equal to zero. We then have for the position of O' in S, _ = misi mi + m2x2 (9_25) + m2 The origin of a zero-momentum reference frame chosen in way is called the center of mass of the two-particle system made up of W] and m 2 You will recognize that Eq. (9-25) also defines what we are accustomed to calling the center of gravity of two objects (cf. Fig. 9-15). In principle, however, these need not be identical points. The center of gravity is literally the point through which a single force, equal to the sum of the gravitational forces on the separate objects, effectively acts. If the values of g at the positions of the two objects were not quite identical, then this . Fgi the forces and Fg2 would not be proportional to strictly m! m 2 in which case the center of gravity and the center of mass would not quite coincide. The difference is negligible for ordinary purposes, but it should be recognized that the center of and , mass, as given by Eq. (9-25), is defined without reference to the uniformity or even the existence of gravitational forces. Perhaps you recognized that we have already used the zero-momentum frame in our earlier section on collisions and frames of reference. When Huygens wanted to apply symmetry arguments to colBut now we have lisions, this was the frame he started with. it in a much more explicit fashion. The introduction of the zero-momentum or center-of-mass (CM) frame leads to a very important and immensely useful way identified of analyzing the dynamics of a system of particles. the essential idea in the simplest possible way we consider a one-dimensional system of two particles. (let us suppose that this is the frame defined the particles have velocities Vi and v 2 frame (S') they have = = l/l V2 where 15 D] V2 is — — velocities v[ develop shall again In frame S by our laboratory) In the zero-momentum and v'2 given by v C defined by Eq. (9-24). can put 337 . To The zero-momentum frame Putting m\ + tn% = M, we » where in any ^±^ - P is = £ the total linear (9-26) momentum, which remains unchanged collision proccss: P = + ntyvi We now HtaD2 = (9-27) const. proceed to express thc total kinetic energy K, as measured in the laboratory frame, in terms of the center-of-mass and the velocity v CM. We the K= velocities v[ ImM + df + im 2 (o'2 2 (£wiid'i + JTJttPa ) + + Kmi + w 2 )0 2 = Now, by and v 2 of mi and m 2 relative to have + vf + m2 (miv'i v'2 )o (9-28) zero-momentum frame, we the very definition of the have miv'i + = miv'y Hence the middle term on the right-hand side of Eq. (9-28) is automatically zero! The first term we recognize as the total kinetic energy K' of the two individual particles relative to the CM; the last of mass term is equal to thc kinetic energy of a single particle M moving with the velocity i; of the CM. Thus we can write K= This is K' n + \Ml - (9-29) such an important result that we shall express it in words also: The kinetic energy of a system of two particles is equal to the kinetic energy of motion relative to the center of mass of the system (the internal kinetic energy), plus the kinetic energy of a single particle of mass equal to the total mass of the system moving with the center of mass. The into great importance of this separation of the kinetic energy two parts — and the way to the possibility to the CM) as a whole. note well that it works only is the the new of analyzing the internal motion of a system {relative without reference to the bodily motion of the system (We shall show shortly that the result holds only for one-dimensional motions, but in general.) 338 if — zero-momentum frame is that it opens a very powerful and simplifying procedure. We have reference frame Collisions and conscrvnlion laws good not One of the implications of Eq. (9-29) of kinetic energy locked up, as is it that a certain is amount were, in the motion of the In the absence of external forces the velocity v center of mass. remains constant throughout the course of a collision process, 2 and the kinetic energy %Md must Iikewise remain unchanged. This means that in the collision of two objects, only a certain fraction of their total kinetic energy, as measured in the laboratory, is available for conversion to other purposes. The amount of this available kinetic energy can calculate K- = = K' is, in fact, identical with K'. We with the help of Eq. (9-29): it $Mo 2 + (imivi — \m2v2) from Eq. Substituting for d \{m\ + /»2)0 (9-20), this leads by simple algebra to the following result: -22?- = K' \ 2 mi + (c 2 - Bl rri2 f (9-30a) The value of K', as expressed by this equation, is thus the kinetic energy of what can be regarded as being effectively a single mass, of magnitude m m 2 /(m + m 2 ), moving at a velocity equal to the relative velocity of the colliding particles. The effective mass l is called the l is given the symbol more compact form reducedmass of the system and Thus we may write Eq. (9-30a) in the K, K' = y.. (9-30b) where u = Prel = m\m2 mi + ni2 02 — 1>1 For example, if a = - -o': moving ary object of mass is 02 1 object of unit, mass 2 units strikes a station- two thirds of the initial kinetic available for the purpose of producing deformations, when TWO DIMENSIONS In general the velocity vectors of plane, is and so on, the objects collide. COLLISION PROCESSES IN and collision 339 energy locked up in center-of-mass motion, and only one third it is two colliding objects define a therefore important to extend our analysis of processes into two-dimensional space. Collision processes in two dimcnsions Actually one +M> +.w Fig. 9-16 (a) Collision seen as occurring in one dimension. (b) As I seen in another refer- I ence frame, the I colli- «M sion appears two- dimensionat. must stand ready to go all the way into three dimensions, because the plane defined by the velocity veetors of two partieles after may not be the same as that defined by the initial velocities. Many collision processes are, however, purely twodimensional, and for simplicity we shall limit ourselves to such a collision cases in diseussing specific examples, even though the theory applies equally well to three-dimensional problems. One thing that is worth recognizing at the outset is that the analysis of purely one-dimensional problems, of the type have diseussed so far, can in faet lead to we predietions about certain types of two-dimensional collisions, because we can turn a two-dimensional one by a Consider, for example, an of view. simply changing our point imperfectly elastic collision between two identical spheres that one-dimensional collision into approach one another along the y axis with equal and opposite velocities, ±u [see Fig. 9-16(a)]. Suppose that as a result of this collision they recoil ±o (symmetry would appear x requires that these final velocities Figure 9-16(b) then shows opposite). the back along the y axis with reduced to how velocities be equal and the same collision —w an observer who had a velocity parallel to This defines a whole class of oblique collision probof which can be solved with reference to the simple axis. lems, all head-on collision of Fig. 9-16(a). One can go even further yet, by viewing the collision of Fig. 9-16(b) from a reference frame that has some velocity parallel to the y direetion. This removes all the apparent symmetry, yct it is All the collisions for which, at the basically the moment of same collision. impaet, the two spheres have a relative velocity ofgiven magnitude along the line joining their centers are dynamically equivalent. 340 Collisions and conservation laws This suggests a important way of simplifying the vcry Imagine yourself analysis of collision problems. which the most simply described; collision is in the frame in be the this will center-of-mass frame, in which by definition the colliding particles approach one another along a straight opposite momenta. Solve the problem line with equal in this CM and frame, and frame by means of a Galilean transfer the result to the laboratory transformation. The grcat simplicity of a collision process as viewed in the zero-momentum frame is illustrated in Fig. 9-17. Since the momentum total in this frame well as approach, with equal Fig. 9-17(a). is zero, the particles separate, as and opposite momenta, Also the magnitudes of the as shown momentum final in vectors are independent of their direction in the CM shown p'2/ can be represented in Fig. 9-17(b), the vectors p[/ and frame. Thus, as with their tips lying at opposite ends of the diameter of a The relative directions of p{,- circle. and p[/ can be anything, depending on the details of the interaction. The relative lengths of these vectors also depend, of course, on the detailed mechanism. In a perfectly elastic collision inelastic collision we have p'f we have p) = = 0. p'i. (In an cxplosive process, for example rocket propulsion, we have the converse p'i = 0, p} processes To > 0.) we may begin with, let And find in p} (a) An less than, equal to, or greater than We are defining an arbilrary collision as described by equal and opposite momentum vectors in the CM frame. (b) In the CM frame, the end poinls of the momentum vectors lie on circles. 341 where p\. us look at one or two examples of perfectly the center-of-mass frame. 9-17 situation, atomic, chemical or nuclear reaction elastic collisions so as to appreciate the beauties Fig. In a completely Collision processes in two climcnsions of a view from elastic collision here as one in which the magnitude of the relative velocity unchanged by the collision. As we have seen, this to saying that the total kinetic energy, as well as the is conserved. tion will later is is equivalent momentum, The definition in terms of kinetic energy conservabecome the dominant one. ELASTIC NUCLEAR COLLISIONS The One nuclear physicist livcs in two worlds. the other collisions is is his laboratory, the center-of-mass frame of the particles and interactions he is By studying. whose learning to skip nimbly from one frame to the other he gets the best of both worlds. Let us see how. Proton-proton collisions Figure 9-18 shows a collision between two protons, as recorded in a photographic emulsion. hydrogen atom in One of the protons belonged to a the emulsion and was effectively stationary before the collision took place; the other entered the emulsion with a kinetic energy of about 5 The most notable MeV. feature of the collision of the two protons after collision is angle of 90° with each make an / 9-18 Fig. collision y Elastic between an incident proton and an initially stationary proton in a photographic emulsion. . {From C. F. Powell and G.P.S. Occhialini, ~ «.„.. / \ Ox- ford Unicersity Press, New York, 1947.) 342 / . -%.,..; «...^... Nuclear Physics in Photographs, Collisions and conservation laws that the paths othcr. This we up get true for all such proton-proton collisions, until is to energies so high that Newtonian mechanics longer adequate to describe the situation. the center-of-mass frame we can By first no is looking into readily understand this. Let the velocity of the incident proton as observed in the laboratory be Then v. this the zero-momentum frame has opposite velocities as shown emerges from the collision is at velocity v/2, and in frame the protons approach and recede with equal and ir — B' on the other in Fig. 9-19(a). in the direction Suppose one proton d', so that the other side of the line of approach. To get back to the laboratory frame we add the velocity v/2 to each proton, parallel to the original line of motion, as shown in Fig. 9-I9(b). But the triangles ABC and A EF are both isosceles, so the directions 0i and 8 2 of the protons as observed in the laboratory are given by $i = 6' /2 02 = (ir - 6')/2 62 = r/2 Therefore, 0i + Moreover, we can easily find the laboratory velocities of the protons after the collision, for we have ui v2 Fig. 9-19 collision = = 2(y/2)cos0i 2(u/2)cos0 2 = = ucosfli wsinfli (a) Elastic between two eaual masses, as seen in the (b) CM frame. Transformation to the laboratory frame, showing a 90° angle between the final velocities. 343 Elastic nuclear collisions two We any such see that in collision the total kinetic energy is conserved, because KE = Final KE = Initial \mv 2 £m(t>cosfli) 2 = £/wt; 2 (cos 2 6i = %mv 2 + \m(o cos + sin 2 0,) 2) 2 Figure 9-20 shows a similar collision between equal macroscopic objects (billiard balls); it is not perfectly elastic, but it is impres- sively close. Neutron-nucleus collisions In nuclear fission processes, neutrons are ejected with a variety of energies, but the average energy Fig. 9-20 collision is of the order of Slroboscopic pholograph of an almost perfectly elastic between equal masses. (Front PSSC Physics, D. C. Heath, Lexington, Massachusetts, 1965.) 344 Collisions and conscrvation laws 1 MeV. These neutrons are however, most effective in causing further reduced to energies of the order of 10~ 2 or fissions if they are 10 -1 eV slow-neutron reactor And Thus an essential feature of every means of slowing down the neutrons. (thermal energies). is a of neutrons with other nuclei (those com- elastic collisions posing the moderator material of the reactor) do most of the job. Suppose that a neutron of mass m makes an elastic collision with a nucleus of mass in the laboratory be v assumed stationary. M. Let the in this ; Figure 9-21 (a) shows the collision as seen in the center-of-mass frame, the laboratory frame given M+ m initial velocity of the neutron frame the struck nucleus will be which has a velocity v »0 (9-31) If the collision turns the neutron momentum vector v frame, shown velocities, as of d', its final and through an angle in Fig. 9-21 (b). measured d' in the zero- velocity in the laboratory frame The magnitudes of the in the laboratory, for are readily calculated. yourself, relative to by Eq. (9-24): (Exercise: Do is the final any given value this calculation for verify that in this case, as with two equal masses, the total kinetic energy, as measured in the laboratory, remains unchanged as a The result of the collision.) biggest energy loss for a neutron as seen oratory frame occurs if it In this case i;(t) Fig. 9-21 = is scattered straight we have D - (v - B) = -(to - {a) Elastic collision between two uneaual masses, as seen in the CMframe. (b) Transformation to the laboratory frame; the angle between the velocities is different from 90°. 345 Elastic nuclear collisions 2v) i n the lab- backward (d' = x). «--«*(i-jjF?y M- m For = e' the neutron loses Thus collision is this?) no energy at all. (What sort of a the kinetic energy of the neutron after the collision lies between the following limits: K mtiX Since = ^mvo Kmax is independent of the mass M of a moderator nucleus, M K is most mKn that tells us what value of neutron average reduction of the greatest lead to the likely to it is the expression for And we energy. see that cannot do as well as M = m makes Kmin equal to zero; any other value of M, whether this for it we be if no other considerations were would make the best moderator, (the proton mass) is equal to m within about Thus bigger or smaller than m. involved, ordinary hydrogen since in this case M 3 Protons, however, also capture slow neutrons rather effectively, thereby making them unavailable for causing further fissions, and it turns out that certain other light nuclei 1 part in 10 . deuterium, beryllium, and carbon) offer a better compromise between moderating and trapping of the fission neutrons. (e.g., INELASTIC AND EXPLOSIVE PROCESSES may be a net loss or gain of kinetic energy as a result of the collision. In We now shall turn to processes in which there analyzing such processes, change of and that the result collision first total kinetic it is energy important to confirm that the is we obtained on does hold good indeed an invariant quantity, the basis of in general. redevelop Eq. (9-29) for two particles directions. Let the particles have masses vj and v 2 in the laboratory frame (5). CM frame (S') _ = is show this, we moving in arbitrary and m 2 and velocities m , Then the velocity, v, of the defined by the equation wiivi + >"2V2 mi + mz If the velocities of the particles as 346 a one-dimensional In order to (9-33) measured Collisions and conservation laws in 5' are v[ and v'2 , we then have Vl = v'i v2 = v'2 We now +v +v write down the total kinetic energy in S, using the fact $mv 2 of a that the kinetic energy as Jm(v particle can also be expressed terms of the scalar product of the vector v y), with itself. Thus we have • K= £mi(vi = }«i(Ti Now in i.e., • + £m (v v2 ) (v| + v) + $m 2 vi) 2 + v) • 2 • (v'2 + v) (v 2 + v) consider one of these scalar products, using the distributive and commutative laws that apply to the dot products of vectors: + v) (v'i Using • (v', this result above expression 8 (Jmii/i K= + v) and for = = v? v', its vi + + 2v', • + v + v v v • 2 counterpart in the second term of the K, we have + im 2 02) + But again we note v 2v'i that, by the + m2v2 7 (miT , ) • v definition of the frame, the second term on the right is + i(m, + m2 )D 2 zero-momentum zero, so that we come back to the simple result of Eq. (9-29): K = K' Applying + this to the states after collision, Ki K, We £A/B 2 = = of a two-particle system before and we have K'i 2 + iMV K} + %Md 2 assume that the total mass remains unchanged and, absence of external forces, so does D. Thus we again in the arrive at the result K/ where - Q Ki is = K} - Ki- Q the amount by which the final kinetic energy exceeds (algebraically) the initial kinetic energy. The actual value of Q may, of course, be negative. The virtually 347 results of the above analysis can be extended, with no modification, Inelastic to such processes as nuclear reactions, and explosive processes : : which the actual identity of the in particies in the final state may be quite different from what one has at the beginning. Suppose, for example, that there is a collision between two nuclei, of masses m m 2 which m 3 and m and 1 masses 4 two react to produce , Then we can . different nuclei, of write the following statements of conservation + my Mass: Momentum: Thus the initial + »14 » ms /W2 = M /wivi + W2V2 = W3V3 + mi\4 wiv'i + »i2V2 = W3V3 + mnv'n and kinetic energies final = Mv = can be written as follows: = Ki Kf = im^f + \m f + £MD ImM? + kmiMf + \Mo Kf - with Ki = 2 Q. Example: The reactions nuclear 2 2 {o'2 D— D (and One reaction. an important of energy generation by nuclear fusion) nuclei of deuterium (hydrogen 2) to of the most famous one is for the process the reaction of two form a helium 3 nucleus and a neutron ' ?H + ?H -» 2He + i/i + 3.27 MeV The 3.27 MeV represents the extra amount of kinetic energy, Q, made available because the masses of the product particies (their rest masses, to be precise) add up to a little less than the masses of the initial particies; the total energy, including mass equivalents, remains constant, of course. Suppose now that a deuteron with a What strikes a stationary deuteron. as viewed in the CM culate the velocity D. 10" 27 about3.34 X frame? is kinetic energy of 1 MeV the final state of affairs (See Fig. 9-22.) First, let us cal- The mass of a deuteron is about 2 amu, or MeV = 1.6 X KT 13 J. Therekg. Now, 1 forc, D, - (HiJ e 1.0 X 7 10 m/sec 'In this equation, the subscript before the letter for number of protons and the superscript shows and ?H both stand for a deuteron. the D 348 Collisions and conservation laws a given nucleus denotes number of nucleons. the total Before m, m, O O After After Fig. 9-22 (o) Reac- tion process, in which the collision ofthe particles m i and m, m2 O'"» leads to the formation oftwo ticles, (6) different par- m z and m,. Same process as CMframe. seen in the Laboratory frame In CM In (b) (a) m2 = Since = D In the ffli = d i/2 m2 and 0.5 frame is initially at rest, we have 10 7 m/sec X CM frame the deuterons have equal and opposite velocities of magnitude equal to X Kinitial = Kfnitiai = hKx = 2 Hence we have C. %miC 2 = £/HU>l 2 i.e., [We MeV 0.5 see here a particular application of Eq. (9-30). If a moving object collides with a stationary one of equal mass, only half the energy initial kinetic is available for their relative motion in the CM frame.] Now consider the result of the nuclear reaction. total kinetic = = Kfinal energy i n the Kinitiol 3.77 + CM frame 3.27 MeV = is way that the momenta m 4 and V4, »»3v'j respectively, + «a «4 349 X Inelastic W4V4 10" ,3 final 3 J He and the neutron in such a are numerically equal. masses and velocities ofthe The given by McV 6.03 partitioned between the This is 3 Denoting the He and the neutron as m 3 and v'3 and we have = ,./ and cxplosivc processes Then = |m 3 (,'3) Putting and so ^(^)W ^+ W 2 m « 3 3 I4) , , ,.2 «> amu and m 4 « 1 amu, this gives us 7 « 2.3 X thus have a full picture of the final situation as viewed U4 We + W3( ffl3 1 = 2 (= 3»!) 10 m/sec. CM frame for any specified direction 9± of the outgoing neutron. To go back to the laboratory frame we have simply to in the add the CM velocity v to each of the vectors V3 and v 4 great advantage of using the CM frame in this . The that, regardless of the final directions as specified by way is the 4, magnitudes of V3 and v^ always have the same values, whereas in the laboratory frame v 3 (and also v 4 ) has a different magnitude for each direction. This does not mean, however, that desirable or necessary to go into the may What wish to answer the question: emitted at some given direction CM frame. 4 in work momenta ft directly always the speed of a neutron the laboratory with respect beam? to the initial direction of a deuteron easiest to is it is For example, one In such a case from the equations for energies it is and as measured in the laboratory: + Q = *8 + K< Pl = P3 + (934) P4 first equation Q represents the amount by which ATflnal from Alitlah so that in the example we have just considered we have Q = +3.27 MeV. Q may be positive, negative, (In the differs or zero; the Note last of these represents an elastic collision.) that Eq. (9-34) represents three independent equations (one for kinetic energy, and two for a two-dimensional problem). In momentum — treating this as the final state there are four unknowns: a magnitude and a direction for each of the vectors The situation is indeterminate unless we put in one v 3 and v 4 more piece of information, as for example the direction of one of . the particles. 350 Q is here taken to be a known Collisions and conservation laws quantity (as are K x and p O, but one could deduce ments of v 3 and V4. WHAT IS it from a complete set of measure- A COLLISION? We are so used to associating the abrupt, violent event that results it may word "collision" with some be well to point out that the we have developed can be applied to other, The only essential features are these: quite gentle interactions. 1. within That the interaction some is confined, for all limited interval of time, so that it practical purposes, can be said to have a beginning and an end. 2. That over the duration of the collision, the effect of any external forces can be ignored, so that the system behaves as though it were isolated. Figure 9-23 shows stroboscopic photographs of a collision between two mounted in the Fig. frictionless vertically pucks carrying permanent magnets, so as to repel each other. There usual sense, but this 9-23 (a) A is certainly a collision "soft" collision, with no contact in the permanent magnets. (b) The identical photographed by a camera moving with the ofmass of the two objects. (From the PSSC "Moving with the Center of Mass", by Herman R. center film, Branson, Education Development Center Film Sudio, Newton, Mass., 1965.) (a) 351 Whal (b) is a collision? no contact collision within the ordinary sense, occurring between two objects (2:1 mass ratio) carrying is f meaning of the term. One can see that the velocities are magnitude and direction except over the physicist's effectively constant in limited region of close approach of the objects. INTERACTING PARTICLES SUBJECT TO EXTERNAL FORCES Having discussed the conservation of linear momentum, and the momentum, we can general relation of force to rate of change of now consider the motion of a system of interacting particles that This are not free of external influence. of course, a very im- is, portant extension of our ideas, because in practice a system never completely isolated from It look will suffice to tension to any particles, and two-particle system. The ex- particles is quite simple but will Let m, and m2 by 1 particle 2, and — 2 i f2 the force exerted the force exerted f i2 be be the masses of the two Fi and F 2 the external forces acting on them, and particle is surroundings. f 12 the internal interaction forces particle by at a number of deferred to Chapter 14. its on i on particle 2 I. Newton's law of motion applied to the particles individually states that Fi F2 + + f21 fi2 = = /wi -7- m (9-35) t/v 2 nt2 —7- Newtonian, and we If the interaetions are case, then the third fl2 Adding + f21 shall consider this the law of Newton requires that = we then the two equations (vectorially), Fi + F 2 = mi Fi + F2 = —+m 2 get -j- or (mi vi + (9-36) /n 2 v 2 ) and f2 equation states that the resultant of all in on which the internal forces f1 2 1 have disappeared. This the external forces acting the system equals the rate of change of the total veetor mentum of 352 -at the system. Collisions and conservation laws mo- We can express this result in another, very compact way by introducing the concept of the center of mass of the system. If the individual positions of the two particles are given by the drawn from some vectors r! and ra, center of mass . _ /rnri m\ This is origin, the position of the given by is +w + 2r2 (g_ 37) rri2 the three-dimensional analogue of (9-25) and corresponds to the vector velocity of the center of mass as already defined by Eq. (9-33), so that + »»lVl where M cordingly, F, and /W2V2 = Mv (= m\ + we have + F2 m 2) is the total mass of the system. M^ di = F = Ac- (9-38) proves the result we are after: this The motion of the center of mass of a system of two particles is the same as the motion of a single particle of mass equal to the total mass of the system acted on by the resultant of all the ex ternal on the individual particles. forces which act The implications of this suggests that a fundamental system of particles is of (1) the motion of result are significant. method to analyze its center of its for treating the saw motion of the system, earlier. is it motion as the combination mass and particles relative to their center of mass. internal First, motion of a (2) the The motions of the latter motion, the one of zero momentum, as we Furthermore, Eq. (9-38) allows us to treat some aspects of the motions of extended objects by the laws of dynamics object for a simple particle. moves In particular, in translation, i.e., when there is when an extended no motion of any particle in the object relative to the center of mass, Eq. (9-38) tells the whole story. We shall return to such questions in Chap- ter 14. Incidentally, Eqs. (9-36) and (9-38) also provide a basis for a criterion as to whether or not a system of colliding particles effectively isolated. lision process holds The conservation of momentum good only to the extent that the external forces can be ignored. If external present, the duration A/ of the collision 353 forces in is a col- effect of any are indeed must bc so short that Intcracting particles subject to cxternal forces F the product At same condition different way of stating this that the forces of interaction between the is colliding particles A negligible. is must be much greater than any external forces which may be acting. THE PRESSURE OF A GAS As our we transfers on the example of last processes and collision momentum of a gas, shall take the calculation of the pressure basis that this is due to the perfectly elastic collisions of the molecules of the gas with the atoms that comprise the wali of the container. We of mass makes assume that the gas shall wi is made up of n particles, each P er urut volume. The most naive possible calculation . the following assumptions: 1. Ali the particles have the 2. The them were, particles at any same speed, instant, traveling parallel to a given direction in space, with the other two thirds traveling parallel to directions, perpendicular to each other The gas 3. is v. can be regarded as though one third of and two other to the first direction. in a rectangular container with perfectly flat, hard walls. None The first container of these assumptions two are we shall an stands. not flat and hard. Nevertheless, the calculamake, using these assumptions, comes very close to with the huge number is it the choice of a rectangular sticky, yielding the correct result. there in the least realistic as very special, and on an atomic scale the walls are is knobbly and tion certainly is false, essential The reason is that, on the average, of particles present in a sample of gas, symmetry in the aggregate motion and an effectively exact conservation of the total kinetic energy. individual molecule may strike the wali, sit there An a short while, a different angle with a different speed— perhaps faster, perhaps slower. But on the average, we may treat the collisions as perfectly elastic because the kinetic energy and jump off again at of the gas as a whole neither increases nor decreases with time. On this basis, we make the following very simple calculation. Consider an element of area vessel (Fig. 9-24). the normal to AA AA in one wali of the containing Resolve the motions of the molecules so that is one of the three mutually perpendicular directions along which the molecules are 354 Collisions and conservalion laws assumed to move. Any [«•— t>Af C 3 Fig. 9-24 Simplest possible approach to calculating of a gas. D uring a time At the elemen! of can be reached only by molecules that lle the pressure AA wali AA cylinder oflengtk v At. inilially within the one molecule approaches the wali with momentum — m Q v; coils with How many all it momentum molecules strike A A in a time Ali of them have the same speed from the wali cannot m v and re- thus provides an impulse 2moU. v, If we assume arrive until after the time Ar has elapsed. our attention is limited to molecules within a cylinder of Thus cross section AA and unit volume. Hence length v A/. number of molecules within There are n molecules per = cylinder nv Ar AA But of these, only one sixth are moving, not just along a perpendicular to AA, but specifically approaching receding from it. impulse communicated total AA line instead of This, then, gives us AA to AA number of impacts with But the average force Ar. that molecules farther than d At AF exerted in A/ in Ar on AA is = \no At AA = 2mov X 5/J0 A; AA = $nmov 2 AtAA the impulse divided by Hence AF = The mean wali, is ^nmov 2 AA force per unit area, exerted normal to the containing what we call the pressure, p. ' Thus we arrive at the result P = Since ^= nm density p, \ nr»o» the total is we can 2 (9-39) mass of gas per unit volume, which is its alternatively put 2 P = hpv or "© 'In this section, 355 The 1/2 p always (<M0) refers to pressure, not pressure of a gas momentum. Thus, our calculation if is we can justified, infer the speed of the invisible molecules from perfectly straightforward measurements of the bulk properties of the ordinary atmospheric pressure and Takung nitrogen gas. room temperature, at for example, we have p p w « 10 5 N/m 2 = 1.15 kg/m 3 kg/m-sec 2 10 Therefore, v « 500 m/sec How is defensible A Eq. (9-39)? is We clearly called for. more careful calculation have seen experimental evidence (Chap- molecules of a gas at a given temperature have a ter 3) that the So the v 2 wide spread of speeds. in our formula should be placed by some kind of an average squared speed, u we could much faith in re- And calculation, —although the rigorous not, place the numerical factor 3 in fact. it, ,. on the strength of our own certainly theory confirms 2 An acceptable treatment of the problem must investigate the consequences of having molecules approaching the wali in quite arbitrary directions; it is accident that this does not change the result in detail. if we considered the numbers of particles striking only an (It would, AA, instead of Nevertheless, the simple analysis that the force they exert.) we have presented gives us a remarkably useful beginning for the understanding of bulk properties in microscopic terms. since the detailed kinetic theory of gases we cern, shall rest is And, not our present con- content with that. THE NEUTRINO At the beginning of this chapter we pointed out servation law or conservation principle in physics but that then its if, in status the face of apparent may failure, it is be greatly strengthened. that any conis provisional, finally vindicated, The most dramatic success of the conservation laws of dynamics took place in connection with the neutrino — that elusive, emitted in the process of radioactive beta decay. neutral The particle prediction stemmed from an apparent nonconservation of energy and angular momentum, but perhaps the most beautiful and direct dynamical evidence for it is furnished by the apparent of its existence nonconservation of linear 356 momentum. Collisions and conservation laws Fig. 9-25 The visible tracks Evidence for the neutrino. of the electron and the recoiling lithium 6 nucleus in the beta decay of helium 6 in a eloud- chamber are not [Front J. collinear. Csikai and A. Szalay, Soviet Physics JETP, 8, 749 (1959).] The situation can be simply stated as follows: It is known that the process of beta decay involves the ejection of an electron from a nucleus, as a result of which the nuclear charge goes up by one unit (if the electron is an ordinary negative electron). If no other particles were involved, the process could be written A-»B + where A is e~ the initial nucleus effectively isolated, momentum and and B the final nucleus. initially stationary, If A were our belief in linear conservation would lead us to predict that, whatever the direction (or energy) of the ejected electron, the nucleus would from inevitably recoil in the opposite direction. this, regardless of involvement of another all Any other details, would B departure demand the particle. Figure 9-25 shows a cloud-chamber photograph of the beta The decay takes place at the position of the sharp knee near the top of the picture. The short stubby track decay of helium 6. pointing in a "northwesterly" direction of lithium 6; the other track another particle — the — neutrino is the recoiling nucleus the electron. if the final There must be momentum — vectors — up to have the same resultant i.e., zero as the stationary °He nucleus. It fails to reveal itself because are to add initially is lack of charge, or of almost any other interaction, allows its escape unnoticed only about 1 — so in 10 12 readily, in fact, that the of its it to chance would be interacting with any matter in passing right through the earth. PROBLEMS 9-1 A particle of mass m, traveling with vclocity vq, makes a completely inelastic collision with 357 Problcms an initially stationary particle of mass Make M. a graph of the final velocity o as a function of the ratio m/ M from m/ M = 9-2 ball how Consider bouncing A m/ to M= 10. momentum conscrvation of linear applies to a off a wali. is placed upon a movements of the box over the table? Just what maneuvers could the mouse make the box perform? If you were such a mouse, and your object were to elude pursuers, would you prefer that the table have a large, small, or 9-3 mouse put into a small closed box that is Could a table. clever mouse control the negligible coefficient of friction? 9-4 In the Phncipia, Newton mentions one that in set of collision experiments he found that the relative velocity of separation of two was objects of a certain kind of material Suppose velocity of approach. mass 3wo, of this material traveling with an an that was struck by a similar object of mass Find the velocity Do- initial of their relative five ninths initially stationary object, final velocities of 2/wo, of both objects. A 9-5 of mass mo, traveling at speed do. strikes a stationary particle mass 2mo. As a particle of through 45° and has a tion of the particle of result, the particle final mass speed of uo/2. of mass mo is deflected Find the speed and direc- Was 2/wo after this collision. kinetic energy conserved ? Two 9-6 skaters (A and B), both of one another, each with a speed of 1 mass 70 A m/sec. kg, are approaching carries a bowling ball with a mass of 10 kg. Both skaters can toss the ball at 5 m/sec relative to themselves. forth two To avoid when they tosses, i. e., are 10 A collision they start tossing the ball m apart. gets the ball but they can throw twice as the entire incident on the is initial back? fast, about weighs half as much If the ball how many back and How one toss enough? tosses do they need? Plot a time versus displacement graph, in which the positions of the skaters are of time Is marked along the abseissa, and the advance (Mark and include the represented by the increasing value of the ordinate. positions of the skaters at space-time record of the ball's motion x = in the ±5 m, diagram.) This situation model of the present view of interaetions (repulsive, An attractive in the above examplc) between elementary particles. exchange supposing that the skaters interaetion can be simulated by serves as a simple a boomerang instead of a ball. sented by F. Reines and J. P. [Thcse theorctical models were preF. Sellschop in an artiele entitled "Neutrinos from the Atmosphere and Beyond," Sci. Am., 214, 40 (Fcb. 1966).] 9-7 Find the average recoil foree on a machine gun firing 240 rounds if the mass of each bullet is 10 g and the muzzle (shots) per minute, velocity 358 is 900 m/sec. Collisions and conservation laws 9-8 Water emerges in a vertical jet from a nozzle mounted on one end of a long horizontal metal tube, clamped at enough thin The to be rather flexible. other end and its of 2.5 jet rises to a height above the nozzle, and the rate of water flow is 2 liter/min. been previously found by static experiments that the nozzle an amount proportional mass of 10 g, hung upon it, causes pressed vertically by and that a m It has is de- to the applied force, a depression of 1 cm. How far is the nozzle depressed by the reaction force from the water jet ? [This problem A 9-9 based on a demonstration experiment described by is Am. E. F. Schrader, "standard 784 (1965).] J. Phys., 33, fire stream" employed by a city 70 on ft a building whose base fire department minute and can attain a height of delivers 250 gallons of water per 63 is ft from the nozzle. Neglecting air resistance: What (a) is the nozzle velocity of the stream? (b) If directed horizontally against a vertical wali, what force would the stream exert? (Assume that the water sprcads out over the surface of the wali without any rebound, so that the collision is effectively inelastic.) 9-10 A helicopter has a total mass M. Its main rotor blade sweeps out a circle of radius R, and air over this whole circular area in from above the rotor and driven i>o- The vertically is pulled downward with a speed density of air is p. (a) If the helicopter hovers at some what must be fixed height, the value of to? One (b) of the largest helicopters of the type described above weighs about 10 tons and has thiscasc? Take p 9-11 A = 1.3 rockct of initial mass rate \dM/dt\ = y. and ^ 10 m. R kg/m 3 What is to for hovering in . Mo ejects its burnt fuel at a constant at a speed po relative to the rocket. (a) Calculate the initial acceleration of the rocket if upward from vertically (b) If vo = its 2000 m/sec, how many kilograms of ejected per second to give such a rocket, of upward it starts launch pad. acceleration equal to 0.5 mass 1000 fuel tons, must be an initial g? 9-12 This rather complicated problem is designed to illustrate the advantage that can be obtained by the use of multiple-stage instead of single-stage rockets as launching vehicles. load (e.g., Suppose rocket (see the figure). the payload — is Nm. — payload, after first-stagc burnout and separation, stage the ratio of burnout fuel) is r, (a) 359 that the pay- mass m and is mounted on a two-stage The total mass both rockets fully fueled, plus The mass of the second-stage rocket plus the a space capsule) has and the exhaust speed Show Problem* mass that (casing) to initial is nm. In each mass (casing plus is vo- the velocity i»j gained from first-stage burn, . . K c C c A>m from starting v ln r /V (b) given by is A' = vi (and ignoring gravity), rest + «(1 - r). Obtain a corresponding expression for the additional velocity, from thc second-stage burn. Adding u i and i>2, you have the payload velocity v in terms of N, n, and r. Taking A' and r as constants, find the value of n for which f is a maximum. V2, gained (c) (d) Show value of and o, scribed by r maximum that the condition for v to be a to having equal gains of velocity in the verify that = and r = it two stages. makes sense corresponds Find the maximum for the limiting cases de- 1 Find an expression for the payload velocity of a single-stage same values of A', r, and vo- (e) rocket with the Suppose that (f) it is desired to obtain a payload velocity of km/sec and r = 0.1. that the job can be done with a two-stage rocket but is im- 10 km/sec, using rockets for which vo Show possible, 2.5 however large the value of N, with a you are ambitious, number of stages. It is single-stage rocket. an arbi- try extending the analysis to (g) If trary = show that once again the initial mass is obtained if the possible to greatest payload velocity for a given total stages are so designcd that the velocity increment contributed stage is by each the same. A block of mass m, initially at rest on a frictionless surface, is bombarded by a succession of particles each of mass bm (<<C m) and of initial speed t>o in the positive x direction. The collisions are perfectly elastic and each particle bounces back in the negative x direction. 9-13 Show that the speed acquired by the block after the «th particle has struck it is given very nearly by v Consider the validity of fluid On (b) 360 « 1 this rebounded model, elastically an object is fluid —» traveling (supposedly struck by the object. would vary as some power, the value of n'! Suppose that a flat-ended object of cross-sectional area at speed v through a fluid oo when the resistive force What 2Sm/m. as well as for an resistive force for n, of the speed o of the object. moving an by supposing that the particles of the initially stationary) (a) — e" an ), wherea = i>o(l this result for 9-14 Newton calculated the through a = of density Collisions and conservation laws p. By A is picturing the fluid / : as composed of n particles, each of mass m, per unit volume (such that nm = p), obtain an explicit expression for the resistive force if each particle that struck by the object recoils elastically from is sphere of radius r, it. of being flat-ended, were a massive (c) If the object, instead medium traveling at speed v through a of density p, what would the magnitude of the resistive force be? The whole calculation can be carried out from the standpoint of a frame attached to the sphere, so that the fluid particles approach it with the velocity -v. Assume that in this frame the fluid particles are reflected as by a mirror— angle of reflection equals angle of incidence (see the figure). You must consider the surface of the sphere as divided up into circular zones corresponding to small angular increments dd possible valucs of A 9-15 various 0. particle of particle of at the mass mi and mass mi. The initial velocity collision that after the collision the particles is «i strikes a stationary perfectly elastic. It is have equal and opposite observed velocities. Find (a) (b) (c) The The The ratio 012/011. velocity of the center of mass. total kinetic energy of the two particles in the center 2 of mass frame expressed as a fraction of %m\Ui (d) A 9-16 The mass final kinetic . energy of 011 in the lab frame. /«1 collides with a mass Define relative velocity 012. as the velocity of 011 observed in the rest frame of 0*2. Show the equivalence of the following two statements (1) Total kinetic energy (2) The magnitude of suggested that (It is is conserved. the relative velocity you unchanged. is solve the problem for a one-dimensional collision, at least in the first instance.) 9-17 A collision pended so appa ratus is made of a with one another (see the figure). f 2 mo, first of n graded masses sus- and not quite in contact mass is jmo, the second is is 3 The first and so on, so that the last mass struck by a particle of mass 010 traveling the third mass set that they are in a horizontal line 0!o, hfma. This produccs a succession of collisions along the line of masses. 361 Problem s The at a speed vo. "». Om o6666666iifm„ A fm„ Assuming that all the collisions are perfectly mass flies off with a speed o„ given by (a) elastic, show that the last 1 + i>o /, Hence show (b) written as ± 1 that, if e (with e / is close to unity, so that <JC 1), this virtually all the kinetic energy of the incident even for large mass to the last one, n. For /= energy of the last (c) can be it system can be used to transfer 0.9, n = 20, calculate the mass, speed, mass in the line in terms kinetic energy of the incident particle. and kinetic of the mass, speed, and Compare this with the result of a direct collision between the incident mass and the last mass in the line. 9-18 A 2-kg and an 8-kg mass collide elastically, compressing a bumper on one of them; the bumper returns to its original length as the masses separate. Assume that the collision takes place along a single line and that you can cause the collision to occur in different ways, each having the same initial energy: Case A: The 8-kg mass has 16 J of kinetic energy and hits the staspring tionary 2-kg mass. Case B: The 2-kg mass has 16 J of kinetic energy and hits the stationary 8-kg mass. (a) Which way of causing the collision to occur will result i n the greater compression of the spring? Arrive at your choice without actually solving for the compression of the spring. (b) 16 J, Keeping the condition of a how should this total initial kinetic energy of energy be divided between the two masses to obtain the greatest possible compression of the spring? 9-19 In a certain road accident (this is based on an actual case) a car of mass 2000 kg, traveling south, collided section with a truck of mass 6000 locked and skidded off the road along a A Southwest. in the middle of an kg, traveling west. line The inter- vehicles pointing almost exactly witness claimed that the truck had entered the inter- mph. you believe the witness? Whether or not you believe him, what section at 50 (a) (b) 362 Do Collisions and conservation laws fraction of the total was converted kinetic energy initial forms of energy by into other thecollision? A 9-20 nucleus A of mass 2m, traveling with a velocity u, collides with a stationary nucleus of mass 10/w. to be traveling with speed Pi at 90° to and B The collision results in its is a change observed original direction of motion, traveling with speed Da at angle 8 (sin 6 is A After collision the nucleus of the total kinetic energy. = 3/5) to the original direction of motion of A. What What (a) (b) are the magnitudes of vi and U2? fraction of the initial kinetic energy gained or lost is as a result of the interaction? A 9-21 particle of mass M with a particle of mass the particle of mass The to «/\/3. m is m initial velocity u collides elastically As a result of the collision through 90° and its speed is reduced deflected particle of mass the original direction of m. in the and initially at rest. M recoils with speed p at (Ali speeds and angles an angle 6 to are those observed laboratory system.) Find (a) M in terms of m, and v in terms of Find also the u. angle 0. At what angles are (b) the particles deflected in the center-ofmass system? Make measurements on 9-22 the stroboscopic photographs of a collision of two magnetized pucks of linear momentum and three time units) (first A (Fig. 9-23) to test the conservation total kinetic and energy between the initial state the final state (last three time units). mass 2m and of velocity u strikes a second particle As a result of the collision, a particle of mass m is produced which moves off at 45° with respect to the initial direction of the incident particle. The other product of this rearrangement collision is a particle of mass 3m. Assuming that this collision involves no significant change of total kinetic energy, calculate the speed and direction of the particle of mass 3/w in the Lab and in the 9-23 particle of 2m of mass CM initially at rest. frame. 9-24 In a historic piecc of research, James Chadwick a value for the mass of the neutron by studying in neutrons with nuclei of hydrogen and nitrogen. fast the maximum was 3.3 X 14 nuclei does 10 was this tell (a) (b) 7 recoil velocity of hydrogen nuclei maximum m/sec, and that the 4.7 X 10 u 1932 obtained elastic collisions He found of that (initially stationary) recoil velocity of nitrogen m/sec with an uncertainty of ±10%. What you about The mass of a neutron ? The initial velocity of the neutrons used? (Take the uncertainty of the nitrogen measurement into account. Take 363 the mass of an Problems H nucleus as 1 amu and the mass of a nitrogen 14 nucleus as 14 amu.) A 9-25 cloud-chamber photograph showed an alpha particle of mass 4 amu with an X of 1.90 initial velocity The nucleus in the gas of the chamber. 10 7 m/sec colliding with a changed the direction collision of motion of the alpha particle by 12° and reduced 10 7 m/sec. The other particle, X of 2.98 10 7 initially stationary, its speed to 1.18 X acquired a velocity m/sec at 18° with respect to the initial forward direcWhat was this second particle ? Was the tion of the alpha particle. fact that these A take account of the results, cloud-chamber measurements of speeds and angles are subject to errors of 9-26 your (In interpreting collision elastic? up to a few percent.) nuclear reactor has a moderator of graphite. atoms of nuclei in the by free to recoil if struck fast The carbon can be regarded as this crystal lattice effectively neutrons, although they cannot be knocked out of place by thermal neutrons. A fast neutron, of kinetic MeV, collides elastically with a stationary carbon 12 nucleus. (a) What is the initial speed of each particle in the center of energy 1 mass frame? As measured (b) carbon nucleus is in the center-of-mass frame, the velocity of the turned through 135° by the collision. What are the speed and direction of the neutron as measured in the lab frame? final About how many (c) collisions, elastic involving random changes of direction, must a neutron make with carbon nuclei mean energy the loss is MeV if its keV? Assume that midway between maximum and minimum kinetic energy is to be reduced from 1 to 1 values. 9-27 A (a) moving M mass particle of with a stationary particle of mass m < collides perfectly elastically M. Show that the maximum possible angle through which the incident particle can be deflected is -1 (/w/M). (Use of the vector diagrams of the collision in Lab and sin CM systems will be found helpful.) A (b) particle stationary particle of of mass mass m collides perfectly elastically with a M > m. The incident particle through 90°. At what angle d with the original direction of more massive 9-28 The is text (p. 348) gives an examplc of the analysis of the dynamics Another possible reac- the following: ?H In deflected m does the particle recoil? of a nuclear reaction between two deuterons. tion is + fH - !H + ]H + this case the reaction 4.0 MeV products are a proton and a triton (the latter being the nucleus of the unstable isotope hydrogen 3 or tritium). Suppose that a stationary deuteron of kinetic energy 364 Collisions 5 is struck by an incident deuteron MeV. and conscrvation laws What (a) (b) maximum and minimum are the proton produced in kinetic energy of the What is maximum the angle (as observed in the laboratory) make that the direction of the triton can incident deuteron? oratory when is it What is its possible values of the this reaction? with the direction of the kinetic energy as emitted in this direction? conveniently handled with the use of measured in the lab- (This problem can be amu and MeV as units throughout.) 9-29 A boat of mass man tionary; a M and m of mass length L is at the is sitting floating in the water, sta- bow. The man stands up, and sits down again. assumed to offer no resistance walks to the stern of the boat, water (a) If the how of the boat, far is does the boat move as a at all to result of the motion man's trip from bow to stern? (b) More realistically, resistance given Show of the boat. assume that the water offers a viscous A: is a constant and o is the velocity by —kv, where that in this case one has the remarkable the boat should eventually return to (c) its initial result that position. Consider the paradox presented by the fact that, according to (b), any nonzero value of k, however small, implies that the boat ends up at its starting point, but a strictly zero value implies that How it do you explain this discontinuous jump in the final position when the variation of k can be imagined as continuous, down to zero? For enlightenment see the short and clear ends up somewhere analysis by 365 D. Problcms else. Tilley. Am. J. Phys., 35, 546 (1967). Conservation of Energy as a principle fundamental role, the quantum mechanics. which is thefirst law . . has played a most fundamental as some would say, of classical physics, and since then in the last half-century in . . . Yet . it is a curiosity that this, of thermodynamics, should have so watered the stonier, the more austere terrain of fruitfully mechanics. C. G. GILLISPIE, It may be discussion The Edge of Objedivity (1960) thought strange that of the foundations of mechanics sofar with hardly a mention ofenergy. . . . desire to slight the significance wished is to we have brought our indeed had no of this concept, but have emphasize that the logical foundation of mechanics quite possible without it. . . . why then should This what we wish to discuss. it The guestion at once have been introduced at all? arises: is We have R. B. LINDSAY AND H. MARGENAU, Foundations of Physics (1936) ' 10 Energy conservation in dynamics; vibrational motions INTRODUCTION of all the physical concepts, that of energy is perhaps the most far-reaching. Everyone, whether a scientist or not, has an awareness of energy and what it means. Energy for in order to get things done. the background, but we is The word what we have to pay itself may remain in recognize that each gallon of gasoline, each Btu of heating gas, each kilowatt-hour of electricity, car battery, each calorie of food value, represents, in one another, the wherewithal for doing what not think in call work. terms of paying for force, or acceleration, or Energy tum. we is each way or We do momen- the universal currency that exists in apparently countless denominations; and physical processes represent a conversion from one denomination to another. The above remarks do not really define energy. No matter. worth recalling once more the opinion that H. A. Kramers expressed: "The most important and most fruitful concepts are It is those to which The it is impossible to attach a well-defined meaning." immense value of energy as a concept lies in its transformation. It is consewed— that is the point. Although we may not be able to define energy in general, that does not mean clue to the 'See p. 62, where cussion ofi we quoted Kramers' remark in connection with our dis- the concept of time. 367 that only a vague, qualitative idea. it is We have set up quantitative measures of various specific kinds of energy: gravitational, which situation has arisen in peared, And whenever magnetic, elastic, kinetic, and so on. electrical, it has always been possible to recognize and define a it a seemed that energy had disap- new form of energy that permits us to save the conservation law. And conservation laws, as we remarked at the beginning of Chapter 9, represent one of the physicist's most powerful tools for organizing his description of nature. In this book we shall be dealing only with the two main categories of energy that are relevant to classical mechanics —the kinetic energy associated with the bodily motion of objects, and the potential energy associated with elastic deformations, gravitational attractions, electrical interactions, and the like. If energy should be transferred from one or another of these forms into Chemical energy, radiation, or the random molecular and atomic motion we lost. This restrict then from the standpoint of mechanics call heat, is a very important feature, because it means it is that, if we our attention to the purely mechanical aspects, the conis not binding; it must not be blindly assumed. servation of energy Nevertheless, as we shall see, there are many physical situations for which the conservation of the total mechanical energy holds good, and in such contexts it is of enormous value in the analysis of physical problems. It is an interesting historical sidelight that in pursuing the we are temporarily parting company with Newalthough not with what we may properly call Newtonian subject of energy ton, mechanics. In the whole of the Principia, with its awe-inspiring elucidation of the dynamics of the universe, the concept of energy is never once used or even referred was enough. But we shall see rooted in F = ma, has its own how to! ' For Newton, special contributions to shall begin with the quantitative F = ma the energy concept, although make. We connection between work and kinetic energy. INTEGRALS OF MOTION In Chapter 6 we briefly presented the basic notion of the work 'As we remarked in Chapter 9, however, some of Newton's contemporaries, particularly Huygens and Leibnitz, did recognize the importance of an energylike quantity— the "vis viva" mo 2 —\a certain contexts, such as collisions and pendulum motions. 368 Energy conservation in dynamics done by a force acting on an object, producing a corresponding We shall now return increase in the kinetic energy of the object. to this topic and develop it considerably further. Let us again take, for a start, the simple and familiar case of an object of mass m acted on by a constant force F (we shall begin by assuming a straight-line motion for which vector symbols are Then by Newton's law we have unnecessary). F = — = m do — — dt do ma di Let the force act for a time from Vi to v 2 Fl during which the velocity changes t Then we have - = mal = m(o 2 - (10-la) ui) This expresses the fact that the impulse of of linear momentum. But suppose we distance over which F is equal to the change multiply the force by the instead of by the time. In this case it acts, we obtain Fx = max = ma = VI — + V2 +u £m(ar)(Di = im(v 2 - . / 2) ui)(ui + Wl) Therefore, Fx = \mv2 2 - 00-2a) J/wi 2 This expresses the fact that the work done by change of kinetic energy. We F is equal to the have no reason to declare a preference between Eqs. (10-la) and (10-2a) as statements of the effect of the force. In fact, they will in general But before amplifying that restriction to constant force sary. last tell remark, let us different things. us note that the and constant acceleration is unneces- For we have F- m d° Multiplying by dt and integrating gives us r »2 '2 F di = m /; do = m(o 2 - ci) J Vl Multiplying by dx and integrating gives us 369 Integrals of motion (10-lb) £ '*—£%* But dv dx di dt —rdx — —rdo = vav Thus *! •a Fdx The vdv = m left-hand side of Eq. (10-1 b) and the left-hand side work done by the total — £j»((?2 is the total impulse of the force, of Eq. (10-2b) force. (10-2b) Ci ) is of course defined as the Each of these integrals can be represented as the area under the appropriate graph of F against x between ccrtain limits (see Fig. 10-1). The general similarity of Eqs. (10-lb) and (10-2b) deceptive, because force, displacement, quantities. The depends very much on its seen, a net force momentum is velocity are vector a given magnitude it acts. Thus in circular motion, as (also a vector) changes continuously, but the situation becomes clear if we return to the proper — dt It is still possible to integrate directly with respect to 10-1 (a) Force Fig. that varies linearly with time. The area under the curve measures the total impulse. (b) The same force plotted as afunction of displacement. The area under the curve measures the total work. 370 Energy conservation in dynamics mag- all. vector statement of Newton's law: F = m however, continually applied to an object, the nitude of the velocity docs not change at The or direction relative to the direction of motion of the object on which we have and result of applying a force of is, / /: ¥dt = m(v2 The — vi) left-hand side of the expression is an instruction that, the if we must form the appropriate vector sum of all the small impulses F A/ applied in sucThat is exactly how cessive time intervals At between t\ and / 2 Newton himself conceived the action of the varying force to which direction of the force changes with time, . a planet, for example, finds around the sun orbit itself Chapter (cf. it moves along its But what about integrating the elements of distance along a path, where these elements F over (Ar) are themselves vectors? This We not a mathematical one. applying a force at some is essentially a physical question, are asking: What the effect of is arbitrary direction to the motion of an Figure 10-2 helps to supply the answer. object? is exposed as 13). applied at right angles to v for a force F! If some very short time At, it changes the direction of the velocity without appreciably changing magnitude. its it If F2 a force is applied along the direction of v, changes the magnitude of v without changing the direction. if we want to fix our attention on changes in the magnitude we should restrict ourselves to forces or force components along the direction of v. If the net force F acts at an angle d to v, we have its component along v given by Thus of v, |F| cos 6 where Av 10-2(c)]. is = m |Av| cos d At the vector change in v, equal to The element of ¥ At/m distance traveled during At is [cf. Fig. given by Ar = vA/ Fig. 10-2 (a) Velocity change due to an impulse perpendicular to v. (b) Velocity change due to an arbitrary change due to an impulse parallel to impulse. 371 Integrals of motion v. (c) Velocity by the Hence the definition of the vclocity vector. ponent along multiplicd by the displacement, v, The above m = cos d |F| [Ar| |v| F we have suppressed we use com- all reference changing the direction of v and are in To information about the magnitude only. neatly force given by |Av| cos 6 a scalar equation; is to the effect of is left with express this more the notation of the "dot product" (or scalar product) of two vectors a b = • |a; |b| In this notation, cos e = ab cos e we have F Ar = m(y Av) • But we can now play a neat (and valuable) trick. Consider the quantity v v. This is a scalar, and its magnitude is just the square • of |v|, i.e., = 2 v 2 o v v • we have, by differentiation, = Av v (This last step is A(u 2 ) vectors i.e., + v - Av = 2v • • Av possible because the scalar product of indepcndent of the order the commutative law holds.) v It is But since simply. in two which the factors come Hence Av = J A(u 2 ) follows that we have F-Ar = Jmift) 2 ) integrating over any path that the So now, lowed under the aetion of the W= 2 F- Jt = lm(02 \ -' force, we body may have fol- find the relation - d') (10-3) r i which deseribes in general terms the relation between the work change of kinetic energy. Figure 10-3 illustrates what is involved in evaluating the work integral of Eq. (10-3), in going from a point A (r0 to a done and point 372 B the (r 2) in a two-dimcnsional displacement. Enerev conservalion in dynamies Fig. 10-3 tion of work done Calcula- along a given path. Equation (10-3) displays the most important property of work and energy: They are if it were traveling horizontally vector An scalar quantities. moving object speed v has cxactly the same kinetic energy as vertically with a momentum would at this same speed, although its be quite different. This scalar property of energy will be exp!oited repeatedly in our future work. WORK, ENERGY, AND POWER This chapter conccrned with dcveloping some general chiefly is dynamical methods based upon the concepts of work and mechanical energy. The methods will, however, practical use of these involve numerical measures of these quantities in terms of appro- The purpose priate units. of this section to introduce is some of these units for future reference. Our unit of 1 If the 1 J basic unit, already introduced in Chapter 6, will be the work or energy = 1 CGS erg = N-m = 1 in the dyn-cm system — the joule: kg-m 2 /sec 2 system of units 1 MKS = 10 Before going any further, is -7 wc uscd, the unit of energy is the erg: J shall introduce a seeming diversion — the concept of power, defined as the rate of doing work: power dW dt (For a mechanical system, putting dW F v.) Power is a importance, because the time that 373 = F • t/r, we have power = concept (and a quantity) of great practical Work, energy, and power it takes to perform some given " amount of work may be a small electric motor may vital consideration. For example, a be just as capable of driving a hoist as a big one (given, perhaps, a few extra gear wheels), but quite unacceptable because the job Power far essentially a practical engineering concept ; is be using would take it it may be too long. we shall not our development of the principles of dynamics. But in one of our accepted measures of work is often expressed in terms of a unit of power the watt. In terms of mechanical quantities, — 1 W 1 W-sec = = J/sec 1 i.e., The most J 1 familiar use of the watt = the relation watts realize that venient it volts energy unit lkWh = 3.6 1 it is through important to ' A con- domestic purposes (especially one's for X 6 10 J and thermal calculations the Standard unit Calorie, defined as the ofwaterfrom amperes, but electrical, the kilowatt-hour (kWh): is In chemical of course, not a specifically electrical quantity. is electricity bill) X is, amount of energy required the is to raise 1 kg 15 to 16°C: Cal = 4.2 X 10 :i J In atomic and nuclear physics, energy measurements are usually expressed in terms of the electron volt (eV) or keV the (10 3 ), MeV 1 V leV = G and ), GeV (10 9 ). The its related units electron volt is of energy required to raise one elementary charge amount through (10 of electric potential difference: 1.6 X 10- 19 J Finally, as Einstein first suggested and as innumerable observa- 'The other familiar unit of power is of course the horsepower. About this say only that 1 hp = 746 W, so that (as a very rough rule of thumb) one can say that it takes about 1 kw to drive a 1 hp electric motor. Most of the power levels of practical importance can be conveniently 6 W, described using power units based on the watt—microwatts (10 3 3 typical of very weak radio signals), milliwatts (10~ W), kilowatts (10 W), and megawatts (10° W, useful as a unit in generating-plant specifications). It is worth noticing, for comparison, that the sun's power output is about we shall 3.8 374 X 10" W! Energy conservation in dynamics TABLE 10-1: THE ENERGY OF THINGS - - 10" J Energy equivalenl of sun's mass 10"J (fi = mc') Daily cncrgy oulpui of our galaxy Kinelic energy of earth* orbital molion (relalive lo sun) Daily energy oulpui of sun - - io^J Kinelic energy of moon's orbital motion (relalive lo earth) Solar energy received per day on earth - - I0»>J World uscof cncrgy in 1950 <I0 M J/y) Kinelic energy of a eyelone Daily energy oulput of Hoover Dam Solar energy per day on 2 square miles Burning of 7000 tons of coal Energy released in complete U 05 fission of kg of Energy equivalent (£ m< *) of g of malter Energy of explosion of I ton of TNT I I - - I0"J Energy conlenl of average One daily diet kilowatt-hour Kinelic energy of rifle bullet I0M Oneerg(IO-'J) --io-"> J Energy cquivalcnt (K Energy produced by Energy equivalent — («<•') of one atomic mass unil fission (slow neutrons) of (£" = »i< *) Average energy lo produce one ion pair --io-»j Energy lo break a DNA molecule Kinetic energy per molecule one nueleus of of one cleclron mass ai in in air ( - 35 e V) two (-0.1 eV) room temperature (0.025 eV) Energy of a photon a I radio broadeast frequencies --10--J Kinetic energy of an electron moving al I m/sec U or Pu tions have confirmed, there is an equivalence between what we customarily call mass and what we customarily call energy. In classical mcchanics thcsc arc trcated as entirely separate concepts, but perhaps worth quoting this equivalence here, so that we it is have our selection of energy measures 1 kg of mass cquivalent to 9 is For the sake of X 10 we show interest, all in 16 one place: J in Table 10-1 some representative physical cxamplcs involving different orders of magnitude of energy, all expressed in terms of our basic mechanical unit, the joule. GRAVITATIONAL POTENTIAL ENERGY Wc shall begin with a simple and very familiar problem. has been thrown vertically upward and losing speed as it is Let us take a y goes higher. An moving under object gravity, axis, positive ward, and suppose that the object passes the horizontal y = yi with velocity u = y2 level with velocity v 2 Thcn the purely kinematic deseription of the motion (Fig. 10-4). is and reaches y t up- given by v2 2 = m2 + 2a0> 2 - .yi) with a- -g leading to the familiar result .„2 V2 _ = ,,,2 2 v\ — 2g(yz - 1 yi) Now let us consider this in terms of work and energy. The changc of the kinetie energy (K) between K2 — K\ = Jmi>2 a — >>i and y 2 Now K\ = — mgiyz — the quantity — mg given by \mv\ z Using the preceding kinematic equation K2 — is is this can be written yi) just the constant gravitational foree, Remember, we have defined g as a positice number equal to ihe magnilude of the local gravitational acceleration in any units that may be chosen. 376 Energy conservalion in dynamics 4 10-4 Fig. -yt u% Velocity change associated with vertkal motion under gramty. on the object as it moves up from y\ to y 2 (see Fig. Thus we see that the change of K is precisely equal to the work donc by the gravitational force, in conformity with the F„, that acts 10-4). general rcsult expressed by Eq. (10-3): K2 - K = x We F,0» 2 - yi) are going to rewrite this equation in a different way, such that the quantities referring to the position y on one side, and the quantities referring at first y = y2 what may seem to be a clumsy fashion, but the reason will quickly appear. Our new statement of the K2 + (-F„y 2 ) rcsult is -Xi + What we have done same valuc at two (.-F yi) here matical statement in such a the y\ appear to the position We shall express the result in appear on the other. = to deliberately frame the mathe- is way (10-4) that the sum of two quantities has different positions. That is the formal basis We define the potential of the statement of conscrvation of cnergy. energy U(y) at any given value of y through the cquation l/GO (thus = -F,y G0-5) U = y = 0). Notice particularly that U(y) is work done by the gravitational force. Sub- making the negalive of the at ' stituting this definition of the potential energy into we Eq. (10-4) thus gct E = K2 + U2 = where E is + tf , (10-6) Ui the total mechanical energy. Putting Fe — —mg in Eq. (10-5), we have the well-known result £/(/,) = mgh for an object at height h above the ground (or other horizontal level that is defined as the zero 'We shall use the symbol V is U for potential energy widely uscd also, but use t's (large 377 symbol of potential energy). For this case and small) to denote we shall avoid velocities. Gravitational potentiaJ energy throughout it this here because book. The we so often Fig. 10-5 Energy diagram for vertical motion above a horizontal surface. the energy-conservation statement can be simply represented as in Fig. 10-5. We shall have more to say about such graphs shortly. undoubtedly be familiar with another way of interpreting a potential encrgy such as U(h) in the last equation. It represents exactly the amount of work that we would have to do You will in order to raise an object through a distance tational puli, without giving it h, against the gravi- any kinetic energy. In order to achieve this, we must supply an exlernal force, Fext if this is insignificantly greater than mg, the object will move upward with negligibly small acceleration (F nct « 0), thus arriving at some ; higher level (Fig. 10-6) with almost zero velocity. Fig. 10-6 1 If the object Use ofan exterrtal force to move an object in the direction of increasing gravita- tionai potential energy. bigger 'To make it even more precise, wc could apply a force just a shade then moving, the object beginning, to get the brief time at than mg for a change to F„ t exactly equal to mg for most of the trip (the object thus continuing to move upward at constant velocity under zero net force), and just before the finally let F„, become a shade less than mg for a brief time point. end, so that the object finishes up at rest at height /i above its starting 378 Energy conservation in clynamics is subsequently released, then the tational force work done on by the gravi- it given to the object as kinetic energy (corre- is sponding to traveling from y = = h to y in Fig. 10-5). Many devices (pile drivers are a particularly clear example) operate in precisely this way. MORE ABOUT ONE-DIMENSIONAL SITUATIONS Equation (10-6) total a compact statement of the conservation of is mechanical energy in any one-dimensional problem for which the force acting on an object, only on the object's position. more result due to To derive this energy-conservation x Equating K2 - F(x)dx this work to Ki = is this given by we have F(x)dx I — but The work done by the change of kinetic energy, (10-7) In order to cast this into the we x. moves from xi to x 2 \ any arbitrary way in has a unique value at any given value of W= environment supplies a generally, suppose that the force F(x) that varies with position force as an object environment, depends its form of a conservation statement, introduce an arbitrary reference point x and express the work integral as follows: r 'z r*i / y F(x)dx = / r 'i F{x)dx- Substituting this in Eq. (10-7), K2 + We F(x) dx \-J F(x)dx (10-8) Jx Jx i, / we have - Ki + -/ 1 F(x) dx\ then define the potential energy U(x) at any point x by the following equation: U(x) - U(x ) = - / F(x)dx (10-9) Jx Notice once again the minus sign on the right. energy at a point, relative to the reference point, 379 More about one-dimensional situations is The potential always defined as the negatioe of the work done by the force as the object from the reference point to the point considered. U(x the potential energy at the reference point ), set equal to zero if we moves The value of can be itself, any actual problem we of potential energy between please, because in are concerned only with differences one point and another, and the associated changes of kinetic energy. we took In obtaining Eq. (10-9) the force as the primary quantity and the potential energy as the secondary one. Increas- however, as one goes deeper into mechanics, potential ingly, and force becomes the derived indeed, because by differentiation of both energy takcs over the primary quantity— literally so, role, sides of Eq. (10-9) with respect to x, = - F(x) we obtain ^ (10-10) This inversion of the roles not just a formal one (although is does prove to be valuable theoretically) for there are it many energy differences which one's only measurements are of between two very distinct states, and in which one has no knowledge of the forces physical situations in direct work function of a metal, example, and for The acting. electronic the dissociation energy of a molecule, represent the only directly observable quantities in these processes of removing a particle to infinity from some may point initial location. not be known How well the force varies — perhaps not at from point to all. We shall often be spelling out the kinetic energy K in terms of m and v, so that the equation of energy conservation in one dimension $mv2 is + written as follows: U(x) = E (10-11) Suppose we choose any particular value of x. Then Eq. (10-11) becomes a quadratic equation for v with two equal and opposite roots: D(x) This is = ± (^^)" 2 dO-12) the expression of a familiar result, which we can discuss an object were observed to vertically upward with speed v, traveling pass a certain point, air resistance could be ignored) then (to the approximation that in terms of motion under gravity. it 380 would be observed, a little later, Energy conservation in If to pass the dynamics same point, traveling downward at the same speed The v. direction of the velocity has been reversed, but there has becn no loss or gain of kinetic energy. In such a case the force from Eq. (10-12) that this result will hold as unique function of x. It through any given point same at it It will certainly direction of long as U(x) is a passes through that point again. Under what conditions does on the can see means that a particle, after passing any speed, will be found to have the kinetic energy every time property? We said to be conseruatiue. is the force have this conservative not be conservative if F(x) depends motion of the object which to it is applied. Consider, for example, the addition of a resistive force to the gravitational force in the vertical object goes upward through a motion of an As object. the on certain point, the net force it (downward) is greater than F„. After it reaches its highest point and begins moving down, the net force on it (again downward) is less than F„. Hence the net negative work done on it as it rises is numerically greater than the net positive work as desccnds. it Thus on balance negative work has been done and the kinetic energy as the object passes back through the designated point less than initially. The done by F(x) should be zero ending at any given value of x\ only is, indeed, that the net if this can one define a potential-energy function. equivalent condition that is F condition is In two- and three- correct. dimensional situations, howevcr, as we shall see 1), the condition that F later be a single-valued function of The condition of zero sary but not sufficient. is satisfied might seem that an It be a unique function of position. In one-dimensional situations this 1 is work over any journey beginning and crucial feature closed path defines a conservative force in all (Chapter r is neces- work over any circumstances and net should be remembered as a basic definition. THE ENERGY METHOD FOR ONE-DIMENSIONAL MOTIONS The use of energy diagrams, such as that of Fig. 10-5, provides an excellent way of obtaining a complete, although pcrhaps qualitative, picture of possible motions in a one-dimensional system. Frequently the information so obtained suffices for obtaining physical insight into situations for which analytic solutions are complicated or even unobtainable. In fact, even when analytic solutions can be obtained in terms of unfamiliar functions, they often are of teristics 381 little help in of the motion. The energy meihocl revealing the essential physical charac- The general scheme is as follows: for one-dimensional motions We . i. — ,— i :< Fig. I„ c6 10-7 -E, , X<f «, Hypothetical energy diagram for a one-dimensional system. plot U(x) as a function of x, and on the same plot draw horizontal In Fig. 10-7 corresponding to different total energies. lines shown such a potential-energy curve and is several values of the total energy. The kinetic energy K of a particle is equal to (E — U), i.e., to the vertical distance from one of the lines of constant energy to the curve U(x) at any point x. For a low energy E u U(x) is E would simply imply a value of v. Such a hence an imaginary and negative value of K although it must mechanics situation has no place in classical greater than for all values of x; this — not be discarded so lightly when one comes to atomic and nuclear systems requiring the use of quantum mechanics. For a higher E2 two regions, between x% represent two quite separate These and x&. and x 4 or between x 7 from one region to the situations, because a particle cannot escape total energy , other as long as seeing this is, the motion can occur in its energy is E 2 One way held at the value . of of course, in terms of the impossible negative value K between x 4 and x 7 But there is another way which is valuhow one "reads" such an energy diagram. example of able as an Suppose that our particle, with total energy £ 2 is at the of . , point x = x3 point is at some Its potential equal to the total energy, for this of U(x) and the line 382 instant. E = E 2 intersect. Energy conservation energy U(x 3 ) at this is Thus the in dynami.cs where the curve particle has zero kinetic energy and hence there on is a force is , The +x force direction. on rest. However, and hence F(x 3 ) is positive Thus the particle accelerates to the negative and hence it, instantaneously at it: At x = x 3 dU/dx in the is — acceleration, decreases as the slope its of the U(x) curve decreases, falling to zero at the value of which U(x) is i.e., right. x at minimum. At this point the speed of the particle maximum, a and as it moves further in the +x direction (with d(J/dx > is 0) now it diagram displays approaches x 4 — but there —V £2 kinetic energy experiences a force in the information before all this direction. us, and shows the The continuing to decrease as the particle Finally, at . —x x4 itself, the velocity has fallen to zero -x direction. What happens? The particle picks up speed again, traveling to the left, until it reaches x 3 with its velocity reduced to zero. This whole cycle is still a force acting in the of motion will continue to repeat total energy does not decrease. motion, of which we can without solving a energy diagram has to is discern single tell as long as the itself indefinitely We have, in short, a periodic many of the principal features equation—just by seeing what the us. The motion between x 7 and x 8 likewise periodic. We can dispose of the other possibilities more indicated the method. For a still higher energy briefly, E3 , having two kinds of motion are possible; either a periodic motion between X2 and x 5 or the unbounded motion of a particle coming in from large , x s then slowing down and x 6 moving ofT to the right and the changes of speed on the way in. Finally, for values of x, speeding up as reversing its duplicating the still it passes direction of motion at all larger energy E4 , , the only possible motion , is unbounded; a particle coming in from large values of x, speeding up, slowing down, speeding up, slowing down again, and reversing its direction of motion at x lt after which it proceeds inexorab!y in the For each of these motions, the speed at any point can be obtained graphically by measuring the direction of ever-increasing x. vertical distance from the appropriate the corresponding point Caution: It line of constant energy to on the potential-energy curve. The curve of U(x) in Fig. 10-7 is slopes and up the peak Iike a roller coaster. 383 almost too graphic. tends to conjure up a picture of a particle sliding The energy method Do down the not forget that for one-dimensional motions it is The a one-dimensional motion that vertical scale is the subject of the analysis. energy, and has nothing necessarily to is do with altitude. After this general introduction, let us consider some specific examples of one-dimensional motions as analyzed by the energy method. SOME EXAMPLES OF THE ENERGY METHOD Bouncing ball Suppose that a ball, moving along a peatedly on a horizontal floor. Let us dissipation (loss) vertical line, first bounces imagine that there is re- no of mechanical energy, so that this energy remains constant at some value E. We shall use>> to denote the (CG) of the ball, and take y = of the ball with the floor. correspond to is 0. For y We > by the first contaet shall take this configuration to the potential energy of the ball given by U(y) = mgy Now y = Fig. {y does not, > in 0) any real physical situation, represent the 10-8 (a) Energy diagram for a ball bouncing ver- lically. wliicli (b) Idealizalion the impaet at wlrich there is 384 U= position of the center of gravity to be defined some y = of(a) is 10 represent a situation in completely rigid bu t in dissipation Energy conservation of energy at each bounce. in dynamics : lowest point rcached by the exert any force An slightly. on the The of the ball. floor does not (the floor) has been it compressed made about the ball. Thus region y < 0. As it does this, equivalent remark can be the ball certainly however, CG ball until moves into the experiences a positive (upward) force that increases it extremely rapidly as y becomes more negative, and completely overwhelms the (negative) gravitational force that exists, of course, at all values of y. This large positive force gives increase of U(y) with y for y < rise to a very steep. [see Fig. 10-8(a)]. For any given value of the total energy, therefore, the ball between positions y t and y-i as shown in the figure. The motion is periodic that is, there is some well-defined time T oscillates — between successive passages the in same direction of the ball through any given point. Now to y2 might in practice For instance, - easily y may be numerically very small compared if a steel ball bounces on a glass plate, we x have y x of the order of 0.01 cm and y 2 of the order many purposes we can approximate the plot of of 10 cm. Thus for < U(y) against y for y by a vertical line, coinciding with the energy axis of Fig. 10-8(a). This represents the physically unreal property of perfect rigidity — an arbitrarily large force is called However, into play for zero deformation. use this approximation, description of the situation. and is we can justifiably The motion is confined to y > defined by i/w, 2 + mgy = E The maximum height h h if then we have a simple quantitative (y> is, (10-13) 0) of course, defined by putting vu — mg = To = 0: (10-14) find the period for the ball to travel T of the from y = motion we can calculate the time to y = h and then double it. That is, we have the following relation [a 2 / J »=0 dy (10-15) Vy because the elementary contribution dt to the time of equal to dy divided by the speed v y at any given point. Now 385 from Eq. (10-13) we have Some cxamples of the energy mcthod flight is OU = 2(£ ± - mgy) "' m *r Taking the positive root, to correspond to upward motion, we have, from Eq. (10-15), r* — r Ti/a * J r=2/ Us-2 /o L2(£ - ««Oj «6- g7o m/ mg)- y]»« can simplify this by noting that £/wg height h [Eq. (10-14)]. Thus we have We r- J*' /Wo This is is just the maximum * (A->-) 1/2 an elementary integral (change the variable to w = h - y) yielding the result You will, of course, recognize the correctness of this result from the simple kinematic problem of an object falling with constant and could reasonably object that this is another of those cases in which we have used a sledgehammer to kill a fly. But it is the method that you should focus attention on, and acceleration, perhaps the use of a familiar example will It facilitate this. should not be forgotten that most motions involve varying accelerations, so that the Standard kinematic formulas for motion with constant acceleration do not But Eq. (10-15), in which v apply. any is defined at any point by the energy equation, can be used for numerically if one-dimensional motion and can be integrated necessary. Before leaving this example, let us use it other instructive feature of the energy diagram. to illustrate one We know that the total mechanical energy of a bouncing ball does not in fact Although there stay constant but decreases quite rapidly. little loss of energy while the ball loss at each bounce. is is in flight, there is a substantial Figure 10-8(b) shows how this behavior can be displayed on the energy diagram. Starting at the point A, the history of the whole motion arrows. 386 The is obtained by following the successive decreases in the Energy conservation in dynamics maximum height of bounce, and the inevitable death of the motion at y apparent by inspection of the Mass on a figure. many physical systems— not chanical systems, but also atomic systems, ones— that can be analyzed by analogy The reason A for this lies in mass analogues 0, are quite spring There are very 1. = two just ordinary and even me- electrical with a mass on a spring. features: typifies the property of inertia, which has its systems and which acts as a repository of in diverse kinetic energy. A 2. spring represents a means of storing potential energy according to a particular law of force that has its counterparts in all kinds of physical interactions. We have already studied harmonic oscillator — of Newton's law, but in this problem — the problem of the some detail (in Chapter7) as an application well worth analyzing the problem again it is from the standpoint of energy conservation — in part as an illus- method but chiefly because the description in terms of energy opens the way to a far wider range of situations. Not only does it provide a pattern for the handling of more complex tration of this problems in oscillatory classical mechanics; it also supplies the foundation for formulating equivalent problems in quantum theory. Our starting point will again be the restoring force of ideal spring as described F = -kx where x is (10-16) the position of the free end of the spring relative to relaxed position, in N/m, and this the negative sign gives the direction of the force, 0- 387 - jo F(x) exerted Within the extension x. stored in the spring U(x) real spring can be exprcssed by a graph such as Fig. 10-9(a), which represents the force its No law over more than a limited range. The propertics of a real spring of its k the "force constant" of the spring, measurable opposite to the displacement of the free end. obeys an by Hooke's law: is, linear by the spring as a function range the potential energy according to Eq. (10-9), F{x)dx= + J kxdx = '^- Sonic cxamplcs of the energy melhod (10-17) Fig. 10-9 (a) Restor- ing force versus displacement for a spring. (b) Potential- energy diagram associated with (a). (c) Graph ofapplied force versus extension in a stalic deformalion ofa spring. where wc havc chosen U= relaxed. Figure 10-9(b) shows x. for x = 0, i.e., this potential when the spring is energy plotted against Since the potential-energy change can be calculated as the work done by a force itself, force FCKl just sufficient to overcome the spring the increase of potential energy in the spring for any given increase of extension can be obtained as the area, between given limits, under a graph of Fcxt against x [sec Fig. 10-9(c)]. F^t can be mcasurcd as the force needed to maintain the spring at constant extension for various values of x. Outsidc the linear region (whose boundaries arc indicated by dashed lines) F(x) can be integrated graphically so as to obtain numerically the potential 388 Energy conservalion in dynamics Fig. 10-10 Mechanical hysteresis. energy for an arbitrary displacement. we examine motion under Before consider briefly what would happen example were made of Compression spring from much very its this spring force, it let us the spring employcd in our As we compress such a lead, for example. original length, in the in Fig. 10-10. if will exert a force that behaves way shown in Fig. 10-9(a) and by the line OQ if we now remove the extcrnal agency, However, the spring will not return to its original length but will acquire a permanent deformation represented by point x r Clearly, such a spring exerts a force that is in Fig. 10-10. not conservative. The value of the force depends not only on the compression but also on the past history. At the value x of the compression in Fig. 10-10, we find two values of the force, one as the spring is x shortened and the other as is called hysteresis Let us now and it is This type of behavior released. results in dissipation consider in more detail the of mechanical energy. way in which the use of the energy diagram helps us to analyze the straight-line motion of energy U= \kx 2 shown the potential and two different total ener- In Fig. 10-11 are a harmonic oscillator. plotted against x, Ei and E 2 For a given energy E\, as we have already discussed, tne vertical distance from the horizontal line E x to the curve U = gies, %kx 2 Fig. 10-11 . for any value of a: is equal to the kinetic energy of the particle Energy diagram for a spring that obeys Hooke's law. 389 Somc examplcs of the energy method This at x. maximum is and the kinetic at x = 0; at this point all the energy maximum particle attains its The speed. is kinetic energy and hence the particle speed decreases for positions on either side of the equilibrium position O and reduced to zero is x = ±.A\. For values of x to the right of +A x or 2 to the Ieft of — A (Ei — U) becomes negative v is negative and there exists no real value of v. This is the region into which the particle never moves (at least in classical mechanics); thus the at the points ; i , positions x = ±A i are turning points of the motion, which is clearly oscillatory. The amplitude A energy E\. x = of the motion i Since the kinetic energy is determined by the total — Ki (=£1 U) is zero at ±Au we have = $kAi2 Ei or ,1/2 - - (10-18) (fJ For a larger energy, E 2 the amplitude , larger in the ratio is (E2/E1) " 2 , but the qualitative features of the motion are the same. It is interesting to note that the general character of the motion as inferred from the energy diagram would be the same for any potential-energy curve that has a minimum at x = and symmetrical about the vertical axis through this point. tions of this sort are periodic but differ All is mo- one from the other in the dependence of speed on position and the de- detail, e.g., pendence of the period on amplitude. Suppose that the period of the motion is Then T. we can imagine for any symmetrical potential-energy dia- up into four equal portions, any one of which contains the essential information about the motion. For suppose that, at t = 0, the particle is traveling gram, this time divided x = through the point particle will at its and u have during maximum = 0. It its — motion. it is At positive displacement (x then retraces its x the positive in velocity at this instant be called v m steps, / = T/4 = (x 390 = —A) and T/4 at / it goes to = T Energy conservation is its its the particle is +A in Fig. 10-11), = after a further rcaching* time T/4 and passing through this point with v further intervals Let direction. the biggest velocity the = — vm . In two extreme negative displacement once again passing through the point in dynarnics Fig. 10-12 Sinus- oidal variation velocity with of timefor a particle subjected to a restoring force proportional to displacement. The motion during the first quarter-period suffices of to define the rest the curve. x = O with v = v m This sequence will repeat itself indefinitely. Furthermore, knowing the symmetry of the problem, we could construct the complete graph of v against t from a detailed graph . for the first quarter-period alone (see Fig. 10-12, in which the basic quarter period is drawn with a heavier line than the rest). we shall go beyond this rather general In the next section examination of the mass-spring system, including nonlinear restoring forces, that, and shall redevelop the rest of the detailed apply to the ideal harmonic oscillator. Before doing results that however, we shall consider one more simple example that illustrates the usefulness of the energy method. Dynamics of a catapult The catapult is an ancient and effective device for launching stones or other missiles with quite high velocities, by converting the potential energy of a stretched elastic cord into the kinetic energy of a mass. I n Fig. 10-13(a) we show the essentials of the arrange- Fig. 10-13 (a) Initial and final f the stages o launching of an object by a catapult, (b) In- termediale state, showing the instantaneous force acting. 391 Somc examples of the energy method An ment. of natural (unstretched) length 2/ elastic cord, attached to fixed supports at the points An / ). mass object of m is and drawn back to point travels along the line flight with speed A and D (A C , is = CD = placed at the midpoint of the cord B. BC, and When it is released, the object at the point C it begins its free o. Let us assume that the cord develops a tension proportional to its increase of length. Then we can use Eq. (10-17) when the stored energy the mass is to calculate at the initial position, B.- Remembering that there arc two segments of cord, each of initial and stretched length l we have / length x , U= X 2 i*(/i - /o) 2 = Wi - 2 /o) In the idealization that the cord has negligible mass and thus does not drain off any of the elastic energy for we can its own motion, equate the final kinetic energy of the projectile to the initial potential \mv* = energy as given above. Thus - /o) 2 (/i - /c(/i we have and so ,1/2 °-(S) /o) This example displays particularly well the advantage of making the calculation with the help of the scalar quantity, energy, instead of the vector quantity, force. calculate the final velocity of the F = For suppose we wanted mass by to direct application of we should have to when the mass = I. The tensuch that AP was at some point P between B and C / ), and the instansion in each half of the cord would be k(l ma. Then, as indicated in Fig. 10-1 3(b), consider the state of affairs at an arbitrary instant taneous acceleration would be obtained by resolving these BC: tension forces along the line m^- = Fx = 2k(l - /o) cos d at This equation for do/dt would then have to be integrated B and C. Or, alternatively, we could calculate work done by the varying force Fx along the path BC. In between the points the total either case, this is a far harder road to the final result than direct application of conservation of mechanical energy. 392 Energy conservation in dynamics is the THE HARMONIC OSCILLATOR BY THE ENERGY METHOD We now shall return to the analysis of the oscillatory motion of an object attached to a spring that obeys Hooke's law. The mass on a spring with a restoring basic energy equation for a forcc proportional to displaccmcnt + £/»w 2 where E is is E \kx 2 = (10-19) some constant value of the Since v total energy. = dx/dt, this can be rcwritten as hm + (4j?\ = E £**2 (!0-20) Equation (10-19) already gives us v as a function of have a full we must description of the motion x (and hence so as to obtain Our way v) as functions of r. of dealing with Eq. (10-20) will appeal to the knowledge of trigonometric functions and one develops x, but to solve Eq. (10-20) at their derivatives that an early stage of any calculus course. We start out by dividing the equation throughout by E. Then we get ni (dx\ 2 We k_ + 2E\dt/ 2E notice that this x2 is a = (10-21) 1 sum of two terms involving the square of a variable (x) and the square of The sum is equal to 1 . its derivative (with respect to Now we can t). relate this to a very familiar relationship involving trigonometric functions: If 5 = sin •/>, then ds and W + s2 = cos2 <p + sin2 p = (10-22) 1 Equations (10-21) and (10-22) are exactly similar must be able to 2E\dt) in form! match them, term by term: W/ 2£ The second of 393 these The harmonic is satisfied by putting oscillator by the energy meihod We X = What '">-*> (t)""' - (t)"'*** We can is ^s? find it by evaluating dx/dt by differentiation of both sides of Eq. (10-24) with respect to = /2EV dx dip cos„- \k) dt But the equation of (10-23) first * m (2g\V ** \m/ dt Comparing d(f> these condition on that by putting U* m (2£\ \m) QM two expressions for dx/dt we Y" is is satisfied find the following <p: = (* J dt \m/ where w t: - u (10-25) the angular velocity (also called the circular frequency) we met in problem (Ch. our previous solution of the harmonic oscillator 7, p. 226), starting from F= ma. Integrating the last equation with respect to (p where = wt <po is + t we thus get ipo the initial Substituting this expression for phase. <p back into Eq. (10-24) then gives us x = fejtj sin (co/ + (10-26) w>) we note that (2E/k) u2 is equal to the amplitude A of the motion [Eq. (10-18)] we arrive finally at the same equation for x(t) that we found in Chapter 7 [Eq. (7-42)]. [If you have some prior knowledge of differential equations, you may regard our method of solution above as being rather If cumbersome. You may prefer to proceed at once to the recognition that Eq. (10-20) leads to the relationship (I) f)' and hence to the following solution by direct integration: 2.1/2 ,2 X) di'"* 4 ~ dx 394 Energy conscrvation in clynamics dx ,dt ut (4» + (po x2)I/2 -1 = = A and so x - sin i sin (ut -J + as before.] ) <p Equations (10-25) and (10-26) tell us something very remarkable indeed: The period of a harmonic oscillalor, as typified mass on a spring, is lude ofthe motion — a result that is not true of periodic oscillations under any other force law. The physical consequences of We depend this are heavily on the use of vibrating systems. If the frequency v (defined as the number of complete tremendously important. oscillations per second, i.e., l/7"or u/2v) varied significantly with would become the amplitude for a given system, the situation vastly more complicated. Yet most some approximation, as harmonic described above. SMALL OSCILLATIONS IN — but if in which an object It is at rest at it to its in is what we — under some point displaced in any direction tending to return it oscillators with properties as GENERAL stable equi1ibrium. force vibrating systems behave, to Let us see why. There are many situations it original position. some experiences a force Such a negligible displacement) will shown marked as x variation with position resting position force, unless is in . Fig. 10-14 offorce with 10-14(a). The normal This force function can be inte- (a) Variadis- placement on either side of the equilibrium position in a one- dimensional system. (o) Potential-energy curve associated with (a). 395 Small oscillations in have the kind of grated to give the potential-cnergy graph of Fig. 10-14(b). tion call no net has pathological properties (such as a discontinuous jump value for Fig. by a complelely independen! ofthe energy or ampli- in general One : UU) Fig. 10-15 Potential-energy curve ofFig. 10-14(b) referred to cm origin located at the equilibrium position. can then form a mental picture of the object as were, of the potential-energy hollow, the it bottom, sitting at the minimum of which x = x Now wc can fit any curve with a polynomial expansion. Let us do this with the potential-energy funetion but let us do it by putting with reference to a new origin chosen at the point x is at . — , + x = xo s from equilibrium. 5 is the displacement where energy curve, now appearing The potential- as in Fig. 10-15, can be fitted by the following expansion + as + = U U(s) ic 2 s + 2 + %c 3 s 3 (10-27) (The numerical faetors are inserted for a reason that will appear almost immediately.) The foree as a funetion of s m = _ is obtained from the general relation so that _ we have = —Cl — F(s) However, by Now, whatever will { = — C2S — F(s) css = = 0, at 5 = 0; this is and so our equation for the equilibrium F becomes (10-28) C3S 2 the relative values of the constants c 2 less than the term in Czs/ci, which can be A small enough. all and for the ratio of the two as small as we c 3 , there it = will 0) be we can be some very is in s 2 is equal to please by choosing s similar argument applies, even the higher terms in the expansion. having c 2 tions 5, made tential-energy funetion has more strongly, Hence, unless our po- special properties (such as sure that for sufficicntly small oscilla- just like the potential-energy funetion of a spring that obeys Hookc's law. 396 2 always be a rangc of values of s for which the term much to — definition, F(s) Hence c position. C2S We can write Energy conservation in dynamics U(s) m %c 2s 2 (• 0-29) which means that the effective spring constant equal to the constant c 2 of We . A; for the motion is shall discuss a specific application — molecular vibration — later in this chapter. this analysis THE LINEAR OSCILLATOR AS A TW0-B0DY PROBLEM So far, in all our discussions of potential energy, we have analyzed we had a single object exposed to given the problems as though forces. A statement of our calculation on an object near the earth's surface could well be in the form: The polen tial energy ofan object of mass the m raised to a height h above the earth's surface is a statement is perfectly legitimate for situations in Such mass of a particle is very small object (or objects) with which of mass of the system A the larger mass. mass is its it effectively mass of the the such a case, the center determined by the position of frame of reference anchored to this larger both a zero-momentum frame and a fixed frame of ence. This near is compared to interacts. In is the case for the earth surface. any two objects It if is mgh. which refer- and an ordinary object moving also the case for interactions between one of them is rigidly attached to the earth. one is analyzing a two-body system (the earth and the object which is One must remember, however, raised): mgh is that, strictly speaking, the increase of potential energy of the system the separation between the earth increased by an amount h. I and the object of mass when m n other words, the potential energy a property of the two objects jointly ; one or the other individually. If it is is cannot be associated with one has two interacting particles of comparable mass, both will accelerate and gain or lose kinetic energy as a result of the interaction between them. basic now It is to this two-body aspect of the potential-energy problem that we turn. Suppose that we have two particles, of connected by a spring of negligible mass axis (Fig. 10-16). masses m, and m2 aligned parallel to the , x Let the particles be at positions x t and x 2 as , shown, referred to some origin O. If the spring is effectively massless, the forces on it at its two ends must be equal and opposite (otherwise it would have infinite acceleration) and hence, accepting the equality of action and reaction in the contacts between the masses and the spring, the forces exerted on the masses are also equal and oppositc. 397 The linear oscillator as Thus, denoting the force a two-body problem c Fig. 10-16 System m "h f« 2 F, 2 Wom 700 of two masses con- /i/l/t nected by a spring, i showing ihe separale coordinates and O forces. exerted on mass 2 by the spring as mass 1 by the spring We equal to is shall relate the F x 2, — F 12 the force F2 i exerted on . changes in kinetic energy of the masses in stored potential to the changes -Vs .i •«'i energy in the spring. The potential energy of the spring First, suppose that The work done by dW = m moves a i the spring Fi2dx2 + F2 1 dx\ = Fi2(dx2 — dx\) - Fi 2 d(,X2 - Clearly the difference distance dx\ while is m 2 moves dx 2 - given by (sinceF2i = — Fia) Xl) x2 — x x , rather than x x and x 2 separately, defines the elongation of the spring (and hence the energy stored in it). r Let us introduce a special coordinate, = X2 — r, to denote this: x\ Then dW = F 12 dr The change of (work done by spring) (10-30) potential energy of the spring is equal to dU = (r)= - —dW. we have Introducing the potential energy function U(r) F12 dr J F12 dr (10-31) The kinetic energy of the masses Our discussion of that we should introduce refer the as 398 two-body systems in Chapter 9 suggests clearly the ccntcr of mass of the system and it. This allows us, in the CM frame with- motions of the individual masses to we have seen, to consider the Energy conscrvalion in dynamics dynamics By Eq. out reference to the motion of the system as a whole. we have (9-29) K = K' where K' v[ two masses as measured frame. Denoting the velocities relative to the CM by and u'2 as usual, we have = imivf K' We K' the total kinetic energy of the is CM in the £A/£5 2 + + \mv£ (10-32) have seen (pp. 338-339) that it is very convenient to express terms of the relative velocity, v„ and the reduced mass, in of the two Vr = — V2 f'l m\m2 M = From the definition of the mio'i Using m , y., particles: + this, CM (zero-momentum) frame, we have = /«2^2 together with the equation for v T , w»2 = + , v,T U ; ; m\ f 02 2 find mi = n\2 we "'i + "r "i2 Substituting these values into Eq. (10-32) one arrives once again at the result expressed by Eq. (9-30a): r.JJHL«l-i 2mi + P»r* 00-33) shall be considering the changes of kinetic energy of the n\2 We masses as related to the work done on them by the spring. the assumption that const., in no external forces are acting, On we have B = which case dK = dK' The motions At this point we can assemble the foregoing results and equate the change of kinetic energy to the evaluated dW [in although, as only 399 The work done by the we saw (and could have onxa — Xi, spring. We Eq. (10-30)] in terms of laboratory coordinates, which is predicted), the result equal to x'2 linear oscillator as a — x[ depends — both being equal two-body problem to the relative coordinate dK = We dK'. we have just seen, can, in fact, put = F\zdr dK' Likewise, as r. (work done by spring) Integrating, K' = Fl2 dr + / And now, const. with the help of Eq. (10-31), we can write this as a statement of the total mechanical energy E' in the K' + For the U(r) = CM frame: E' (10-34) specific case of a spring of spring constant k and natural length r , we can + r = r U(s) = $ks2 put s Also dr ds Thus the equation of conservation of energy [Eq. (10-34)] becomes J* (f) +ito8-i? 00-35) where M = This is mim-2 mi + m2 exactly of the angular frequcncy -It is ($)"* form of the w and r- its period linear oscillator equation; T are given by 2 (, *(?r to be noted that the reduced mass /x is less one of the masses is °- 36) than either of the individual masses, so that for a given spring the period free oscillation than if its clamped is shorter i n tight. COLLISION PROCESSES INVOLVING ENERGY STORAGE With the help of 400 the analysis developed in the last section, Energy conservation in dynamics we can gain further insight into certain inelastic or explosive collisions of the type discussed in Chapter We 9. problem by imagining a shall introduce the chanical gadget that could be constructed without little much me- trouble. The gadget is a spring equipped with a buffer that slides along a guide and becomes lockcd iri place if the spring is compressed by more than a object of For amount certain m mass 2 , and that an object of m simplicity, let us take mi approach If Mi will is it Suppose that 10-17(a)]. [see Fig. (assumed to be of negligible mass) this device mass is m t attached to an collides with to be initially stationary, 2 and it. let with a speed «i. small, the collision be compressed a is The spring perfectly elastic. when m^ little strikes it, but it will return and at this instant the colThroughout the time that the to its original unstretched condition, comes to an end. lision process spring is at all compressed, the mass and tive force to the right m i is m 2 is subjected to an accelera- subjected to a decelerative force amount of work is done onm 2 and an equal amount of negative work is done on m\. The total kinetic energy is the same after the collision as beforehand, but to the it left. A certain positive has been reapportioned. If ui the is increased, the situation maximum compression At locking position. to one another, is finally mi and this instant and because the spring them apart again, reached in which of the spring just brings is m 2 +m 2 ; the way they remain. In other words, move on as a single composite object of collision has suddenly become completely this is the inelastic. Fig. 10-17 (a) Collision energy-storage device. inooking one object with an (b) Same collision of-mass frame. The collision is elastic if netic energy in this frame is less than the to compress the spring 401 to the prevented from pushing they would continue to mass mi it are at rest relative by the crilical in the center- the total ki- work needed amount. Collision processes involving energy storage What changes from frame the masses have a CM collision We can most easily discuss this elastic to inelastic? from the standpoint of the section) which the defines the critical condition at frame total kinetic 10-17(b)]. [Fig. energy given (cf. In this the previous by 2 K' = ivu>T with = u - —— + m\ vT we denote the initial as Ku we can put If mi = — "i = (because uz 0) n\2 + kinetic energy of m Lab frame in the x m2 at the critical value of K x all the energy K' is used up in compressing the spring to the Iocking position; this requires a Now amount of work equal to the energy, spring in this configuration. Thus we can put well-defined the K' m 2 = K, = U , stored in f/o or + m2 _ mi (10_ 37) Uq n\2 If U is given, the above equation defines at which inelastic collisions become a threshold value ofKu and below which possible the collisions can only be elastic. If in we consider then returns to that reextension velocity it and a pushes is compressed beyond critical length m Ku higher values of still which the spring i and m and we stops. 2 apart, giving kinetic energy of relative obtain situations Iocking point but its During them a motion this partial final relative in the CM frame. The collision is still inelastic but only partially so. A well-defined amount (C/ ) of the originally kinetic energy of the colliding masses has been locked up in the spring, collision can be analyzed in these terms. and the dynamics of the Our of inelastic and explosive collisions was, analysis in Chapter 9 in fact, made precisely K way. The quantity Q, representing the value of K/ that we introduced there (p. 347) is in this case simply equal to in this -t/„- 402 Energy conservation jn dynamics — purely classical dynamical situation described above The is by many collision processes in atomic and in which an incident particle (e.g., an electron closely paralleled nuclear physics, or a proton) strikes a stationary target particle that has various sharply defined states of higher internal energy than "ground The sharpness and state." understood teristic states is that is we need here is in normal its discreteness of these charac- terms of quantum theory, but the knowledge of their existence. If all a study the distribution of kinetic energies of the electrons or made of protons after they have collided with target particles of a given about the excited energy species, the results give information Some examples of this levels involved. kind of analysis are shown The higher the bombarding energy of the incident particle, the larger the number of excited states that can be stimFigure 10-18(a) shows data ulated and detected in this way. from the scattering of electrons by helium atoms, and Fig. 10— 18(b) shows some results for the scattering of protons by the in Fig. 10-18. nucleus boron 10. massive that the available kinetic energy insignificantly different and one must use Eq. (10-37) to effect so (K') is that perfectly elastic collisions may In the considerable energy U difference between and the these atomic or nuclear scattering processes is is infer the excitation from the threshold value of Ki> One important ones CM the in from the electron energy K\. however, the center-of-mass case, latter is In the former case the target particle still classical occur, with a certain probability, even after the threshold energy for inelastic processes has been exceeded. feature of This an example of the general is quantum mechanics that one only has relative probawhen several outcomes are possible. of events bilities return briefly to the classical situations once again, one To can of course imagine mechanisms that would provide for an increase rather than a decrease of kinetic energy as the result of a One collision. could, for example, have a spring already pressed that would be released by the initial impact. if com- a trip mechanism were activated — an explosive collision Such a process would, like the inelastic collision with energy storage, require a certain minimum comparable to threshold energy to this require the firing of a detonator. It experiment toy pistol) in is make go. Something very is possible to do a quantitative which a simple percussion cap (as used mounted on a mass that is free nated by the impact of an incident object 403 it happens with chemical explosive systems that to recoil, (e.g., in a child's and on an Collision processes involving energy storage is deto- air track). 180 190 Scattered electron energy, e' a) ">B<0.72 MeV) Elastic scattering peak • ">B(0) w B*(2.15MeV) •C(0) •"B*(1.74 MeV) A 250 Relative proton momentum, ~ 300 nJll .350 units of Br (b) Fig. Experimental resulls on elastic and inelastic collision proshowing Ihe production of characteristic excited states ofsharply 10-18 cesses, defined energy: (a) Scattering of electrons by neutral helium atoms. [AfterL. C. Van Alta, Phys. Rev., 38, 876 (1931).] (b) Scattering of Some carbon was also present as an protons by nuclei of boron 10. unavoidable contaminant. [After C. K. Bockelman, C. P. Browne, W. W. Buechner, and A. Sperdmo, Phys. Rcv., 92, 665 (1953).] 404 Energy conservation in dynainics kinetic energy required under these conditions The threshold significantly greater than if the cap is mounted on an unyielding is support. THE DIATOMIC MOLECULE 1 The diatomic molecule, as or Cl 2 , typified by such molecules as HC1 a system to which the methods developed in this chapter is can be very effectively applied. In such molecules, the atomic nuclei play the role of point masses (which of course they are to an extremely good approximation, being so tiny in comparison to an atomic diameter) and the outer clectron structure of the atoms plays the role of a spring system that can store potential energy. We thus have the physical basis for characteristic vibrations of the nuclei along the line joining them, possible mode of internal the oscillations as such, and this will be a motion of the molecule. Before discussing we must considcr the shape and energy scale of the potential energy curve for a molecule of this type. We there is begin with the knowledge that in a diatomic molecule a fairly well defined distance betwecn the nuclei of the two atoms. order of This is called the angstrom (10 1 -10 bond length and m). The fact that is always of the such an equilibrium distance exists implies that the potential-energy function U(r) minimum at the separation r equal to the bond length. One way of defining such a potential, while still leaving plenty of room for empirical adjustment of the constants to experihas a mental data, U(r) = is to assume the following potential-energy function: d_ * where A and B (10-38) are positivc constants, and a and b are suitably chosen positive exponents with a positive potential energy that falls > b. away distance and represents a rcpulsive force. The term A/r" is a rapidly with increasing The term - B/r6 is a negative potential energy, also falling off with increasing r but less rapidly than the first term, and it represents an attractive force. These funetions, and the U(r) curve obtained by adding them, are shown Now The but it in Fig. 10-19. from Eq. (10-38) we can define the equilibrium remainder of this chapter can be omitled without loss of continuity, does introduce some quile interesting fcatures in addition to the atomic physics as such. 405 dis- The diatomic molecule Fig. 10-19 Simple model oflhe potentialenergy diagram ofa dialomic molecule, defining an eguilibrium separation ro of the nuclei. tance; At r it is the distance at which dU aA dr f<>+\ = r , bB ro- r 0: bB ' (10-39) fi + 1 (10-40) " experimentally known, this defines one connection between the parameters A, B, energy = we have aA If r o is dU/dr is a, and b. Also, at r = r , the potential given by and substituting for A/r a from Eq. (10-40), this gives ™--£(«-l) The energy U(r ) is a measurable quantity, for dissociate the molecule completely we must add energy, the dissociation energy D, equal (cf. Fig. 10-19). 406 It is typically a few eV. Energy conservaiion in dynamics if we are to a quantity of and opposite to U(r Thus we can put ) 3H) (10-41) Equations (10-40) and (10-41) can be used to narrow the choice of parameters — we have we must Clearly, however, unknowns. further, independent information, or tions, or both, to down two equations and four either appeal to make some specific some assump- have a quantitatively defined situation. The molecular spring constant The quantity we really need to know, sidering molecular vibrations, is for the purpose of con- the effective spring constant of the molecule for small displacements from equilibrium. We discussed this situation in general terms earlier in this chapter and arrived (p. 395), « t/(s) at the result [Eq. (10-29)] \C2S 2 where s = — r The constant ro can be c2 identified with spring constant be deduced from the given form of k and can U (r): *-(3L-(3)L Applying sides of this to the present problem, we shall differentiate both Eq. (10-39) with respect 2 d U _ a(a+\)A dr2 ~ r o+2 + b(b to r: \)B UW"«J r 6+2 This Iooks complicated, but for the particular value r we can reduce it to 2 (d V\ = = r one term with the help of Eq. (10-40): a+ 1 bB _ 6(AJ+ \)B Therefore, This 407 still looks rather forbidding, but The diatomic molecule if you compare Eq. (10-44) with Eq. (10-41) you will see that the equation for the dissociation energy taking D contains a very similar D as known we can greatly value of k, as follows. (a - b)B From combination of factors, and simplify our statement of the Eq. (10-41) we have = aD ro" Substituting this result in Eq. (10-44) k D we find = abD (10-45a) ro 2 Beyond this point we cannot go, even with a knowledge of and r without assigning a value to the product ab which can , a single adjustable parameter, so that be taken to play the role of we content oursclves with a final semiempirical equation: k = C- where C is Fig. 10-20 (10-45b) r 2 an empirical constant (=ab) greater than unity. Empirical potential-energy diagram for the molecule HCI, based on the Morse potential, with energies expressed in 408 wave numbers (cm~ '). Energy conscrvaiion in dynamics Actually a quite different analytic form of the potential function finds favor with spectroscopists. Morse M. potential (after P. features as the one Morse). we have used It It known is (see Fig. 10-20) but difference of exponentials rather than of simple Like the one we have used, it as the has the same general is based on a powers of r. has adjustable constants that can be Indeed, once the theory of deduced from spectroscopic data. molecular vibrations has been one uses observations to set up, feed back the numerical values into the theoretical formulas. a two-way It is traffic, in other -1 (cm ) on the ordinate of back to them shortly. units Notice the unfamiliar words. Fig. 10-20; we shall be coming The molecular vibrations The period of vibration of a diatomic molecule with atoms of masses /«i and m 2 is given by combining Eqs. (10-36) and (10-45b): T=2xr (&)' (-£-) where/i = (10-46) reduced mass in vibrations = m\m%l(jn\ + m2 ). The frequency v, per second, proves to be a more interesting quantity: -U¥T Let us see what this equation might suggest with an actual mole- We cule. shall take carbon monoxide (CO) for which the following data apply: /m( l2 C) = 12amu = m2( lc O) = 16amu = 2.7 A- 1.1 X lO-^m r « D« First 10- 2C kg -26 10 kg X X 1.1 lOeV = 2.0 1.6 X K) -18 1 ^ we have _ _m1 m3_ /Mi +- M»2 _ 12 ô x 2? x _ ]0 20kg _ j 16 x _2 o 1() kg 'These values come from a tabulation of molecular constants in the Smithsonian Physical Tables (published by the Smithsonian Institution, Washington, D.C.) 409 Tlic diatomic molecule Therefore, DI « 1.4 y. X 10 8 m 2 /sec 2 whence j- If « X lO^C'^sec- 1.7 we ignored C the factor *1.7X 1' 1 2 we would have , 10 13 sec-' Since molecular vibrations are studied mainly through spectroscopy and the measurement of wavelengths of absorbed or emitted radiation, let us calculate the wavelength X corresponding to v: The 7 visible X 10 -7 spectrum extends from about 4.5 That is mental vibrational line of We -7 m to we have calculated would be well exactly where we find the spectral lines associated with molecular vibrations. -6 10 m, so the wavelength into the infrared. 10 X CO Actually there is a funda- with a wavelength of about 4.7 X m— about one quarter of the value we have just calculated. would have obtained This value of Eq. (10-47). having a = 5 and b = this C value by putting C" 2 = 4 in could correspond, for example, to 3 in the original expression for U(r) This specific form of potential would result from repulsive force varying as l/r°, and a longer-range a short-range 4 We should emphasize, however, attractive force varying as l/r [Eq. (10-38)]. . that our model of the potential is a very crude one and should not be regarded as a source of precise information about intramolecular forces. Our purpose in introducing it is simply to illustrate the general way in which vibrations about an equilibrium configuration can be analyzed. There with is it, is a feature is this is that —a —that we vital feature The energy of a quantum of the radiation Planck's constant) and is provided by a transition quantized. hv (where h is between two sharply defined energy hv have ignored; molecular vibration, and the radiation associated states of the molecule: = Ex - E2 (10-48) We shall not go into this here, except to point out that the smallest possible 410 jump between successive vibrational energy levels does, Energy conservation in dynamics as happens, correspond exactly to the classical oscillator it quency we have calculated. It is the basic process, as expressed in Eq. (10-48), that quantum nature of the makes the frequency v rather than the wavelength X the important quantity. however, spectroscopists measure wavelengths and since - 1 simply. custom of To convert this to an equiva- measurement we must use the lent energy Since, in the first instance, v is inversely proportional to X, there is a giving results in terms of X fre- relation, applying further, and used any photon, that to E= hv = hc\~ l This scheme of units has been extended still spectroscopy as a measure of energies generally, whether or in not photon emission D the dissociation energy D= cm- = 1 36,300 Using the above J-sec, we have D= For example, involved. is HC1 of 3.63 relation, is 10 6 X and in Fig. 10-20, given as m" 1 substituting h 6.62 X 10- 34 in this case X (6.62 X 10 8 )(3.63 X 10°) = 0, 10- "J = 7.2 = 4.5 A particle X 10" 34 )(3 eV (approx.) PROBLEMS — 10 / F whose value at becoming zero at of mass m, at / = t = rest at 10-2 What T. is An object of mass 5 kg is subjected to a force linearly with time, the kinetic energy of the particle acted on by a force that varies with the position of the object as shown. at the point x = 50 1 is = r? at t 41 t Fq and which decreases is m? Problems x = 0, what is its If the object starts out speed (a) at x = from rest 25 m, and (b) at 10-3 (a) F force A particle of mass m, which increases relationship between F and acted upon by a initially at rest, is F= linearly with time: Deduce Cl. the and graph your the particle's position x, result. (b) How initial velocity 10-4 is the graph of F x altered if the particle has made use of the "mechanical versus uo? (a) Since antiquity man has advantage" of simple machines, defined as the ratio of the load that is to be raised to the applied force that can raise work-energy relation dW = F • t/s, it. and appealing By analyzing the to conservation of energy, give a reasoned definition of "mechanical advantage." (b) In terms of your definition, calculate the particular "mechanical advantage" offered by each of the devices shown. Radius b 10-5 (a) The Stanford linear accelerator ("SLAC") of approximately 200 to electrons at the rate IceV/ft. delivers energy How does this compare to the energy per unit length imparted to electrons by a cathode-ray tube? By a television tube? (b) The effects of relativity are extremely pronounced in this examplc that is (after 2 miles of travel, the electrons in _7 within 10 % that of light). How SLAC attain a far velocity would the electrons travel in attaining the speed of light if their kinetic energy were given by the classical value 10-6 A while it railroad car travels which the coal 412 KE = is £mo 2 ? loaded with 20 tons of coal in a time of 2 sec through a distance of 10 is m dischargcd. Energy conservation in dynamics beneath a hopper from (a) What average this loading process to extra force must be applied to the car during keep (b) How much work (c) What moving it 10-7 A car between (b) and (c). being driven along a straight road at constant speed is passenger in the car hurls a ball straight ahead so that his hand with (a) is Of on what A common at a given rate answers in the road? (a) to the Satisfy yourself that car. forces are acting 10-8 objects, over work done by the passenger you understand exactly what what distances. device for measuring the power output of an engine of revolution is known as an absorption dynamometer. small friction brake, called a Prony brake, is clamped to the output shaft of the engine (as shown), allowing the shaft to rotate, held in position by a spring scale a (a) v. leaves the gain of energy of the ball as measured in the (b) Relate the and by the it a speed u relatiue to him. What reference frame of the car? A constant speed? the increase in kinetic energy of the coal? is (d) Explain the discrepancy A at does this force perform? known distance R and is away. Derive an expression for the horsepower of the engine in terms of the quantities R, F and 03 arm R as the torque exerted (the force recordcd at the spring scale), (the angular velocity of the shaft). (b) You will recognize the product of the force F times its lever by the engine. Does your relation between torque and horsepower agree with the data published for automobile engines? Explain 10-9 An why electric measures 30 by 20 pump ft may there and is be discrepancies. used to empty a flooded basemcnt which is collected water to a depth of In a rainstorm, the basement 15 ft high. 4 ft. (a) Find the work necessary to (b) Supposing that the pump the water out to ground level. a 50% efficiency, how (c) If the of 15 ft, Problems driven by a 1-hp motor with depth of basement below ground werc 50 how much work would a 4-ft flood? 413 pump was Iong did the operation take? ft instead be required of the pump, again for Look up the current world records for shot put, discus, and javelin. The masses of these objects are 7.15, 2.0, and 0.8 kg, respectively. Ignoring air resistance, calculate the minimum possible 10-10 (a) kinetic energy imparted to each of these objects to achieve these record throws. (b) What force, exertcd over a distance of 2 m, would be required to impart these energies? (c) Do you think that the answers imply that air resistance imposes a serious limitation in any of these events? 10-11 A down an ascending escalator, so as same vertical level. Does the motor driving do more work than if the man were not there? perverse traveler walks to remain always at the the escalator have to Analyze the dynamics of this situation as 10-12 The Great Pyramid of Gizeh when lost a certain amount of its length 230 block of stone of density about 2500 What erected (it has since m. kg/m 3 m high It is effectively a solid . gravitational potential energy of the the total is first outermost layer) was about 150 and had a square base of edge (a) you can. fully as pyramid, taking as zero the potential energy of the stone at ground level? (b) Assume that a slave employed in the construction of the pyramid had a food intake of about 1500Cal/day and that about 10% of this energy was available as useful work. How many man-days would have been required, at a minimum, to construct the pyramid? (The Greek historian Herodotus reported that the job involved 100,000 10-13 men and It is took 20 years. If so, it was not very claimed that a rocket would less chute model in it had been allowed to —see the figure. To slide a greater height if, were ignited at a lower rise to instead of being ignited at ground level (A), level (B) after efficient.) it from rest down a friction- analyze this claim, consider a simplified which the body of the rocket is represented by a mass M, the fuel is represented by a mass m, and the chemical energy released in the burning of the fuel is represented by a compressed spring between M and m which stores a sufficient to eject 414 Energy m definite amount of potential suddenly with a velocity corrservation. in dynamics V relative to energy, U, M. (This — corresponds to instantaneous burning and ejection of i.e., the fuel Assuming a value of g independent of height, calculate rise if fired directly upward from rest at A. Let B be at a distance h vertically lower than A, and suppose (a) how all an explosion.) Then proceed as follows: high the rocket would (b) that the rocket What is fired at is B after sliding the velocity of the rocket at leased? just after the spring To what (c) is down the frictionless chute. just before the spring is re- released? A height above higher than the earlier case? B will the rocket rise now? Is this By how much? Remembering energy conservation, can you answer a claimed that someone had been cheated of some energy? (e) If you are ambitious, consider a more realistic case in which the ejection of the fuel is spread out over some appreciable time. Assume a constant rate of ejection during this time. (d) skeptic 10-14 who A neutral What vacuum. hydrogen atom is electron oolts at the number = 6 10 23 X 10-15 A from rest bottom? (1 eV = 1.6 X through 100 m in kinetic energy in its 10 -19 Avogadro's J. .) spring exerts a restoring force given by = -kiX- F(x) where x falls the order of magnitude of is k 2 x* the deviation from its unstretched length. The spring rests and a frictionless block of mass m and initial velocity v hits a spike on the end of the spring and sticks (see the figure). How far does the mass travel, after being impaled, before it comes to rest? (Assume that the mass of the spring is negligible.) on a frictionless surface, 10-16 A particle moves along the x is as shown in the function of position figure. sketch of the force F(x) as a function of curve. 415 Indicate Problcms on your graph Its potential axis. x Make energy as a a careful freehand for this potential-energy significant features and relationships. M and /0-77 A rically ovcr a frictionless horizontal peg of radius R. uniform rope of mass length L is draped symmetthcn dis- It is turbed slightly and begins to slide off the peg. Find the speed of the rope at the instant it leaves the peg completely. 10-18 Two (a) m, masses are connected by a massless spring as shown. Find the minimum downward force that must be exerted on m\ such that the entire assembly will barely leave the table this force is suddenly removed. (b) Consider problem this when in the time-reversed situation: Let the assembly bc supported abovc the table by supports attached tonil. Lower the system until m z barely touches the table and then release /M2 How the supports. knowledge of (c) tuilion by m\ drop far will this distance help Now that (1) letting you have mz you bcfore coming to a stop? solve the original the answer, check be zero and (2) letting it Does problem? against your mi be zero. in- Espccially common in the second case, does the theoretical answer agree with your sense? If not, discuss possible sources of error. 10-19 A particle moves in a region where the potential energy is given by = U(x) (a) 3x 2 - x3 What is the maximum is value of the total mechanical energy possible? In what range(s) of values of x in the positive 10-20 J) x. such that oscillatory motion (c) U in Sketch a freehand graph of the potential for both positive and negative values of (b) (x in m, x is the force on the particle direetion ? A highly elastic ball (e.g., a "Superball") is released from rest a distance // above the ground and bounces up and down. With each bounce a fraelion /of its kinetic energy just before the bounce is lost. Estimate the length of time the ball will continue to bounce and / = h = 5 m rtf. 10-21 The elastic cord of the catapult shown in the diagram has a its ends are attached to fixed supports a total relaxed length of 2/ ; distance 2b apart. 416 if Energy conservation in dynamics Show (a) that if the tcnsion developed in the cord tional to the increase in direction F» its length, the is propor- component of force in the x is = 2kb \-}—n - -r-sr \sin 6 sin cos J O/ Noting that the position of the stone in the catapult is cot 6, derive the expression for the work done in (b) x = —b given by moving a distance to find the total dx. Then integrate this expression between do and 9 work done in extending the catapult and compare with the result that is obtainable directly by considering the energy stored in the strctched cord (p. 392). (After carrying out the calculation that prescribed above, you will better appreciate the advantages way in the may come from the use of energy conservation instead of a work integral that involves the force explicitly.) A 10-22 which What the ball = (Take y tion of height y. (b) dropped upon r is ky. a graph of the potential energy of the ball as a func- Make (a) M and radius exerts a force proportional to the distance F= of deformation, when mass perfectly rigid ball of a deformable floor is as the undisturbed floor simply resting on the floor? is level.) the equilibrium position of the center of the ball (Note that minimum of the potential-energy curve.) increased By how much is the period of this corresponds to the M (c) over its period of bouncing on a perfectly rigid floor? 10-23 A at time f newtons x = at position acts m= of mass particle = 2 kg on a A 0. on the particle from / = to (a) Plot the resulting acceleration x as functions of t = Otot = (b) 10-24 A What t is frictionless table is at rest force of magnitude t = Fx = 4 a x , velocity o„ and position in this period. the total work done by the force during the period sec? 1 spring of negligible mass exerts a restoring force given F(x) sin(ir/) 2 sec. + = -kix by k 2x 2 (a) Calculate the potential energy stored in the spring for Take displacement x. (b) It is stored energy U = at x = found that the stored energy for x for x = +b. What is a 0. ki in terms of = —b Ari is twice the and bt (c) Sketch the potential-energy diagram for the spring as defined (d) The spring in (b). end fixed. A mass in the positive 417 Problems x lies on a smooth horizontal surface, with one the other end and sets out at x = 2 with kinetic energy equal to k ib' /2. How m is attached to direction fast is it moving (e) x = +6? at What x are the values of at the of oscillation? [Use your graph from part 10-25 An extreme ends of (he range (c) for this.] mass m, moving from the region of negative x, it experiences a x = with speed vo. For x > object of arrives at the point force given by = -ax 2 F(x) How far +x axis along the does it get? 10-26 The potential energy of a particle of mass m as a function of (The discontinuous jumps in the value of U are not physically realistic but may be assumed to approximate a real situation.) Calculate the period of one complete its position along the oscillation if x axis is as shown. the particle has a total mechanical energy E equal to 3l/o/2. 2t/„ — U 10-27 Consider an object of mass 'h, m constrained to travel on the x axis (perhaps by a frictionless guide wire or frictionless tracks), atlached to a spring of relaxed length /o and spring constant k which has its \ other end fixed at i. (a) Show x = 0, y (see the figure). / that the force exerted on m in the x direction is o ~- kx -( l+ $) [ l (b) For small displacements (x x 8 and hence « /o), show that the force is proportional to U{x) What is A ~ Ax* (x « /o) in terms of the above constants? The period of a simple harmonic oscillator is independent of its amplitude. How do you think the period of oscillation of the above motion will depend on the amplitude? (An energy diagram (c) may be helpful.) 10-28 Consider a particle of mass 418 Energy conservation in m moving along dynamics the x axis in a force field for which the potenlial energy of the particle given by is U = Ax 2 + Bx* (A > 0, B > 0). Draw the potential-energy curve and, arguing from the graph, determine something about the dependence of the period of oscillation T upon the amplitude xo- Show that, for amplitudes sufficiently small so that compared tude is to Ax i , Bx A is always very small the approximate dependence of period upon ampli- given by ,_ A (i-g„) A 10-29 between The of mass particle potential energy tangular hump is and energy shown, is slightly of height period of oscillation [It is m vertical walls as i.e., E is changed by introducing a AU (« E) and width Ax. U= 0. tiny rec- Show that the changed by approximately (m/2£3 ) 1/2 (Ai/AAr). worth noting that the effect of the small irregularity in potential energy depends simply on the product of individual values of these quantities. disturbing effects bouncing back and forth over a region where This AU is and Ax, not on the typical of such small —known technically as perturbations.] Ut - U= blocks of masses m and 2m rest on a frictionless horizontal They are connected by a spring of negligible mass, equilibrium L, and spring constant k. By means of a massless thread Two 10-30 table. length connecting the blocks the spring The whole system is held compressed at a length L/2. moving with speed y in a direction perpendicular spring. The thread is then burned. In terms of m, is to the length of the L, k, and o find (a) (b) (c) (d) How do The total mechanical energy of the system. The speed of the center of mass. The maximum relative speed of the two blocks. The period of vibration of the system. the quantities of parts (a) through (d) change velocity c is if the initial along, rather than perpendicular to, the length of the spring? 10-31 The mutual potential energy of a Li + ion and an function of their separation r 419 Problems is I - ion as a expressed fairly well by the equation __ 2 A = -Ke + __ £/(,) where the values of K= term arises from the Coulomb interaction, and the first constants in its X 9 . 10 The equilibrium Lil molecule 9 MKS units are N-m 2 /C 2 On How much work (a) 1.6 X KT 19 C distance ro between the centers of these ions in the about 2.4 A. is = e , the basis of this information, eV) must be done to tear these ions (in completely away from each other? ion to be fixed (because (b) Taking the I what is the frequency v small amplitude? 2 value of d U/dr 26 as 10kg.) 10-32 2 -7 , two at r = ro—see -' p. 407. Take the k as the mass of the Li + to the van der Waals attractive force, which with "W-7T + I M experience atoms of mass identical force proportional to r and graph your so massive), it is n Hz) of the Li + ion in vibrations of very (Calculate the effective spring constant (a) If in addition varies as r (i <" / > 7, >6 show a repulsive that > result versus r. (b) Calculate the equilibrium separation ro in this molecule in terms of the constants by requiring dU(r) = dr (c) — t/(r ). The dissociation energy What is its D of the molecule should be equal to value in terms of A, n, and r ? (d) Calculate the frequency of small vibrations of the about the equilibrium separation M, r . Show that it is the constant n, the equilibrium separation, molecule given by the mass and the dissociation energy of the molecule, as follows: „ \2nD 10-33 The potential energy of an ion in a crystal charged ions may lattice of alternatcly be written where A, B, and n are constants, e is the elementary charge, and r is the distance between closest neighbors in the lattice, called the lattice constant. 420 This potential arises from the l/r 2 force of electrostatic Energy conservation in dynamics attraction, together with a short-range repulsive force as two adjacent ions are brought close together. Make a graph (a) of U(r) versus what can you say about the correctly, r. Making stability sure to identify r of the crystal structure? (b) Calculate the lattice constant ro in terms of A, B, e, and n, from the equilibrium condition dU(r) dr (This defines the inter-ionic distance for which the energy of the lattice as a whole u is minimized.) Show (c) that the binding energy per No = = —NolKro) (d) 2.8 1.7 (dimensionless) if esu (e 4.8 X value for u and 421 crystal, A, and the exponent n = 10 23 mole X result of part (b), determine For NaCl Problems 10 -10 r is is esu). -1 u in terms of r . the equilibrium lattice spacing ro is about 10. expressed in Using compare with is Avogadro's number ~6 Using the mole of molecules The constant A centimeters and e is equal to is given in these values, calculate a theoretical the experimental value of 183 kcal/mole. A technique succeeds in mathematical physics, not by clever trick, or a happy accident, but because some aspect ofa physical o. G. it a expresses truth. sutton, Mathematics in Action (1954) 11 Conservative forces and motion in space EXTENDING THE CONCEPT OF CONSERVATIVE FORCES throughout chapter simplification motion in or we consistently applied one important restriction, by confining our discussion to 10 one dimension only. This clearly prevented us from studying some of the most interesting and important problems i n dynamics. In the course of the present chapter ourselves of this restriction method of To analysis to still and in the process begin the discussion, let us consider a problem in motion Suppose we have two very smooth tubes connecting two points, same shall free the energy greater advantage. under gravity near the earth's surface. levels in the we show A and B, A vertical plane (Fig. 11-1). at different small particle, placed in either of these tubes and released from rest at A, slides down and emerges If the at B. tubes are effectively frictionless, by them on the particle are always at right angles to the particle's motion. Hence these forces do no work the forces exerted Path2 Fig. 11-1 Alternatice paths between two gtoen points for motion in Path 1 a vertical plane. 423 on the particle may occur in its kinetic energy What these forces do achieve is to whatever changes ; cannot be ascribed to them. compel the at B particle to follow a particular path so that traveling in a designated direction. If it follows emerges with a velocity v t as shown; emerges with a velocity v 2 Of course . if it it emerges path 1, it follows path 2, it the energy of the particle it moves along either tube; the gravitational work on it. But we observe very interesting fact: doing a does change as force is Although the directions of the different, and velocities Vi given to the particle by the gravitational force paths beginning at How does A and ending this the particle travels work dW done on dW = where |F„| \ds\ it is same for all given by = cos 6 F„ ds • between the directions of F„ and a constant force is the is at B. come about? It is not difhcult to see. As through some clement of displacement ds the B is the angle the force F„ v 2 are quite The kinetic energy magnitudes are the same. their (i.e., But ds. the same at any position) with the following components: Fx F„= -mg = The element of (we take the y axis as positive upward). Now placement has components (dx, dy). tion of dW as dW = Fx dx + To cos 6 |F| \ds\ from the basic dis- defini- we have F„ dy any two vectors see this, consider A and B in the xy plane, making angles a and /3, respectively, with the x axis (Fig. 11-2). Then if the angle between them is 6 we have, by a Standard trigonometric theorem, that cos 6 The = cos(/3 — scalar product S = Fig. 11-2 |A| |B| (cos a) = cos /3 cos a 5 (= A B) • cos /3 a + is + sin/3sina thus given by sin/3sina) Basis for obtainiiiR the scalar product in terms of individual components, as in calculaling the work F • ds in A B • may be convenient an arbilrary displacement. 424 Conscrvative forces and molion in space But |A| cosa = A x and similarly for the • generally, sina if = A„ components of B. Thus we have + A u By A B = A X BX More |A| (two dimensions only) the vectors also have nonzero z components, A B = A X B X + A y B u + A,B, • Thus in general dW = In F • present the Fy = —mg, dW Hence less ds we shall = Fx dx have + Fydy + F, dz (11-1) two-dimensional problem, with Fx = and us this gives = —mgdy change of for a vertical coordinate from y\ to y 2, regard- of the change of x coordinate or of the particular path taken, we have a change of K2 — -K l kinetic energy given = /. JdW= -mg(y 2 - which exactly reproduces the for purely vertical by (11-2) yi ) result that we derived in Chapter 10 motion and which permits us again to define a gravitational potential energy U(y) equal to mgy. This result makes for a great extension of our energy methods we have to situations wherc, in addition to gravity, so-called "forces of constraint," which control the path of an object but, because they act always at right angles to to change its energy. Let us consider its some motion, do nothing specific examples. ACCELERATION OF TWO CONNECTED MASSES We problem which, if tacklcd by the dircct Newton's laws, requires us to write a statement of F = wa for each of two objects separately. The problem is to find the magnitude of the acceleration of two connected masses, mi and m 2 moving on smooth planes as shown in Fig. 11-3. shall begin with a application of , The 425 accelerations of the masses are different vectors, a! Acceleration of two connected masses and a 2 , Fig. 1 1-3 masses Motion ofiwo coimected —a simple example of the use of energy-conseruation methods. even though they have the same magnitude statement of F = ma will not Thus a a. single Using the scalar quantity suffice. we can exploit the fact that the magnitudes of the displacements and their time derivatives are common to both masses. Thus we have |vi| = [v 2 = v, so that the total kinetic energy of the system is given simply by energy, however, | K= J(mi +m and its Now u dt surface slope in a time dt is given change dK = = 2 2 )v + + («u {m\ mi)odv mi)oadt moved by each mass the distance is on which it (y positive upward), is and rides, means a change dy potential energy by for in vertical The m2 parallel to the the distance v dt coordinate equal to associated change dU down the —v dt sin d in gravitational thus dU = -m2gvdtsin6 But, given conservation of total mechanical energy, assuming friction to dK be absent, we have + dU - Thus (mi + m£)va whence the dt — m2go dt sin 6 = familiar result rri2 m\ OBJECT MOVING IN + gsintf /712 A VERTICAL CIRCLE Suppose that a particle P of mass m of a rod of negligible mass and length 426 is /, attached to one end the other end of which Conservative forces and motion in space ' / / I I \ \ \ Mo Fig. 11-4 t ion ofa simple s ^- pendulum. is pivoted freely at a fixed center C. 'r "~ is 1 1-4). =O O Then at the conveniently described in terms of the single angular coordinate 5 y Let us take an origin the normal resting position of the object (Fig. position of the object a ^ B, or, if along the circular arc of its we prefer, path from O by the displacement (assuming the rod to be of invariable length). angular displacement If the object is B, the y coordinate of the given by is y = /(1 and hence its = 1/(0) - cos 0) by potential energy - mgl(\ (11-3) cos 8) Using our basic energy-conservation statement we have Ki + K2 + U 2 = t/i any two points on the path. Substituting for and v and for U by Eq. (11-3), we have for + \mui 2 mgl(l - cosfli) = \mv 2 2 + mgl(\ K in terms of m - cos 02) Therefore, V2 2 = oi Clearly, if 82 2 + > 2gl(cosB 2 6 If the object velocity v , U then v 2 - < (11-4) cos 0,) V\, were started out at the lowest point with a we should have In a sense, therefore, the motion is one-dimensional, even though the one dimension is not a straight line. But we shall not stress this aspect of such a system at 427 this point. Object moving in a vertical circle 02* = P0* - With the help of What initial -COS0 2) 2^/(1 this result velocity of the circle? and: we can answer such questions as: needed for the object to reach the top is What position does it reach if started out with less than this velocity? Notice the great advantage that the energy method has over the direct use of F the object radial is = ma in this The problem. velocity of changing in both magnitude and direction, it and transverse components of acceleration, there unknown push on or puli the object from the rod is has an — yet none of these things need be considered in calculating the speed v at any Once v has been found by Eq. (1 1^) [or perhaps if that is more convenient], then one can proceed to deduce the acceleration components and given (or y). by going back to Eq. (11-2) other things. There may be subtleties in such problems, however. pose, for example, that instead of a a string to constrain the object. works only one way; push radially outward. velocity Fig. it is more or This is less rigid now One may have it cannot a situation in which the not great enough to take the object to the top of the Jl-5 (a) Path of the bob of a simple pendulum Up to = w. (b) Stroboscopie photograph of a bati falling away from a circular channel at the point where the contact force becomes zero. (Photograph by Jon Rosenfeld, Edueation Research Cenler, M.l.T.) (b) (a) 428 Sup- we had a constraint that can puli radially inward but launched with insufficient velocity to maintain a tension in the string rod Conservalive l'orccs and molion in spacc : circle (although it might be possible with a rigid rod); the object The breakaway will fail away point reached when the tension in the string just is in a parabolic path [Fig. 1 l-5(a)]. to zero falls and the component of F„ along the radius of the circle is just equal to the mass m times the requisite centripetal acceleration v 2 /L An exactly similar situation can arise if an object moves along a circular track made of grooved metal, and Fig. ll-5(b) shows a stroboscopic photograph of an object falling away from such a track at the point where the normal reaction supplied by the track has which The angular fallen to zero. this occurs [cf . Fig. 1 l-5(a)] is position, m , at defined by a statement of Newton's law mg cos(tt - ' <L) = nv 2 assumes v <e, < r ( 1 where u* = vo 2 - 2*/(l - cos 0„) This leads to the result COS0m = 2 2 oo 3 3gl can thus deduce that a particle that starts out from O with less than VJgl will fail to reach the top of the circle, whereas We o with a rigid rod to support suffice. it an initial speed of 2\/gl would Notice, therefore, that the energy-conservation principle should not be used blindly; one must always be on the alert as to whether satisfied at every stage Newton's law can be with the particular constraining forces available. AN EXPERIMENT BY GALILEO It was (1638), Galileo, in his Dialogues Concerning who first Two New Sciences clearly stated the result that the speed attained by an object descending under gravity depends only on the vertical distance traveled. He applied this result to the uniformly accelerated motion of an object slopes, as indicated in Fig. from Galileo's own book. 1 down smooth l-6(a), He demonstrate the corrcctness of which is plancs of different based on a diagram could not, however, directly this proposition using inclined planes as such; instead, he performed a very clever experiment 429 An experiment by Galileo Fig. 11-6 (a) Speed atlained by a block frictionless plane depends only traveled, not on the slope. on the sliding down a verlica! disiance (b) Galileo's pendulum experiment to demonstrate the properties ofmotion on idealized inclined planes. with a mass swinging in a circular arc, and applied his remarkable what he observed scientific insight to in this situation. Here is an account of the experiment, taken with only minor changes of notation from modernity of Imagine own Galileo's his presentation this description; the clarity and quite striking: is page to represent a vertical wali, with a nail driven and from the nail let there be suspended a lead bullet of one or two ounces by means of a fine vertical thread, OB, say into it; from four own to six feet long [see Fig. ll-6(b), based sketch]. On draw a horizontal this wali angles to the vertical thread OB Now breadths in front of the wali). the attached ball into position on Galileo's AA', at right(which hangs about two finger- OA observed to descend along the arc line bring the thread and set it free; first ABA' . . . till it OB it with will be almost reaches the horizontal AA', a slight shortage being caused by the resistance of the air and infer that the ball in its momentum on reaching through a similar the string. B which was arc BA' Having repeated From this we may dcscent through the arc to the this same cxperiment AB just sufficient to carry it projects out some five it height. many times, let us drive a nail into the wali close to the perpendicular N, so that rightly acquired a or six now OB, say at finger-breadths in order that the thread, again carrying the bullet through the arc AB, may strike thus compel 430 it upon the nail N when the bullet reaches B, and BC, described about N as to traverse the arc Conservative forccs and motion in space ' center . . Now, . gentlemen, you will observe with pleasure that thc ball swings to the point you would at see the some lower C in the horizontal line same thing happen point. if . . . way Galileo convinced himself In this quired by an object in descending any it different path. He that the speed ac- vertical distance is suf- equal vertical distance by a up through an ficient to carry AA', and the obstacle were placed then added to this the reversibility of the motion along any given arc and could reach his main conclusion that the speed attained by his suspended object must be the same whether descends along the arc it on the arc beginning level AB A A' and Finally, he could visualize such or the arc with its C B, or any other lowest point at B. motion as taking place on a continuous succession of inclined planes of different slopes, and so formulate his proposition about uniformly accelerated motion This was an enormously important along different inclines. because he was then able to make inferences about free under gravity by observing the motion of an object rolling result, fail down an inclined plane with a very gentle slope. mitted him to stretch thc time scale (a = gsin 6!) This per- of the whole phenomenon and make quantitative measurements of distance against time, using as his clock a watcr-filled container with a hole in it that he could open and close with his finger— a brilliantly simple device. MASS ON A PARABOLIC TRACK Suppose we bend a piece of very smooth metal track so that it has the shape of a parabola curving symmetrically upward with its vertex at point O (Fig. 1 1-7). Let the equation of the parabola be y = JCx2 where C is (11-5) a constant. Imagine an object of mass m, ferred to U(y) an arbitrary zero at O, is along the track potential energy, re- given by = mgy Quoted from Sciences (H. the English translation of Dialogues Concerning Crcw and A. de Salvio, translators), York. 431 free to slide Its gravitational with negligible friction. Mass on a parabolic track Dover Two New l'ublications, New Fig. 11-7 Motion of a particle on a parabolic track in a vertical plane. which, by Eq. (1 1-5), we can alternatively write as a function of x: U(x) where k = \mgCx 2 = \kx 2 (11-6) a constant of the same dimensions (newtons per meter) is At any point {x, y) the mass has a speed v, and by conservation of energy we as a spring constant. necessarily along the track, have + U= E \mv 2 (= const.) Denoting the x and y components of Eq. (11-6), we have %mv x 2 This has + by vx and Imv,* of the and using a marked resemblance (11-7) to energy-conservation the Only the presence equation of the linear harmonic oscillator. 2 term \moy u„, = E + ikx 2 v spoils things. This suggests to us that if we had a situation in which vy were very small compared to vx the motion of the mass on this track would closely approximate that of a harmonic oscillator. What do we need to achieve this? , Common is sense tells us more or less immediately that only very slightly curved, so that y vertical motion is « x if the track at each point, the very small compared to the horizontal and approximately harmonic motion will occur. Such motion can be achieved and beautifully demonstrated 1 in which a metal glider rides on a with a "linear air track" cushion of air blown up through holes in the track. Because of the low friction, the oscillation can take place even if the curvature is extremely small. To take an example, here is a specification of the shape of the track in an actual demonstration (the setting was by machine screws at 1-ft spacings; hence the lapse from MKS): 'R. B. 432 Runk, J. L. Slull, and O. L. Anderson, Am. J. Phys., 31, 915 (1963). Conservativc forces and motion in space X, ft ±1 ±2 ±3 ±4 ±5 ±6 ±7 & A A 1 4 ff & M y,in. Taking any pair of these values, one establishes the value of the constant C in Eq. (11-5), e.g., x = 4ft = 1.22 m y - i in. = 6.35 X 10"» m Thus C = 2y/x 2 = 8.55 What would be mass m K)- 3 X 1 For the cxpected period of the motion? and "spring" constant = mgC. from Eq. (11-6), k m" k, we have T = 2r(m/ky 12 and, , Therefore, « 11.9sec 2 2»vTT9« 21.6 sec m/k = (gC)"' 7"- The measured period was within a few tenths of a second of this and independent of amplitude to the accuracy of the measurement. In this example (and the others of motion under gravity) the moving object, i n its two-dimcnsional motion, really does ride a contour that corresponds against x. Chapter form to the graph of U(x) These are truc two-dimensional motions; they arc physically distinct in in 10. from To the onc-dimensional situations discussed appreciate the difference between them, consider two quite realizable physical systems. [Fig. ll-8(a)] resting Fig. is a mass on a smooth m table. 11-8 (o) Mass (b) Same mass on a smooth parabolic track. (c) Arrangements and to (b) (a) can be made have identical variations ofpotential energy with x, bui the motions are different. 433 Mass on first one attached to a horizontal spring and The second attached to a horizontal spring. The a parabolic track [Fig. ll-8(b)] is an equal mass m free to slide on a parabolic track. The strength of the spring and the curvature of the track are adjusted so that the systems have the same parabolic potential energy curve of U(x) against x, as shown Fig. in the same magnitude of If projccted horizontally ll-8(c). same speed v from the origin with the velocity at , both masses will have any other value of into oscillation, both systems will be periodic. But x. If set their periods same show a progressive change of period with amplitude (which way do you think it will be The mass on different. period at all will be?). amplitudes It is ; the spring will have the the other system will only in special situations, such as that with the very slightly curved track, that the two motions become essentially the same. THE SIMPLE PENDULUM In diseussing the simple pendulum, we are returning to the problem of an object moving in a vertical circle. This time, more carefully. It is an important type of physical system, over and above its use in eloeks, and it The simis not quite so simple as its traditional name implies. idealized as a point mass on plicity is primarily in its structure howevcr, we shall go into it — a massless, rigid rod. As we saw earlier [Eq. (11-3)], the potential energy, U(6), expressed in terms of the angle that the supporting rod makes with the vertical, can be written in the form U= mgl(l - cos 0) = 2mg/sin 2 Ifwe plot this expression for Fig. 11-9 "- (11-8) I U as a funetion of d, we Energy diagram for a rigid pendulum, using the angle 6 as the coordinate. is The pendulum trapped in oseil/a- tory motion about 6 = Oi/Eis/ess than 2mgl. 434 Conservative forees and motion n space i get Fig. 11-9. In Fig. 1 1-9, 6 can take values of between 6 —t all values from pendulum ever, all positions of the and -\-k. — oo to + cc . Howby in space are described Any value of d outside this range corresponds to one and only one angle 6 inside the range. The obtained by adding to or subtracting from the latter is former a whole-number multiple of 2r. Two is less kinds of motion are possible, depending on whether E than or greater than 2mgl (referred to a zero of energy defined by an object at rest at the lowest position of the pendulum bob). If the total energy figure), there are is sufficiently great (e.g., as for no turning points in the motion; E2 in the 8 increases (or decreases) without limit, corresponding to continucd rotation of the pendulum rather than oscillation. dulum bob minimum is maximum = at 6 = at ±tt (or ±3tt, ±5ir, (or The speed of the pen- ±2ir, ±4ir, etc.) and etc.). It is clear from Fig. 11-9 that to produce such rotational motion the pendulum must have a kinetic energy at least equal to 2mgl at the lowest point of the swing. If E< 2mgl, say angle B changes from However, the motion amplitudes. E y, is oscillatory and the — 6q and back again (Fig. 11-9). not harmonic except for sufficiently small +6q is the motion to In the neighborhood of O, one can very nearly match the potential-encrgy curve of Fig. 11-9 by a parabola (Fig. 11-10), so that for these small amplitudes the oscillations are just those of the linear oscillator. statement is One way of justifying this made in Chapter 10, to recall the general argument, almost any symmetrical potential-energy curve can be approximated by a parabola over some limited range of small that displacements. Fig. 11-10 Another way Potential- energy diagram of a simple pendulum. 435 The simple pendulum is to note that in Eq. (1 1-8), if d is small, one can set sin(0/2) approximately equal to 0/2, so that we have « U(8) A Imgie1 (11-9) way depends upon using the same approximation that Newton used when he described the moon as a falling object. third Applying to the present problem, it the circular path x2 + - (l is y) 2 and taking the origin at O, given by the equation = / a (seeFig. 11-11). Therefore, y 2 - y = + 2ly /-(/ 2 x2 = -* 2 )" 2 2\l/2 = If x2 / <5C l ->('-i) 2 we can use , approximate constant result 1 £ 2 l [Compare C the binomial expansion to yield the (11-10) Eq. (11-5); you will then realize what the in that equation represents. In the actual experiment this with described, the motion was like that of a simple pendulum nearly 400ftlong!] Equation (11-10) suggests a somewhat different approximation for small oscillations of the pendulum, describing its motion angular one. terms of its horizontal displacement instead of its But whatever analysis we adopt, it is clear that the period must depend on the amplitude. We can also argue qualitatively which in way it varies. Looking at Fig. 11-10, we can say that the curve l-y Fig. 11-11 Geomelry ofilisplace- ments of a simple pendulum. 436 Conservalive forccs and motion in spacc pendulum of U(6) for the describes a kind of spring that gets Compared "softer" at large extensions. would hold that an for Thus one might guess tively less at larger displacements. the motion increases. room to the parabolic behavior ideal spring, the restoring force is rela- that becomes more sluggish and hence that the period Certainly there is one extreme case that leaves" no for doubt. If the = exactly arrives at n pendulum sit would practice this is energy is such that the pendulum just there is no restoring force jr, upside down indefinitely at all; the — although of course an unstable equilibrium. amplitudes short of this the increase of period is in Even If drastic. at you are interested, the section after next describes the analysis of larger-amplitude motion. But we first shall make a detailed study of the small-angle approximation. THE PENDULUM AS A HARMONIC OSCILLATOR The speed of pendulum bob the any point at so that the cncrgy-conservation cquation 5^/ 2 (^) + U{6) = E is equal to / dd/dt, is (exact) '(D' Using the approximate expression for U(6) from Eq. (11-9), we have iml 2 {^) + $mgl0 2 = E (approx.) or Tt) (f) + 2 ]° ° const By now we have met can identify (g/l) 1 ' 2 (n-u) - this form of equation scveral times and as being the quantity w that defines the period of the oscillation: -2r-*(i)' [Caution: It is /2 (11-12) the angular displacement itself that undergoes a simple harmonic variation, described by the equation 0(0 437 = O sin(oo/ 4- v,,) The pendulurrms a harmonic (11-13) oseiliator This can be confusing; the actual angular displaccment of the pendulum is deseribed in terms of the sine of the purely mathematical phase angle (w/ the pendulum + <p dB/dt— not is The ), w the actual angular velocity of in (ut + <p ), which serves merely to define the periodicity.] The behavior of harmonic tion to a ment on it its so important that is moment First, let us take a we shall com- to consider the equation of motion by a direct application of We Newton's law. to whether oscillator further. derivation of the simple pendulum as a elose approxima- can do we analyze this in two different ways, according the problem in terms of linear or angular displacements. Linear motion We consider the horizontal foree aeting on the when it [Fig. 1 is a horizontal distance 1— 12(a)]. x from If the tension in the its pendulum bob equilibrium position suspending string is FT , we have 2 m— = -Fr sin0 = -FT -. at* I If the vertical small, we Ft component of bob is negligibly SO that FT « mg acceleration of the can, however, put cos 6 « F = mg a For small angles 6, cos 8 ~ 1 - d 2 /2 w I, and the equation of horizontal motion becomes Fig. 11-12 Basis for analyzing the motion of a simple pendulum (a) in terms of horizontal displacements, and (b) in terms of angular displacements. 438 Conservalive forees and motion in space 2 m dJ x mg W~~T x leading once again to Eq. (11-12) for the period T. Angular motion In (= case we consider the transverse acceleration ae d 2 8/dt 2 ), which is produced by the component of F„ in the this l direction of the tangent to the circular arc along which the pendulum bob moves (see Fig. ll-12b). The magnitude of the tension force F^ (which actually varies with time) does not enter into this treatment of the problem. We have, simply, 2 rf w/—- = — F„sin0 = — mg s'm 6 For small « 0, sin which leads to Eq. (1 9, and so we have 1-13) for as a function of time. It can be seen that the analysis in terms of the angular variables is a "cleaner" treatment than the other, involving only the one approximation sin 9 « 9. The isochronous behavior of that its period is over a wide range the simple pendulum — the fact almost completely independent of amplitude — provides a striking example of this remarkand unique property of systems governed by restoring able forces proportional to the displacement from equilibrium. Sup- pose we had two identical pendulums, each with a string of ft, suspended from a high support. Each would have a period of about 6 sec. And if we set one swinging with an amplitude of only an inch or two, so that its motion was almost length 30 imperceptible, and set the other swinging with an amplitude of 5 so that ft, about 5 to put it swept through the central position at a speed of ft/sec, the difference in their periods them significantly out of step in less would be too small than a hundred swings or so. This same isochronous property of the pendulum was of and perhaps crucial help to Newton, Huygens, et al. when their fundamental experiments on collision phenomena (cf. Chapter 9). AH these experiments were done with masses suspended from equal strings. As long as the small-amplitude great they 439 made The pendulum as a liarmonic oscillatoi Fig. 11-13 Two pendulums ofeqnal (a) Icngth are re- leasedfrom arbitrary posltions al ihe same instan/, The isochronous properly of ihe pendulum ensures that ihe collision occurs w/ien bolh masses are al Ihe (b) boltom politis of iheir swing. approximation two masses released at the same instant is valid, from arbitrary positions points at thc same [Fig. ll-13(a)] will rcach thcir lowcst instant [Fig. ll-13(b)], so that the collision occurs when each mass is traveling horizontally with vclocity— thc magnitude of which (b) simplc harmonic motion. enough large = t'Lx Even ' is its maximum given by the equations of for angular amplitudes that are to require the use of the exact equation for 2gl{\ - cos pm „ , O) the colliding masses will reach their lowest points at almost the same instant as long as O is less than 90°, which it would have to be for an object attached to a string rather than to a rigid rod. section briefly discusses the detailed dependence of The next period on amplitude for a simple pendulum. THE PENDULUM WITH LARGER AMPLITUDE 2 To how find pendulum departs from the period of a simple small-amplitude value, its ideal we write the equation for conservation of energy in the exact form ©' + where w,, 2 = 2 2<oo (l - cos 0) g/l and #o > s tne - 2 = 2coo (l maximum cos 8 ) (11-14) angle of dcflcction from the vertical. The period of 7X0,,) 'You can oscillation thcn given by is dO = fs/2 J-t (co$8 wo" easily vcrify that the - (11-15) cos<?o) 1/2 magnitude of this velocity is proportional, in thc small-amplitude approximation, to the horizontal distance through which a mass is initially drawn aside before bcing released. This makes analysis of thc cxperiments very simple. 2 440 This section may be omitted without loss of continuity. Conservative forces and motion in space Fig. 11-14 Precise measuremenls on the period of a simple pendulum, showing the quadratic increase of period wilh amplitude, and the close approach to isochronism obtained by the top damping endofthe supporting wire between cycloidal jaws, so that the free length shortens as the amplitude increases. For small amplitudes, we of course have a period 2w/(j)o, as discussed the previous section. iri T The equal to integral of Eq. (11-15) cannot be carried out exactly; one has to resort to numerical methods or to a series expansion of the integrand which approximation to the period, the following gives, as a next result: 1 +£sin 7tyo)« To ['+**>'(?)] If O is (11-16) not too large, another acceptable form of this terms of the horizontal amplitude T(A) « To A (= /sin 6 ) is result, in the formula (11-17) (+tf) some precision measurements The pendulum had a length of about Figure 11-14 shows the results of that verify Eq. (11-17). 3 m, and the values of 6 1 greatest amplitude used up to about worth noting that the greatest values of d 10° (« over-all studied is was about range of O- The graph 'M. K. Smith, Am. 441 (11-17) is The pendulum with m, so that less It is the least to the than 2 parts in 1000; the in1 part in 10 5 , and the very nicely demonstrated over this also J. Pftys., 32, T from change of dividual points are accurate to about validity of Eq. 0.5 0.17 rad) are represented. shows the results of using a special 632 (1964). larger amplitude pendulum support that shortens the pendulum such a way that the tendency for so that the path of the pendulum be shaped a portion of a cycloid curve. is method of making an the almost O >s Ideally, the support should completely compensated. (Incidentally, T as 6 increases, in to increase with exactly isochronous pendulum, through the use of a cycloidal suspension, was discovered by Huygens. In 1673 he published a great book, the HoroJogium Oscillatoriwn, results were practical first first in which many important dynamical presented within the framework of the very problems of clock design.) UNIVERSAL GRAVITATION: A CONSERVATIVE CENTRAL FORCE The problems of energy conservation we have that discussed so far have involved only the familiar force of gravity near the earth's surface— a force which, as experienced by any given same magnitude and the same direction object, has effectively the in a given locality, regardless of horizontal or vertical displacements (within wide tional interaction distance between their centers. limits). is But, as we know, the basic gravita- a force varying as the inverse square of the two It is 1 and exerted along the particles line joining an example of the vcry important class of forces— central forces— which are purcly radial with respect to a given point, the "center of force". It has the further property of being sphcrically symmetric; that force depends only on the and not on the direction. We shall the magnitude of the is, from the center of force, radial distance show that all such spherically symmetric central forces are conservative, and shall then consider the special features of the l/r tion and force that holds for gravita- electrostaties. If a partiele is on aeting 2 it exposed to a central force, the force veetor has only one component, spherically symmetric, then Fr F T . F If the force is also can be written as a funetion of r only: F r We is (11-18) =/(r) shall prove that any such sphcrically symmetric central force conservative by showing that the work done by the force on a •Recall, howcvcr, that small local variations are in Tacl detectable by sensitive gravity survey instruments, as deseribed in Chapter 8. 442 Conservative forces and motion in space __M—_______ ___) (a) 7%. 77-75 (a) Diagram for eonsideration ofpotential- energy changes in a Central force field. (b) Analysis of an arbitrary pai h into radial elements along which work is done, and transverse elements along which no work is done. (c) Closed path in a conservative central force field. test particle, as the latter B another point ticular path connecting If this result holds, changes its position II-I5(a)], does not [Fig. then from a point A to depend on the par- A and 73, it will be true that the particle, but only on these end points. if it went to B by this path, and returned from B to A by any other would have no net work done on it by the central force and would (if no work were done on it by other forces) return to A with the same kinetic energy that it had to start with. This from A path, then corresponds exactly to the conservative property as defined for one-dimensional motions. Let the center of force be at the work done by the O [Fig. ll-15(a)], central force F on and consider the test particle as undergoes a displacement ds along the path as shown. work is where a r, This given by dW = of it is F • ds = F ds cos a (11-19) the angle between the direction of F, and the direction of ds. From i.e., the direction the figure, however, we see that ds cos where dr a = is dr the change of distance from O, resulting displacement ds. Inserting this value of ds cos a from the into Eq. (1 1-19), we have dW = 443 Fdr Universal gravitation: a conservative central force where the magnitude of the force [F as indicated in Eq. F on (1 We 1-18). the test object as /(/)] depends only on r then have for the moves from it = A work done by to B, r 'b W= •> Because (11-20) f(r)dr / 'A has a value that depends only on this integral we can conclude and not on the path, metric central force is its limits that the spherically sym- This result can be arrived conservative. at in slightly different terms by picturing the actual path as being up from a succession of small steps, as shown in Fig. One component of each step is motion along an arc built ll-15(b). at constant so that the force r, ment and the contribution is to W perpendicular to the displace- is zero, is and the other component a purely radial displacement so that the force and the placement are work F dr by in the same direction, resulting in the dis- amount of the central force. [A converse to the result we have just derived the following is important proposition A central force fleld that is also conservative must be spherically symmetric. To show this, suppose that a center of force Imagine a closed path in Fig. ll-15(c). short portions of circular arcs central, from it O around two radial lines BC and DA. definition, the force is purely has no component perpendicular at any point. ABCD, it Thus, O by very drawn from O, and by the two by Since, exists at the point ABCD, formed if to the radial direction we imagine cxperiences no force along a particle carried BC and DA. The condition that the force be conservative thus requires equal and opposite amounts of are of equal length, the same on each. work along AB and CD. Since these lines the mean magnitude of the force must be If we imagine the lengths of these elements of path to become arbitrarily small, we conclude of the force at a particular scalar value of r direction in which the vector r is drawn. is that the value independent of the The most important sources of such spherically symmetric force fields are spherically symmetric distributions of mass or It is electric charge.] important to realize that a force may be without being central or spherically symmetric. conservative For example, the combined gravitational effcct of a pair of concentrated unequal masses, separated by some distance, has a complicated 444 Conservative forces and motion in space dependence on position and direction, but we know that it is conservative because it is the superposition of two individually conservative force fields of thc separate masses. Given the by Eq. (11-20), we can proceed rcsult expressed to define a potential energy U(r) for any object exposed to a spherically symmetric central force: r 'd VB - Ua = If the kinetic KA KB and J (11-21) f(r)dr f (r) / 'A energy of the object at points , respectivcly, then (if A and no work is B has the values done by other we have forces) W Kg - Ka = Ka + UA - KB + and Ub -B (11-22) Thus we have established an energy-conservation statement an object moving under the action of any central force. For an inverse-square F(r) = force, for we have dl-23) ^2 In such a case, therefore, Eq. (11-21) gives us r 'n UB - Ua 2 C r" dr J 'A from which we get Ub- Ua- Cl — -—a) Vb r (11-24) J There remains only the choice of the zero of potential energy. It is far and convenient to set U= for r = ao, i.e., at points infinitely from the source at O, since thc force vanishes at these points, terms one can say that the existencc of either no consequence to the other one under these conWe now apply Eq. (11-24) to the case where rA = y-, and the potential energy of thc test particle becomcs in colloquial particle is of ditions. UA = Ub = 0, C/rji — or, if we drop the now redundant subscript B, there follows 445 Universal gravitation; a conservative central force = - V(r) (11-25) f for the potential energy of a test particle as a function of its position. Equation (11-25) is valid pulsive force, the constant and C m (which re- being negative for attractive forces positive for repulsive forces. mass an attractive or a for either In particular, for a particle of under the gravitational attraction of a point mass we M suppose to be fixed at the origin O) we have shall F(r)=-^ (.1-26) DW--2? 01-27) Note were points two objects — i.e., refer only to the inter- that can be regarded as though they their linear As we saw their separation. most two equations that these last action between dimensions are small compared to in Chapter 8, however, some of the and important gravitational problems concern interesting the gravitational forces exerted by large spherical objects such as the earth or the sun. lems, we saw In our earlier discussion of such prob- that the basic problem the interaction between is a point particle and a thin spherical shell of material. In Chapter 8 we presented a frontal attack on problem, going directly this to an evaluation of an integral over all contributions to*the net force. Now we shall approach the problem by way of a consideration of potential energies, the great value of the and in the process we shall see potential energy concept in such calculations. A GRAVITATING SPHERICAL SHELL Suppose, as in our earlier treatment of the problem, that we have a thin, uniform shell of matter, of radius 11-16). Let a particle of mass distance r from m R and mass M (Fig. be placed at some point P a we deal directly in the center of the shell. If terms of forces, then, as we saw, the force from material in the A must be resolved along the line OP. In other words, a vector sum of all the force contributions is necessary. If we deal in terms of potential energy, however, we vicinity of a point such as can cxploit 446 its most important propcrty: Potential energy Conservative forces and motion in space is a Fig. 11-16 Diagram for considering the gravitational potential energy due to a circular zone ofa spherical shell thin of matter. We scalar guantity. all can just add up the contributions to The parts of the shell. from the F tr As force on m is U from then obtained simply relation --™ (11-28) dr in our direct calculation of the force, we take advantage of the symmetry of the system by considering the zone of material marked on the sphere between the angles off For (Fig. 11-16). area = 2ir R this and B + dB sin B dd 2wR 2 sm9d6 mass B zone we have 4jt/?2 M i.e., dM = iMsinddd All of this material in Thus the contribution toEq. (11-27), by dU = The ... . GMm 2 A energy of GMm 2 447 same distance, s, from P. makes to U is given, according at the s = - This integral is dM GmdM total potential U(r) dM that is to f J sin 6 m sin 6 dB s is thus dB (11-29) s be evaluated, keeping «ravilating spherical shell R and r constant, by from zero to sweep allowing to ir as to include the contributions of calculation is made (s changing accordingly) so parts of the shell. The simple by the fact that, in the triangle AOP, all we have 2 s = R2 + r 2 - 2Rrcos0 Differentiating both sides with respect to and r are constants for the (11-29), 0, remembering that R purpose of the integration in Eq. wehave 2*^do = 2Rrsin6 Thereforc, sin 6 dd ds^ Rr But the left-hand side of this is just the integrand of Eq. (1 1-29), which thus becomes U(r) = - GMm (11-30) ds 2Rr 7»-o Equalion (11-30) is the key result in cakulating gravitational potenlial energies andforces due lo spherkal objecls. In evaluating Fig. 11-17 (o) Gravi- tational potential energy of a point parlicle as a function of its distancc from the center of a splierical shell radius R. tion (b) Varia- of F with riaed from thin of r, de- (a). 448 Conscrvative forces and motion in space we must the integral of ds, however, Case 11-16). The 7r R, as in Fig. (point o«to<fe shell) (11-31) (i.e., r s are defined as follows: s,„i n = r giving smnx = r giving two cases: > outside the shell on limits = = P Point 1. distinguish — R +R Hence Smnx Anln = 2*1 U(j.)--2MH Point Ca.se 2. P inside the shell (i.e., r < The R). limits now become 8 = giving s mi „ = /? - r 9 = x giving smsx = /? + r Hence U(r) These two from them = - results are, ^^ are disarmingly alike. however, quite calculate the forces Case If The forces derived we proceed now to from O, is a constant. that can vary. It is only r, R the distance Hence we have 1. F = j— r Case m different. from Eq. (11-28), we must remember that (the radius of the shell) of the mass (11-32) (point inside shell) (point outside shell) (11-33) (point inside shell) (11-34) 2. Fr = In Fig. 11-17 we show the graphs of potential energy and forcc as functions of the radial distance r of the test the center of the shell. The discontinuities at r mass = R m from are easily seen. Once we have obtained these results, to consider a solid sphere of material. 449 A gravitaling spherical shell it is a simple matter A GRAVITATING SPHERE Our primary assumption is that the sphere can be regarded as made up of a whole succession of uniform spherical shells, even though the density may vary with radial distance from the center. Granted this assumption of symmetry, we can at once draw some conclusions about a sphere, of total Case the gravitational force cxerted by such M and radius R. mass Observation point oulside the sphere 1. each component spherical shell acts as though (r > R). Since whole mass its were at the center, the same can be said for the sphere as a — = F(r) we have moon and accelerations of the already used in Chapter 8 in famous comparison of the gravitational discussing Newton's Case the apple. Observation point inside sphere 2. must be more (11-35) (point oulside sphere) the result that is which the density of the main with radius, we have terial varies This way Regardless of the whole. (r < Here we R). but we can at once make two clear state- careful, ments: a. For all the spherical shells lying oulside the observation point, the contribution to the force b. For all shells byEq. zero, by Eq. (1 lying inside the radius defined mass servation point, the is is 1-34). by the ob- effectively concentrated at the center, (11-33). To specify the force in how much mass is enclosed this case, therefore, we must know drawn within the sphere of radius r through the observation point. Case l', (special) The same as case 2, but with the extra proviso that the density of the sphere sphere is homogeneous. In this case, of the mass contained within radius this is the ratio of the partial the amount of mass r is i.e., equal to r 3 /R 3 , because volume to the whole volume. Hence to be considered tively at the center, — the is the same for all r we know that the fraction a distance r is equal to Mr 3 /R 3 , effec- from the position of the test mass. This then gives us F 450 ^ = W r (P°' nt inside homogeneous sphere) Conservative forces and mol ion in space (11-36) Fig. 11-18 (a) Force on a poinl particle as a function of ils dislancefrom the center ofa uniform solid sphere of radius R. {b) Variation of mutual polential energy with r, obtained front {a) by integration. The approximately linear increase ofU with r jusi ouiside r = R corresponds to gravily as observed near the earih's surface. The combined results of Eqs. (11-35) geneous sphere are shown struct the in Fig. ll-18(a). graph of potential energy versus distance of a particle of mass This is and (11-36) for a homo- shown in Fig. m ll-18(b). One can r for all also con- values of the from the center of the sphere. For interior points (r < R) we have »-w"j," GMm (R2 - r2) 2R3 But we already know, by putting U(R) = Thus we in Eq. (11-31), that - GMm R </(,)= In particular, at r A = R get (r<R) 451 r = 0, -^(3K2-,2) we have gravitating sphere (11-37) 3GMm = - 1/(0) If 2R one imagines starting out from r = 0, then, as Fig. ll-18(b) shows, the potential energy increases parabolically with distance up to = R r The and then goes over smoothly into the continued U increase of with r that is variations of F(f) good, as wc be homogeneous. no requirement is On it in Fig. 1 1-18 hold that the sphere should the other hand, one that Eqs. (11-36) must be careful to and (11-37), for to the special case of a sphere of the Thus > R r have seen, for any spherically symmetrical distribution of matter; there remember described by Eq. (11-31). and U(r) for r < R, refer only same density throughout. does not correctly describe the variation of gravitational force with radial distance inside such objects as the earth and the sun, which have drastic variations of density with r (see Fig. 11-19). This invalidates a favorite textbook cxercise: that a particle would execute simple harmonic motion "Show in a tunnel bored along a diameter of the earth." The practical impossibility of making such a tunnel may, however, be considered as a far more powerful objection. used to thinking that the potential energy of an object of mass m, a distance h above the earth's surface, is given If Fig. one 11-19 earth. is (a) Radial variation of density inside the (After E. C. Bullard.) (b) Calculated variation of density wiih radial distance inside the sun (I. Iben and Z. Abraham, M.I.T.). 452 Conservative forces and motion i n spacc U= by the formula mgh, may seem hard it U(r)=-™^ There no really is (r>R) however, once one recognizes that difficulty, the zero of potential energy linear to reconcile this by Eq. (11-31): result with the result expressed arbitrary is and that the simple formula applies only to objects raised through distances compared to the earth's radius. We more general formula [i.e., Eq. (11-31)], that are exceedingly small have, in fact, from the U(R h)-U(R)=-^R + h + ^R + GMm h R2 G Mm/ Since, however, on m, 2 the gravitational force is mg exerted this gives us U(R + - h) « U(R) By putting U(R) = energy, we see that mgh as an arbitrary reference level of potential U = mgh is an acceptable approximation that applies to small displacements near the earth's surface. This approximatcly linear increase of potential energy with distance just outside a sphere is indicated on Fig. ll-18(b). ESCAPE VELOCITIES Suppose we have an object sphere, such as a planet, and outer space so that This problem At is it at the surface of a large gravitating we want to shoot the object What speed must we never returns. easily considered in terms of the surface of the planet (r = R) off into give it? energy conservation. the potential energy is given by U(R) = At r = co _ ^ R U = The particle, to reach /=<», must some kinetic energy .K(oo). Thus at must have a kinetic energy K(R) defined through the we have 0. survive to this distance with launch it conservation equation 453 Escape velocities Therefore, GMm K(R)> The critical R condition is reached if we take the equality in the above statcmcnt; the object would then The minimum zero residual speed. just reach infinity with escape speed at radius R is thus given by GMm , , Therefore, ^ *= (11-38) In calculating the escape velocity from the earth's surface, may be convenient to state the result of Eq. (11-38) form force of w in it another makes use of our knowledge that the gravitational on a mass m at r = Re can be set equal to the magnitude that times the local acceleration due to gravity, g: GMm ~R^ =mg Therefore, GM and so, from Eq. (11-38), we have = VgR E ) v2 DO Putting 10 G in the (H-39) familiar values g ~ 9.8 m/sec 2 , RE ~ 6.4 X m, we have bo « 11 .2 km/sec Notice once again the remarkable implications of energy con- For the purpose of calculating complete escape, wc in which the eseaping servation. need spccify nothing about the direetion object is anything fired. in It could be radially outward, or tangentially, or between ; the same value of v u applics to all cases [Fig. ll-20(a)]. It is interesting is 454 that the magnitude of the escape speed v a exactly \/2 times as great as the orbital speed that a partiele Conservative forees and motion in space Fig. 11-20 (a) possibility from a The ofescape gravitating sphere depends on the magnitude of the velocity, not direction. on (b) its The escape speedfrom a double-star system depends on the position of the start ing point P. would have if could skim around the earth's surface in a it circular orbit of radius The preceding Rj,;. calculation assumes that the object which the escape takes place is isolated from from other objects. all More complicated escape problems would arise if, for example, we wanted to calculatc the escape velocity from the surface of one member of a double-star system [Fig. 1 l-20(b)]. We should then have to consider where the Iaunching point P was in relation to the centers O and t 2 of the two stars. But the scalar still property of potential energy makes the calculation relatively simple. The forward sum potential energy of a And arately. mass m at P is just the straight- of the potential energies due to the two stars sep- again the direction of launeh for complete escape at the critical velocity is immaterial (as long as the trajectory misses the other star!) A is, particularly important example of 11-21 (a). schematically in Fig. mass m, its If we This shown is consider a partiele of potential energy as a funetion of position along a straight line joining the centers of earth by the type of system this last of course, the earth-moon combination. ll-21(b). full line in Fig. and moon Mathematically is it is shown as given by the formula U(r) = - GM.\ r m GMf.iti r where and D is ' (11-40) r the distance between the centers of earth r is the distance of the partiele This potential energy tributions from earth 455 D- Escape velocilies is simply the and moon. and moon from the center of the sum of earth. the separate con- Effi BBS ;[.-''- 'v Moon Fig. 11-21 (a) Sche- malic diagram of the earth-moon system. (b) Form ofvariation Ofthe total potential energy of a particle along Ihe line joining the centers of earth and moon. The maximum value of U(f) occurs = where (dU/dr) rm This 0. is the attractive forccs due to earth and posite, so that at the point r = r„, clearly also the point at which moon are equal and op- we can put GMp.m CMsttn - (.D rm y This gives 1 Now + Wm/Mk) 11* the value of This represents a center. MM/ME about SV. so that rm w 0.90D. point about 24,000 milcs from the moon's The value of U(r) is at this value of r is given, according to Eq. (11-40), by U(r,„) = - GMK m GM M m 0.923 0AD G Me>» 0.9D 1. + 3 1 ) 23GME m D If a spacecraft 456 is to reach the moon from Conservative forccs and motion in the earth along the spacc direct line considered, energy hump. an after If, initial it would have to surmount this potential were to be done by simply coasting further, this rocket blast at the point of departure from the earth's surface, the necessary minimum speed at the earth initial would be defined by the following equation: GM^— Em , 2 fwoo Uh = ... l.23GMB m = U(rm ) D This then gives us 2GMk ( l.23RK \ factor 2GMe/Re «s just the square of the speed needed for complete escape from the earth, as given by Eq. (11-39) and The evaluated numerically as I1.2km/sec. 2 the above expression for v , one Putting finds u « R E /D ~ g\y 11.1 km/sec. in At minimum velocity the traveling object would just barely surmount the potential "hill" between earth and moon and would then fail toward the moon, reaching its surface with a this (You should check thesc speed of about 2.3 km/sec. results for yourself.) MORE ABOUT THE CRITERIA FOR CONSERVATIVE FORCES 1 In our discussions of forces and potential energy, we have pointed out that the fundamental criterion for a force to be conservative is work done by the force be zero over that the net any closed path. As an aside we pointed out that in one-dimensional problems, but only in one-dimensional problems, condition is of position. how, automatically met We shall now two dimensions, in if give a simple example to illustrate this latter condition A force is F(x, y) that is a unique function of x and theless be nonconservative. Our example in the xy plane, is this: finds itself this the force is a unique function Suppose that a not sufficient. y may particle, at never- any point exposed to a force F given by Fx = -ky F„ = +kx where k is 'This section 457 a constant and x and may More about y are the coordinates of the be omitted without loss of continuity. ihc critcria for conservative forces Fig. 11-22 Rectangular path in a plane. (b) Smooth tube shaped to (a) force a parlicle tofollow the palit shown in (a), (c) Closed path up oftwo different paths between force is conservalive, the particle. llte net work A and B. giuen points made If the is zero. Such a force evidently dcpends only on the position of Now let us calculate the work done by this force the particle. on the particle if the latter closed path shown Fig. in moves counterclockwise around the ll-22(a). would never follow that the particle of the force F and nothing else, but (as It this we may well be objected. path under the action did in our first to potential energy at the beginning of Chapter F approach 10) we can by another force, supplied for that the object can be freely moved example by a spring, so around without any net work being donc. Or, alternatively, we can imagine a very smooth pipe, as in Fig. 1 l-22(b), which comimagine that is exactly balanced pels the particle to travel along the sides of a rectangle and yet pipeis always at right angles to the particle's motion. kinetic energy to carry away from it. Now Starting at O, a to the point by F during W\ since = y = the particle by F along 458 P this / We should proviso, in this case, that the particle has add the also by the particle because the force exerted does no work on the it enough around the path even ifFistaking energy for the caleulation. let with move along the x axis a distance coordinates x = a, y = 0. The work done the particle motion is Fx dx = -ky everywhere on moves from P j dx = this portion of the closed path. to Q{x = a, y = b). Next, The work done this path is Conservative forees and motion in space w» = Fy dy = +ka / ly-0 Jv-0 For the path from W = 3 and W4 The this (0, b), the is done as the 4 is dx = +kab / zero, since W\ force round in the particle x and hence moves from Fv is R zero every- a F given by Eq. (1 1-41) If k were taken in this direction. traveled clockwise, amount each Ikab is * not conservative, although positive, the kinetic would be increased by 2kab of the particle artificial trip is therefore + W% + Ws + W = depends only on position. circuit work done path work done W= it O = kab = total and the W work to the origin where on R to Fx dx = -kb / finally, the back Q dy Jo its for each complete the other hand, if the particle kinetic energy would be decreased by may seem — and This time. On indeed is (1 same this —a very this non- example; nevertheless, forces having precisely conservative property [although not the described by Eq. energy analytic forms as 1-40)] play an important role in physics, especially in electromagnetism. There is that is another way of looking at the analysis of such a The situation. physical criterion for the force to be conservative should do no net work, either positive or negative, as it the particle to which around the path of it Fig. is applied makes a complete circuit ll-22(a) in either direction. consider a more general situation [Fig. Let us ll-22(c)] in which a from a point A to a point B along one path and returns from B to A along a dirferent path (again imagine that particle travels constraining forces, doing no work, are applied as necessary). We shall assume that the force we are studying any constraining forces) ticle only. F-ds+ Path More about l not including a function of the position of the par- If this force is conservative, W=j 459 is (i.e., we then have F-ds = P«th 2 the critcria for conservative forces It follows from this that F ds = - / / F • Path But now, Path 2 1 if the force if change the Hence, cis is a function only of position, we can interon the right and reverse its sign. limits of the integral F is conservative, we must have rB •B Fds 'A Path Path 2 Pi 1 1 If this condition is satisfied we can put Wab=I Fds (ll-42a) And without reference to the particular path. same for all paths from A to B, we can conclude if WAB the is that the force is conservative and that the potential-energy function can be defined through the equation Vb -Ua--Ja In evaluating F work (ll-42b) ds integrals such as that one may wish to resolve the force F into F„, and resolve the element of path ds and dy, thus getting W If F,dx+ \ one uses (for a this equation, its in Eq. (ll-42a), Fx and components into its components dx two-dimensional problem) F„dy however, one should always remember that, basically, the integrals are defined as being taken along actual designated path of the particle. an This might seem to be obvious, because the force must necessarily be applied wherever the particle is. However, there are many situations in which an it happens to be (gravitational object experiences a force wherever force, for example). One can then set up a purely mat hemat ical statement that defines a value of general each component of F unless one already 460 knows is F for any given x and a function of both x and that the force Conservative forees and motion in is y. conservative, ^pace In y. And it is 11-23 Fig. Consideration of work done dong a specified path between two points. essential to f'ollow the given path, as in Fig. 11-23, the simpler finding the integral of (y — = y\ along the It is that, in and not take but perhaps unjustified course of (for example) Fx from by the const.), followed line CB (x = x2 = line y>\ AC to y2 const.). perhaps worth ending many x 2 along the integral of Fy from to jci with the remark this discussion one may usc the concept of potential situations, energy even when additional nonconscrvative forces act on the particle. For example, a of the earth field may satellite moving in the gravitational be subject to the frictional drag of the earth's upper atmosphere. This drag is a dissipative force, nonconservative, and the total mechanical energy of the motion will not be constant but will decrease as the motion proceeds. Nevertheless, one may still properly talk of the gravitational potential energy that the satellite possesses at any given point. FIELDS 1 The forces that we have labeled gravitational, magnetic are "action-at-a-distance" forces. contact between interacting objects comes No into play. kind of interaction, the idea of a field of force To electric, and apparent physical For most is this useful. introduce this concept, consider the gravitational attraction of the earth for a particle outside it. The puli of the earth both on the mass of the attracted particle and on relative to the centcr of the earth. its depends location This attractive force divided by the mass of the particle being pulled depends only on the earth and the location of the attracted object. We can therefore assign to each point of space a vector, of magnitude equal to This 461 section Fielcis may bc omitted without loss of continuity. the earth's puli on a particle divided by mass, and of direction its Thus we imagine a identical with that of the attractive force. collection of vectors throughout space, in general different in magnitude and direction at each point of space, which define the gravitational attraction of the earth for a test particle located at a any arbitrary position. The such vectors totality of and the vectors themselves are called the field, or intensities of the In this field is called strengths field. example, the gravitational field strength g at a point Pis F The magnitude of 2 (e.g., m/sec ) and M where point P is the (11 _43) r mass of the earth, r the position vector of the and relative to the center of the earth, in the direction of r. side the surface of One can of acceleration in units given explicitly by the law of gravitation as ._ C £E»_ 6 £e g measured this field is is Equation (11-43) is e r the unit vector valid at all points out- an ideal spherical earth. generalize this for the field produced by an arbitrary distribution of matter, for which the field strength g as a function of position describes quantitatively the gravitational field. The gravitational force exerted field is then given by This field on an object of mass m by this F = mg. description of forces ing electromagnetic forces. The is especially useful in specify- electric field produced by a charged particle or by a collection of such charged particles is described by the electric field strength or intensity S, where Z = F q, is * the vector force acting on a positive test charge of magnitude and £ depends on position. charges at rest, As an frequently arbitrary the situation is For electric fields aid to visualizing the character of a force field, use is made of the concept of a field line. Starting from an point, we draw an infinitesimal line element in the direction of the field at that point. neighboring point 462 produced by similar to the gravitational case. in the field, Being thus brought to a we draw another Conservative forces and motion in line space element in new the direction of the field at the making a smooth curve, limit of of a force of such and so on. the tangent to which, at any point, In the we the line elements vanishingly short, tion of the field at that point. line— has no point, This construct— which obtain the direc- is the field is but a vivid picture of the properties real existence, can be obtained by drawing a whole collection field lines. Hand-in-hand with the concept of a of force goes that line of a line or surface of constant potential energy. We shall therefore discuss this briefly in the next section. EQUIPOTENTIAL SURFACES AND THE GRADIENT OF POTENTIAL ENERGY 1 Besides its utility in connection with the dynamics of conservative systems, the concept of potential energy enables us to describe conservative fields of force quantitatively in a relatively simple The reason fashion. we can use that is the simple scalar field of potential energy to calculate the relatively complicated vector of force. field we know This calculation proceeds as follows. the potential energy of space; i.e., we have a U Suppose of a test particle at each point form relation of the U =f(x,y,z), where the single-valued function f(x, y, z) depends on the particular field of force under consideration. If we wish to know at what points of space the test particle will potential energy, say £/ we , set U = U have a given value of and obtain an equation of the form f(x, y, z) This is = const. = Uo the equation of an energy eguipotenlial a surface, and surface. There this surface is called exists a whole family of these equipotential surfaces, one for each value of U(,. by definition, it requires no work to move our one point to another on the same equipotential Since, test particle surface, it from follows that the lines of force are everywhere perpendicular to the equipotentials. We should draw attention to a distinction between two quantities here. Just as (or charge, etc.) so we we define field as the force per unit define potential per unit mass (or charge, etc). This 463 section may To (<p) mass as the potential energy take the specific case of gravi- be omilled without loss of continuity. Equipotential surfaces; poiential-encrgy gradicnts tation, force F we thus have and gravitational (mVsec 2 ). The complete (or, in the following paired quantities: gravitational (newtons) with gravitational potential energy field g (m/sec array of 2 ) may <p and cquipotential surfaces field lincs two dimensions, equipotential sets (joules), with gravitational potential picture of a complete field pattern. two U lines) provides a graphic we In two dimensions see of curves which, however complicated their appearance be, are everywhere orthogonal to one another. The gravita- which tional field of a spherical object has a simple pattern, in the field lines are radial lines and the equipotentials are a set of concentric spheres [Fig. ll-24(a)]. two spheres close together is far less The field pattern (For one way of constructing such a diagram, see Problem In making drawings of Fig. 1 1-24 the equipotentials, (a) Equipotenlials and field lines due it is 1 1-26.) often convenient due t o a sphere with a \/r 2 force law. (b) Eauipolenlials and field lines due lo a syslem oflwo nearby spheres, 464 of masses 2 to simple [see Fig. ll-24(b)]. M and M. Conservative forces and motion in space to draw them for equal successive increments of the potential makes (or potential energy); this just like a contour To map — which the picture very informative, in effect (see Fig. 11-24). it is obtain a complete specification of the force must be able to get the magnitude of the force as direction, at every point of space. its P and at a point point. Its let it may F* - we Consider a test particle be displaced an amount ds to a neighboring potential energy will change dU = -F-ds This field, vector, as well by an amount F,ds be written in the form - ~ (11-44) as In words, the component of the force F in any direction equals the negative rate of change of potential energy with The position in that direction. hand its side of Eq. (1 1-44) is spatial derivative on the right- called a direclional derivative, because value depends on the direction in which ds is chosen at the we should be using the notation of partial derivatives: F, = —dU/ds.) If we move from P to a neighboring point on the same equipotential surface as that on which P lies, then dU/ds is zero for this direction. If, however, we move to a neighboring point not on the same equipotential point P. (Strictly speaking, surface as that containing P, dU/ds will be different from zero. That particular direction for which dU/ds has maximum its value at a given point defines the direction of the line of force at that point, dU/ds and the magnitude of maximum this value of the magnitude of the vector force at the point in question. is maximum This direction, is value of dU/ds, together with its associated called the gradient of the potential energy; it is vector directed at right angles to the equipotential surface. symbols, we write F = -grad U To a In (11-45) help clarify the idea of the gradient, consider a conservative field in two dimensions. Here the equipotentials are than surfaces as they would be in three dimensions. 1-25 we show two equipotentials, one for = U and lines, rather In Fig. 1 one for U= equipotential 465 U l/ -f AU. U + M/ Starting at P on U we by any of an , infinite can move to the number of Equipotential surfaces; potential-encrgy gradienls dis- _^\pyui* Fig. 11-25 StAT^l^' Different 1 >u. paths between two neighboring equipotentials. IH^HHHHHH placements. However, for a given change of potential energy one moves along a shortest possible displacement As. for this direction this the direction of the gradient of potential energy. is 1 is 1-25, where three directions AU, in the change of po- rate of energy with position from P and This is are shown. that As is shorter in length than either As! or As 2 and hence that AU/As is larger than It is clear IN The maximum tential indicated in Fig. MOTION change line of force to attain this AU/As or x AU/As 2 . CONSERVATIVE FIELDS We now turn our attention to the problem of the motion of particles in a conservative field of force. We have of course number of problems involving motion under gravity, but here we shall try to indicate the value of the energy method as it applies to more complex situations. already discussed a If the only force acting on a particle is that due to the the law of conservation of mechanical energy provides a step in solving for the motion. $mo 2 + U(x,y,z) where v 2 = vx whole story. + 2 It is vu + v, 2 first This law, expressed as = E 2 field, (H-46) , does not, however, contain the the result of combining three statements of Newton's law as applied to the three independent coordinate directions, and the synthesis of these three vector equations into one scalar equation involves the discarding of information that in principle is still available to us. Given that U is a conservative potential, we can always find its gradient and hence the vector With this, plus the initial conditions (values force [Eq. (I l—45)]. of r and v at of the problem Our t is = 0) everything that one needs for a solution provided. interest here, however, lics in taking Conservative forees and motion i n. advantage of the spacc energy-conservation equation as far as possible, and supple- menting it with whatever other information This extra information may may be necessary. take the form of the explicit use of one or two of the Newton's Iaw statements, as we the example about to be discussed. Or it may, shall see in in suitable circumstances, be contained in a conservation law for quantities besides energy — in momentum. angular particular, ploitation of this latter property The ex- one of the principal concerns is of Chapter 13. As a good example of the methods motion of a charged field. particle in a Suppose we have a pair of inside an evacuated tube Fig. we combined parallel shall consider the electric and connected to a battery as shown 11-26, so that a uniform electric field 8 between the magnet plates. The and magnetic metal plates mounted plates are placed We shall exists between the poles of a that produces a uniform magnetic field perpendicular to the page. (= V/d) in B in a direction assume that electrons start out with negligible energy and velocity from the lower plate. This could, for example, be arranged by giving the lower plate a photosensitive coating and shining light onto it, as in a commercial phototube. Consider one electron of charge q will = —e emitted at O. It be accelerated vertically by a constant force equal to eV/d directed along the positive y axis, path indicated in the figure. and will be deflected into the Since the magnetic deflecting force, always acting perpendicular to the particle's velocity, does no work and hence does not directly affect the particle's energy, the statement of energy conservation can be written simply as follows: Fig. 11-26 Motion of an electron in vacuum under the combined action of electric and magnetic fields. Adi Motion in conservative fields B into the page Fig. 11-27 *U4- Analysis ofthe magnelic force into componenlS associated with the separate x and y components 0/7. [= — (eV/d)y] equal to zero for y = the bottom plate where the kinetic energy at y = O where we have chosen i.e., at V O, is negligible. As statcd at the beginning of this section, we need more Since the electric force information to determine the motion. is directed along the y axis, the only x component of force on the electron is the x component of the magnetic force. To evaluate this force, think of the velocity vector of the electron at point of The component of magnetic situation some path resolved into x and y components (Fig. 11-27). its parallel to the is y force associated with vx in this axis and does not concern us here, but the velocity component v u gives rise to a magnetic force component Fx given by and Newton's law of motion for the x component of motion is accordingly evv B=m^ (11-48) Equations (11-47) and (11-48) can be solved for the motion of the particle, as follows: Since v„ eB dy = = dy/dt, Eq. (11-48) can be written as m do x or, integrated, mv x = eBy since vx = O when y (\\-49) = 0. Now we use the value of vx from Eq. (11-49) in the energy equation (11-47) and get 468 Conservative forees and mol ion in space nwy2 This is + T®V y = (11-50) of the form of the energy equation in one dimension; the total energy is eV zero and the effective potential energy U'(y) is given by e = - -jy U'(y) + hm ®" This effective potential-energy curve is precisely of the form of a harmonic on a point at a certain angular frequency O) = co is 1 shown in Fig. 11-28. It positive value of y) and its characteristic given by eB — m This implies a very interesting Fig. is oscillator potential (centered result. Equation (11-50) and 1-28 show that, as y increases from zero, the kinetic energy of the electron increases to some creases again. maximum value and then de- Mathematically, Eq. (11-50) defines a value of y given by putting v u = maximum 0: m V y»** - If this is less eB2d than the separation d between the plates (a condition that can be obtained, as the making B large enough), the above equation shows, by y displacement of the electron will perform simple harmonic motion with the angular frequency w and with an amplitude A equal to £.ymax . instant the electron leaves the lower plate, Fig. 11-28 Effective potential-energy curve for the y component of motion in the arrangement of Fig. 11-26. 469 Mol ion in conscrvativc fields Taking t we can put = at the Fig. 11-29 Cycloidal path ofelectron between charged parallel plates in a magnetic field. = y - /1(1 (ll-51a) cosco/) What about the x component of the motion? If we look at Eq. (1 1-49), we see that vx is proportional to y. Since y is always positive, vz is always in the positive x direction. When one couples this with the harmonic oscillation along y, one sees that the path of the electron shown in Fig. 11-29. dx — eB y y where w is the we have, from Eq. (11-49), oiy m dt a succession of rabbit-like hops, as is Specifically, same angular frequency, eB/m, that we introduced Thus, substituting the explicit expression for y from earlier. Eq. (ll-51a), we have dx = a>/4(l — cos tor) dt Integrating this and putting .v = A(ut - x = O at f = O we then find that (11-51 b) sinwr) Equations (11-5 la) and (ll-51b), taken together, show that the path of the electron is in fact a cycloid—just as rim of a wheel of radius A rolling along the x if it were on the axis. THE EFFECT OF DISSIPATIVE FORCES We field mentioned may be earlier useful how even forces are also present. the conservative properties of a force in circumstances in The slow decay which dissipative of the orbits of artificial earth satellites provides a particularly interesting example of this. We all know that a satellite placed in orbit a few miles above the earth's surface will eventually 470 Conservative forces and motion in hundred come down. The spacc Fig. 11-30 (a) Earth satellite spiraling inward as energy. it (b) loses Small element of the path sliown in (a). descent is, many thousands however, spread over Thus, although the actual path is as shown is very nearly a closcd orbit, which in Fig. for simpiicity. 1 l-30(a), the motion during a we single revolution will take to The statement of Newton's law thus given, with extremely little of revolutions. a continuous inward spiral, at be circular any stage is by the usual equation for error, uniform circular motion: F= GMm _ mo r r* Thus the kinetic energy at any particular value of r is given 2 „ G Mm K = %mu = by . (11-52) 2r This exposes a very curious feature. smaller, the kinetic energy increases speeds up. This happens the satellite is in spite — As the orbital radius gets in other words, the satellite of the fact that the motion of continually opposed by a resistive force. If there were no such force the orbital radius and the speed would remain constant. We know, however, that the and the amount of when we consider satellite the potential energy: U= - GMm 471 The eflcct must have this loss is well defined. of dissipative forccs lost energy, This becomes clear The total energy, E, when the orbit radius given by is r, is E-K+V--^ As r decreases, E becomes (11-53) more strongly negative. We may note the following relationships in any such circular orbit: E= -K=\U But (11-54) one may wonder: still Why does the the face of a resistive force? Figure The AB line of the The starting point A lies on a B lies on a circle of radius r + Ar satellite. We negative). sake of it We path. a feels then see that as the component of the of radius GMm — — r; (Ar being actually satellite travels from gravitational attraction along If the resistive force is R(p), the total force acting the satellite in the direction of As _ F= circle greatly exaggerate the magnitude of Ar, for the clarity. to B, its satellite accelerate in l-30(b) suggests the answer. represents a small segment, of length As, of the path the end point A 1 cos a — is on given by R(v) Thus, in the distance As, the gain of kinetic energy, equal to the work F As, given by is AK = F As = ^Âscosa In the first term on the right, R(v)As we can substitute — Ar As cos a = In the second term on the right, we can substitute As = v Al If we also put »10 AK = mu Av, we arrivc GMm — Ar — Ao = However, by . Usingthisresult, ( . , R(u)u At diffcrcntiation of both sides of Eq. (1 1-52) nwAv = - 472 „, at the following equation: we have GMm Ar . -2rT~ wecan substitute for the value of —(GMm/r 2 )Ar onscrvairvc forees and motion in space previous equation, and in the nwAv = 2mv&o — we get R(v)vAt Hence Ao = At R(v) m a most intriguing The This is The rate of increase of speed of the satellite magnitude of the tional to the motion! an inward GAUSS'S LAW is is not an error. directly propor- is resolved opposes its when we recognize by changing the orbital path from a circle to an agent that allows the gravitational work on the satellite, the amount of which is spiral, acts as do positioe numerically twice as great as the by the positive sign resistive force that The seeming paradox that the resistance, force to result. amount of negative work done resistive force itself. 1 The particular case of the inverse-square field of force point mass or charge is so important that brief discussion of Gauss's Iaw, which is we statement of the inverse-square law. Since cerned with mechanies here, we we are primarily con- shall diseuss the terms of the gravitational rather than the electric begin with a very elementary observation. gravitational field g M field acts at M. centered on field. As we have We seen, the given by the equation every point of the surface of a sphere of radius r If we use the convention that the positive direction is radially outward, then the radial by the plied is problem in GM .. This due to a point mass due to a append here a a compact and powerful will total surface area component of g multi- of the sphere gives us a quantity <f> defined as follows: <t> We = 47rr £r = -4irGM 2 shall call <t> the flux of the gravitational field g through the surface of the sphere. This 473 seetion may Gauss's law be omitted without loss of continuity. 11-31 (a) Gaussian surface enclosing a parlicle of Fig. mass M. There is a net gravitational ftux due to (b) Gaussian surface nol enclosing due to M' M. M'. The net flux is zero. Now a noteworthy property of is <j> that it is indcpcndent If we had two spheres of different radii, r and r 2 centered r. on M, the flux of g would bc the same through both. We can of , x extend this result to the case of a closed surface of any shape surrounding the mass [Fig. 11-31 (a)]. If at an arbitrary point outward normal to the surface makes an angle B with r, as shown, we can rcsolve g into orthogonal components, parallel to the surface and along the normal. If we think of the on this surface the flux literally as a is kind of flow of the g field aeross the surface, only the normal component that contributes. this it Multiplying by the element of area dS, we thus have d<$, = - — cos e dS can be equated to the projection of dS perpen2 and the quoticnt dS cos O/r is the element of solid Now dS cos 6 dicular to r, angle, dil, subtended at d<f> The M by dS. Thus we can put = -GMcKl total flux of g through the surface, as defined in this way, is then obtained by integrating over all the contributions dQ. But this means simply including the complete solid angle, Air. Thus we have -I 474 ,','o -GM dtt = -4irGM Conservative forees and motion in spacc exactly as for a sphere. M' mass If we consider the situation for a point outside the surface, the result quite different. is This time a cone of small angle dti intersects the surface twice; the contribution to the flux d<t> = -G M' is 'dSi cos given by 6i dS2 cos It is 2 r& n* not hard to verify that the two terms inside the parentheses — dQ are equal to which and -\-dQ, respectively, so that follows that the total flux it One can <f> is d<t> = 0, from also zero in this case. formalize these calculations a little by representing an element of surface area as a vector, dS, with a magnitude equal to dS and a direction along the outward normal, as The element of 11-31. Fig. to the scalar product g dS. • flux, d<f>, It is is mz, . , . • inside a surface of Their gravitational tional field g The = gi g at + fields = J and we see g2 g • + that this 11-32 (a) Basis gravitational flux due an arbitrary collec- of masses inside a given surface. (b) The tion addition of external masses does not change the net flux, although it will alter the local field intensities. 475 combine to produce a mu m 2< 1 l-32(a). resultant gravita- g3 -\ by dS of calculating the to kind, as in Fig. any point P: is the individual masses: Fig. some total gravitational flux is then given <t> in then apparent that, in Fig. dSi and g 2 dS 2 are of opposite sign. Suppose now that we have a number of masses, 11-31 (b), gi • shown then defined as equal Gauss's law simply the sum of the contributions from gi / • + dS <t> If, how it = — 4irG A/to • t may t»i + / g3 • • • rfS + •) total enclosed mass, be distributed (11-55) ai we suppose as in Fig. ll-32(b), m'2 H- depends only on the this total flux regardless of g 2 dS + ntz + = —4irG(mi Thus I that additional masses m[, are placed outside the surface, our previous calculation shows that these contribute nothing to the flux through the surface. pletely general result; and its force is validity it total gravltational Thus Eq. (11-55) emerges as a comis known as Gauss's law (or theorem), depends completely on the an exact inverse-square law. fact that the An holds for electrostatics, in which context Gauss, in developed it. If a physical system has certain theorem leads We at once to law of exact parallel to fact, it first obvious symmetries, Gauss's important conclusions about field a few examples, some of which we have already treated by other methods. strengths. shall consider APPLICATIONS OF GAUSS'S THEOREM Field outside a sphere Suppose we want to know the gravitational outside an isolated sphere of mass center [Fig. Fig. 1 1-33 1 1—33(a)]. If the M field at a distance Use of Gauss's law to calculale the gracitalional field of a solid sphere (a) at exterior points, and (b) at interior points. 476 a point P, r from its mass distribution within the sphere at Conservative forccs and motion in space is symmetrical— i.c, if the density of matter same the is at all points at the same distance from the center of the sphere— then the gravitational field g itself has the same strength at all points on a spherical surface of radius r. Thus, if we draw a "Gaussian surface" in the form of a sphere of radius tional flux # = is 4irr r, the total gravita- given by gr = -A-kGM 2 whence gr = The is GM - -5- at once result is of course entirely familiar, that, but the point to notice once we have Gauss's general theorem, the process of carrying out an explicit integration over the mass distribution, as we did in Chapter 8, becomes unnecessary. Field inside a sphere An exactly similar treatment holds for the other familiar problem of the gravitational distribution [Fig. that, if we field inside a spherically 11— 33(b)]. symmetric mass Gauss's theorem tells us at once imagine a spherical surface of radius r < R, the material outside this surface contributes nothing to the flux of g through the surface. surface of radius r Thus, m(r), is if mass enclosed within the the we can at once deduce that Gm(r) The assumption of spherical symmetry throughout outside as well as inside the radius Otherwise the value of g spherical surface, may be r, is, the system, however, essential. different at different points and even though the total flux still on a has no contribution from exterior material, the field strength at individual points will be affected. first example, that the sphere is Thus we must assume, effectively isolated, i.e., as in the far from any outside masses. Field due to afiat sheet Suppose we have an infinite sheet of matter, of surface density (mass per unit area). Symmetry in this case tells must be everywhere normal to the sheet and that 477 Applications of Gauss's theorem <r us that the field it has the same Use ofGauss's law to calculate the gravi- Fig. 11-34 talional field ofaftat sheet ofmatler. magnitude on the two sides Gaussian surface for this (Fig. 11-34). We can construct a problem in the form of a right cylinder, Then with ends of arbitrary shape but the same area A. no there g through the curved sides of the cylinder. The enclosed mass is just a A, so the total gravitational flux passing is flux of through the surface is equal to the field is — AtrGaA. strength But, by Gauss's theorem, this multiplied by the total area of the g two ends. Thus we have 2 Ag = -A-kGiA Therefore, — 2ttGV g = The (11-56) strength of the field thus entirely independent of the is distance from the sheet of matter, assuming this to be indeed of unlimited extent. In terms of gravitational systems, this example But the realistic. is not very result is highly relevant in electrostatics, where distributions of electric charge over large plane areas are fre- The quently met. and the constant parallel-plate capacitor field at all is a prime example, points between the charged plates Millikan experiment, for example, can at once be under- in the stood in the light of the above calculation. PROBLEMS 11-1 A particle of mass m slides curved into a vertical circle (see the can be described in lerms of the vertical distance h that (a) tion of 0, The position or in terms of the arc length has is of the particle s, or by fallen. Consider the force that acts on the particle along the direcpath at any point, and show that the work done on the its particle as it without friction on a wire that figure). it moves through an arc length ds dW = F t ds = mgR is given by sin d dd By integrating this expression from d = to 6 = do, show = mgR(\ — cos 0o), and express this result work done is (b) that the W in terms of the vertical distance h that the particle 11-2 478 Two blocks, of masses m and 2m, rest has descended. on two frictionless planes Conservative forccs and motion in space inclined at angles of 60° an string through and 30°, respectively, and are connected by a Using eyelet of negligible friction (see the figure). energy methods, find the magnitude of the acceleration of the masses, the tension in the connecting string. and deduce 11-3 down Show that a mass on the if in a circular horizontal (and the mass swing the tension starts A at rest), then at the lowest point is (This result book on pendulum 11-4 is in the string simply hangs there. his end of a string is allowed to swing in which the string is initially arc from a position three times as great as is quoted by Huygens at the mass end of clocks, published in 1673.) pendulum bob of mass m, at the end of a string of length /, rest at the position shown in the figure, with the string at At the lowest point of the arc the previously stationary block, of mass nm, that zontal surface. The What What What (a) (b) (c) (d) In collision is A released. velocity is • • • mi and m% momentum" pulled aside a horizontal distance Xi is sticks to it, and then decreases again to that energy is not conserved. in this experiment, so that experi- are suspended from strings you and which the combined mass after The student obviously not conserved in this proccss. is increases, after the collision, .) swings out to some final position x/. mentum pendulum does the string make? (Obtain to the vertical m? and It strikes a given to the block by the impact ? assigned a "conservation of is strikes the impact occurs? the tension in the string at this instant? Mass m\ /. bob just before is which two masses, of length bob frictionless hori- perfectly elastic. the oscillations of the student on a the specd of the your answer in the form cos 6 = in is is what maximum angle 6 ment of the the from 60° to the vertical. 11-5 when zero. objects that It is zero The student mo- initially, also claims Analyze carefully what actually goes on are in a position to answer the student's objections. 11-6 A small ball of putty, of mass m, is attached to a string of an upright on a wooden board resting on a horizontal table (see the figure). The combined mass of the board and upright is M. The friction coefficient between the board and the table length 479 / fastened to Problems The is fi. position. the ball swings How (b) What far does the is the 11-7 M, which An board move after the collision? minimum value that n must have to prevent the while the ball swings lcft gives the critical condition at 6 object of mass m on slides down? Assume ~ 45°. a frictionless loop-the-loop object is released from rest at a height h above the The apparatus. While down, the board does not move. (a) m <3C with the string in a horizontal rest the upright in a completely inelastic collision. board from moving to the that from ball is released It hits top of the loop (see the figure). (a) What the is magnitude and direction of the force exerted on the object by the track as the object passes through the point (b) Draw A? an isolation diagram showing the forces acting the object at point B, and find the magnitude on of the radial acceleration at that point. Show (c) that the object must start from h > r/2 to successfully complete the loop. < For h (d) fail away from the Show that this happens at where a is the angular dis- r/2 the object will begin to track before reaching the top of the circle. a position such that 3 cosa = 2 -|- 2h/r, tance from the (upward) vertical. 11-8 A mass ball of m hangs at the end of a string of length /. It is struck so as to start out horizontally from this position with a speed vo- What must (a) Do be if the ball is just able to travel through a complete circular path ? (b) If height 2/ if the vertical (c) no l-o = \/4g/ (so that the ball could just rise through the projected vertically), what angle does the string when the ball begins to fail away from its make with circular path? In situation (b), analyze the subsequent motion, assuming losses of total mechanical energy. A mass slides down a very smooth curved chute as shown. At what horizontal distance from the end of the chute does it hit the 11-9 ground ? 480 Conservativc forces and motion in spacc 11-10 A dome (see figure) daredevil astronomer stands at the top of his observatory wearing down velocity to coast roller over thc skates, dome and Neglecting friction, at what angle (a) starts with negligible surface. does he leave the dome's surface? (b) If would he he were to leave the start R = your answer for 11-11 peg figure. A is 8 break his m, and pendulum of length L The pendulum Ifa = is 60°,/3 (b) 481 use is g fail, =* far from thc base should in situation (a)? 10 m/sec held at Evaluate 2 . an angle a from the vertical. shown in the < Show 30°, is the maximum that the string will bucklc during thc bob's aseent (rcos/3 — buckling oecur? Problcms from rest. and r = Ly/3/4, what bob will reach ? released = angle with the vertical that the this what angle located in the path of swing of the string as (a) when initial velocity t»o, at For thc observatory shown, how (c) his assistant position a net to A with an dome ? Lcosa) < 3(L — r)/2. At what angle B does A carnival skill game involves such a pendulum and peg A prize is awarded for causing the bob to strike the peg. (c) apparatus. In the game, L, rest at r, and are fixed ; the contestant releases the any desired a. Find the value of a for which a prize (Express your answer in the form cosor 11-12 A frictionless airtrack y = \Cx 2 An . = • • • bob from will be won. .) deformed into the parabolic shape is object oscillates along the track in almost perfect simple harmonic motion. 30sec, by what distance (a) If the period is x = ±2 m (b) If the amplitude of oscillation x direction ? of uniform circular motion 11-13 A how many height of 20 m T = above 2 m, how long does it is x = —0.5 useful for this.) pendulum clock keeps proximately is m to x = +1.5 m traveling (The description of SHM as a projection take for the object to pass from in the positive the track at is = 0? higher than at x perfect time at seconds per week would = level. Ap- gain or lose at a (Assume a simple pendulum, with this level? 2Tr\/I/g. Earth's radius ground it 6.4 X 6 10 m. 1 week ~ 6 X 10 5 sec.) 11-14 Two identical, and equally charged, ping-pong balls are hung from the same point by strings of length L (see the figure). (a) Given that the mass of each is m and that the equilibrium position is as shown in the figure, what is the charge on each? (Recall that the electrostatic force between two charges at a separation r is given by F= (b) m.q m,q 2 kqiq2/r' .) Suppose that the other by the same amount. balls are displaced slightly Make describe the subsequent motion. toward each appropriatc approximations and Find an expression for the frequency of oscillations. (c) On the other hand, if the ping-pong balls are displaced same direction and by the same amount, what will the subsequent motion be? In this second case, does it matter whether they are charged or mcrely held apart by a massless rod? slightly in the 11-15 (a) Show that free fail in the earth's gravitational field infinity results in the same would be achieved by a 482 from velocity at the surface of the earth that free fail from a height H= RB (= Conservative forces and motion in space radius of earth) under a constant acceleration equal to the value of g at the earth's surface. that the speed at the earth's surface of Show (b) dropped from |= v height,/; - « RK H (2*A)i« (c) Verify the i) statement in the text (p. 454) that the speed needed for escape from the surface of a gravitating sphere where co is t'o\/2, skimming the surface of the sphere the speed of a particle is object given approximately by ) is (/i an in circular orbit. 11-16 A physicist plans to determine the mass of the earth by ob- pendulum as he deseends a mine radius R of the earth, and he measures the den- serving the changc in period of a He knows shaft. the ps of the crustal material that he penetrates, the distance h he deseends, and the fractional change (AT/T) in the pendulum's period. sity (a) What is the mass of the earth in terms of these measurements and the earth's radius? (b) Suppose that the mean over-all density of the earth the mean density of the portion above 3 km. is twice (This supposition is actually rather accurate.) How many seconds per day would a pendulum clock at the bottom of a deep mine (3 km) gain, if it had kept time accurately at the surface? 11-17 The figure shovvs centric spherical shells. a system of two uniform, thin-walled, con- The smaller shell has radius R and mass M, and mass 2M, and the point P is their point mass m is situated at a distance r from P. the larger has radius 2/? A center. (a) in What is the gravitational foree F(r) these shells exert each of the three ranges of (b) mass What m when is infinitely it is far r: < can pass what at < R, P? (Take Problems < 2R, r on m > 2R? when m speed when it from rest very far away from the reaches P'! (Assume that the particle through the walls of the shells.) moon to be a sphere km and mass 7.3 X 10 22 kg. of uniform density with moon so as to connect any two smooth tunnel 483 r the potential energv to be zero 11-18 Assume the radius 1740 R < from P.) is its freely r the gravitational potential energy of the point is (c) If the particle is released spheres, common is bored through the Imagine that a straight on points its surface. Show (a) that the motion of objects along this tunnel under the action of gravity would be simple harmonic. (b) Calculate the period of oscillation. Compare (c) moon the with the period of a this satellite traveling 11-19 The escape specd from the surface of a sphere of mass R radius around n a circular orbit at the moon's surface. i is ve = (2GM/R) 1 '2 ; the mean speed M and v of gas molecules of mass m at temperature T is about (3kT/m) l/2 , where k (= 1.38 X 10 -23 J/°K) is Boltzmann's constant. Detailed calculation shows that a planetary atmosphere can retain for astronomical times (10 9 ycars) Using the data below, find only those gases for which D J$ 0.2y e which, if any, of the gases H2, NH3, N2, and CO2 could be retained . for such periods by the earth; by the 6.0 X 10 24 kg, RE = M = Mu/Si, R = R = 0.53*s , and 11-20 A 6.4 0.27Rn, and T - 0.8TE ME moon; by Mars. = TB = 250°K. For the moon, T = TK For Mars, M = O.llAfe, 10 3 km, X . . double-star system consists of two stars, of masses 2M, separatcd by a distance D center to how the gravitational potential energy center. Draw a M and graph showing of a particle would vary with position along a straight line that passes through the centers of both What can you stars. about the possible motions of a particle infer along this line? 11-21 A of mass double-star system M and distance of 5R. is composed of two identical stars, each radius R, the centers of which are separated by a A particle leaves the surface of nearest to the other star and escapes (a) Ignoring the orbital to an one star at the point effectively infinite distance. motion of the two stars about one another, calculate the escape speed of the particle. (b) Assuming one another (like the that the stars always present the moon same face to to the earth), calculate the orbital speed of the point from which the particle is emitted. How would the existence of this orbital motion affect the escape problem? 11-22 The earth and the 3.84 X 10 8 m. moon are separated by a distance Ignoring their motion about their common mass, but taking into account both of their gravitational D = center of fields, answer the following questions: (a) How much work must km above to reach a height 1000 of the be done on a 100-kg payload for it the earth's surface in the direction moon ? (b) Calculate the diiTerence in potential energy for this mass between the moon's surface and the earth's surface. (c) Calculate the necessary payload to the moon. 484 It initial kinetic energy to get the must be greater than the potential difference Conservalivc forces and motion in space given in (b) because the payload must overcome a "potential Find the location of the top of difference 11-23 from the earth Two stars, to hill." the potential it. M, each of mass orbit about their center of mass. planetoid of common orbit is mass m (« M) happens system (the line The radius of and compute this hill r (their separation their to move along is A 2r). the axis of the perpendicular to the orbital plane which intersects the center of mass), as shown in the figure. (a) Calculate directly the force exerted on the planetoid if it is displaced a distance z from the center of mass. (b) Calculate the gravitational potential this displacement z and use Find approximate expressions for the potential energy and (c) zy> the force in the cases Show (d) r and z « r. that if the planetoid center of mass, simple harmonic TP is displaced slightly from the motion occurs. Compare the period of this oscillation with the orbital period 11-24 A energy as a function of to verify the result of part (a). it frictionless wire is 7"o of the binary system. stretched between the origin and the point x = a, y = b in a horizontal plane. A bead on the wire starts out from the origin and moves under the action of a force that varies with position. point Find the kinetic energy with which the bead arrives at the (a, b) if The components of the force are Fx = k\x, Fy = kiyThe components of the force are Fx = kiy, F„ = (a) (b) A: 2* (with ki, *2 >0). In one case the force is conservative, in the other case considering a different path (a, 0) to (a, b) — e. g., — verify which is from (0, 0) to (a, 0) it is not. By and then from which. 11-25 The nuclear pari of the interaction of two nucleons (protons or neutrons) is described pretty well by a potential V(r) is greatcr than 1 F (1 F = = 1.4 F and X = 70 MeV-F. (The /-o proposed by H. Yukawa and is named after him.) when = —\e~ r/ro/r 10~ ls the separation r expression, m). In this potential was Find an expression for the nuclear force that acts on two nucleons separated by r > 1 F. (b) Evaluate this force for two protons 1.4 F apart, and com(a) (each of) pare this with the repulsive (c) 485 Coulomb force at that separation. Estimate the separation at which the nuclear force has Problem s — 1% dropped to of value a t r its = What 1.4 F. is the Couiomb force at this separation? The (d) why last result indicates the Note part of our macroscopic experience. range of distance over which the interaction potential is not is important. In contrast and Couiomb forces have been to the nuclear force, gravitational called forces of infinite range. Yukawa that ro characterizes the Indicate why such a description is appropriate. is shown a plot of gravitational equipotentia!s two spheres of masses 2M and M. One way of 11-26 In Fig. ll-24(b) and field lines for developing this diagram as follows. is P is given any point potential at n and equipotential r 2 are An measurcd from the centers of the spheres. by defined is that the gravitational G(M) G(2M) where We know by <pp = such equipotcntials, draw a set of In order to construct constant. circles, with the sphere of mass M as center, such that the values of 1/ra are in arithmetic progression IA2 = e.g., 2M 1, 1.5, 2, Then with 2.5, 3, 3.5 the sphere of mass as center, draw a set of circles with radii r\ twice as great as the ~ + (2/n) (IA2), the corresponding circles P and M. The in each pair represent equal contributions to <p by values of r 2 Since . <p 2M two sets of circles have a number of intersections, and each corresponds to a well-defined value of 2.85 units But 5.5. and 10n = <pp ~ circles as follows: /•2 = n have <pe ~ we (2/1.0) + 5.5 is obtained also at the intersections of 0.67, 10 units, <fp. = value of (2/n) 0.50. + r2 = n = 0.33, intersection For instance, with 10r2 0.80; r 2 = (1/0.285) = = of other pairs 0.40, n = 0.67; same and other equiOnce the equipo- Joining these intersections having the (1/ra) gives us an potentials can be constructed in the tentials are obtained, the field lines equipotential, same way. can be drawn as lines everywhere normal to the equipotentials. 11-27 A science fiction story concerned a space probe that was re- trieved after passing near a neutron star. that rode in the probe was found The unfortunate monkey The conclusion to be dismembered. reached was that the strong gravity gradient (dg/dr) which the probe had experienced had pulled the monkey apart. Given that the probe had passed within 200 km of the surface of a neutron star of mass 10 33 g (about half the sun's mass) and radius 10 km, was this explanation reasonable? (The probe was in free fail at the time.) 11-28 A straight chute is to be used to transport articles a given /. The vertical drop of the chute can be freely The articles are to arrivc at the top of the chute with negligiblc horizontal distance chosen. 486 Conservativc forces and motion in. space and minimum. velocity the chute is to be chosen such that the transit time what (a) If the surfacc is frictionless, for which the time (b) tion is What a the angle of the chute is minimized? is the corresponding angle if the coefficient of fric- is m? (c) If a playground were designed to give minimum dura- slide tion of ride for given horizontal displacement, of friction of the child-slide surface make is 0.2, and if the coefficient what angle would the slide (Ignore the curved portion at the lower with the horizontal? end of the is slide.) (d) If the optimizaiion problem of (a) is encountered and if curved chutcs are allowed, can you guess roughly what form the best design would have? 11-29 The magnetic force exerted on a particle of charge q and mass traveling in a uniform magnetic field B (which is constant in time) m is F = kqs X B (where the constant k = 1 in the MKS system). (a) Show that the work done by such a force on the particle is given by zero for arbitrary particle motion. (b) Situations which a moving charged in arise to the instantaneous velocity. particle is FD and of direction opposite (For example, the ionization of atoms slowed by a force of constant magnitude along the track of a charged particle produces an almost constant energy loss per unit distance, corresponding to a constant retarding force.) its Show speed will that a particle has speed eo at if be given by u(t) = vo — will of a magnetic not affect this (c) an Fnt/tn until moo/kgB with 0, t = thcn for / > /huo/Fd, after result. Under the action of a magnetic initial velocity = remain motionless. Note that the presence which time the particle field will / force alone, a particle with vo (normal to B) describes a circle of radius r a circular frequency o = kgB/m. Show subject also to the force described in (b) moves inward along a and that the number of in the spiral before circuits it makes = that a particle it spiral comes to rest is given by {kqBoa/2irFti). 11-30 (a) Obtain an expression for the gravitational thin disk (thickncss d, radius R, the center of mass of the disk. surface density a = pel to and density field due to a p) at a distance In calculating the field, h above assume the be concentrated in a disk of negligible thickness a distance h beneath the test point. This will give an accurate result whenever h^> Note that for and reduces the complexity of the d, R —* « , Gauss's theorem, as it caleulation. the caleulation agrees with the predietion of must. (b) Express the field obtained in part (a) as a fraetion of 2irGcr R = 2h, R = 5h, and R = 25h. How many seconds per year would for the cases (c) 487 Problcms a pendulum clock gain when suspended with bob the 5 cm above a lead floor 1 cm thick, if it keeps correct time in the absence of the floor? 11-31 Any mass of mattcr has a gravitational "self-energy" arising from the gravitational attraction among its parts. (a) Show that the gravitational self-energy of a uniform sphere of mass M and radius R is equal to -3GM 2 /5R. You can do this by calculating the potential energy of the sphere directly, integrating over the interactions between all possible pairs of thin spherical either shells within the sphere (and remembering that each twice in this calculation) or, perhaps more shell is counted simply, by imagining the sphere to be built up from scratch by the addition of successive layers of matter brought in from infinity. (b) Calculate the order of energy of the earth. magnitude of the gravitational Check where this lies on the displayed in Table 10-1. 488 Conservativc forces and motion in self- scale of energies space Part III Some special topics Whenfirst studying mechanics one has the impression that everything in this and sett/edfor existence three branch ofscience all time. is simple, fundamental One would hardly ofan important clue which suspect the no one noticedfor hundred years. The neglected clue is connected with — that ofmass. one of the fundamental concepts of mechanics A. EINSTEIN AND L. INFELD, The Evolution ofPhysics (1938) 12 and Inertial forces noninertial frames imagine that you are You on a very smooth road. sitting in a car are holding a heavy package. The car cannot see the speedometer from where you is moving, but you sit. Ali at once get the feeling that the package, instead of being just a you dead weight on your knees, has begun to push backward horizontally on you as well. Even though the package anything except yourself, the effect applied to it is as is if not in contact with a force were being and transmitted to you as you hold it still with respect to yourself and the car. If you did not restrain the package it would in fact be pushed backward. You notice what happens to a mascot that has been hanging at in this way, that this is the end of a previously vertical string attached to the roof of the car. How do you intcrpret these observations? If you have any previous experience of such phenomena, you will have no hesitation in saying that they are associated with an increase of velocity of the car — with a positive acceleration. Even if this were you had a well-developed acquaintance with Newton's laws, you could reach the same your first i.e., experience of this type, but conclusion. An acceleration of the car calls for an acceleration of everything connected with requires, through F = by your hands. Nonetheless, itself is if somehow it; the acceleration of the package ma, a force of the appropriate it does size supplied feel just as if the package subjected to an extra force— a "force of in- 493 —that ertia" change the comes into play whenever the motion of an object. effort is made to state of class. They can be phenomena as the motion of a Foucault These extra forces form an important held responsible for such pendulum, the effects in a high-speed centrifuge, the so-called on an astronaut during launching, and the preferred g direction of rotation of cyclones in the northern and southern forces These forces are unique, however, hemispheres. some other that one cannot trace their origins to as was possible in the sense physical system, for all the forces previously considered. tional, electromagnctic, Gravita- forces, for example, and contact have their origins in other masses, other charges, or the "contact" of But the additional forces that make another object. pearance when an object is physical objects as sources. Are these their ap- being accelerated have no such inertial forces real or not? That question, and the answer to it, is bound up with the choice of reference frame with respect to which we are analyzing the motion. Let us, therefore, begin this analysis with a feminder of dynamics from the standpoint of an unaccelerated frame. MOTION OBSERVED FROM UNACCELERATED FRAMES An unaccelerated reference frame belongs to the class of reference frames that we have called inertial. We saw, in developing the basic ideas of dynamics in Chapter 6, that a unique importance and which Galileo's law of one such frame has been interest attaches to these frames, in We inertia holds. have seen how, if any other frame having an arbitrary constant velocity to the first is also inertial, and our inferences about the identified, relative forces aeting on an object are the same To a good first in both. approximation, as we know, the surface of the earth defines an inertial frame. So also, therefore, system moving at constant speed over the earth. was the first does any Galileo himself person to present a elear recognition of this faet, and one aspect of it that he diseussed is useful as a starting point for us now. In his Dialogue Concerning the Two World Systems, in which he advocated the Copernican view of the solar system in preference to the Ptolemaic, Galileo pointed out that a rock, dropped from the top of the mast of a ship, always lands just at the foot of the mast, whether or not the ship is moving. Galileo argued from this that the vertical path of a falling object does not compel one to the conclusion that the earth 494 Inertial forces and nonineriial frames is stationary. Fig. 12-1 (a) Para- bolic trajectory under grauity, as observed in the eanh's reference frame. The velocily v o zontal, (b) initial is hori- Same motion observed from a frame wilh a horizontal velocity greater than vo. (c) Same motion observed from The comparison here is between an object falling from rest relative a frame having bot h horizontal and vertical to the earth and another object falling from rest relative to the ship. If we considered only an object that starts from rest relavelocily components. moving tive to a shtp, frame and parabolic its path would be vertical in the earth's considered an object projected with relative to the earth, its in the ship's More generally, if we some arbitrary velocity frame. subsequent path would have diverse shapes as viewed from different inertial frames (see Fig. 12-1) but of them would be parabolic, and all lyzed, would show that the all of them, when ana- had the vertical acfrom the one force F„ (= mg) due to Let us now contrast this with what one finds if the falling object celeration, g, resulting gravity. reference frame itself has an acceleration. MOTION OBSERVED FROM AN ACCELERATED FRAME Suppose that an object is released from rest in a reference frame that has a constant horizontal acceleration with respect to the earth's surface. Let us consider the subsequent motion as it appears with respect to the earth and with respect to the accelerating frame. We x up two shall take the direction of the positive axis in the direction of the acceleration and will set rectangular coordinate systems: system S, at rest relative to the earth, and S', fixed i n the accelerating frame (Fig. 12-2). Take +v (relative to 5) s- Fig. 12-2 ordinales frames Relationship ofa particle that are in accelerated relative motion. 495 in of coiwo T —fP O Motion observed from an accelerated frame +x the origins of the frames to coincide at S at the velocity of S' with respect to The x' = 0, y = / = What undergoing 12-3(a)]. = t 0, and suppose that upward, which x = positive from a point for h. S and will the trajectories in observer in S, is released at is = two systems are taken as vertical axes of the and the object t this instant is equal to Vo- we already know For an S' look like? the answer. To him, the object free fail with initial horizontal velocity v [Fig. Thus we have {x = (As observed i n S) vot y- h- 2 i** These two equations uniquely define the position of the object at time t, but to describe the motion as observed in S' we must express the results n terms of the coordinates x' and y' as meai To sured in S'. x = x y = y where x s is and — transform to the S' frame, = the separation along the vot We know + §at x axis of the origins of .au a-in S') ca (As observed we K= W \ find Vot " = h - (" ' + 2 %a,2) = ~% a ' „ , \gi- Thus the path of the particlc as observed given by the equation -"{h- in S' is a straight line y') 8 Fig. 12-3 (a) Para- bolic trajectory of a . . . . : particle under gravily, ^s as observed in the earlh's reference frame S. (b) frame N \ Same motion observed in S' that has - \ a O a Seen constant horizontal acceleration. in S (a) 496 Inertial forces S that 2 Substituting these values x'= substitute x, S' (see Fig. 12-2). x, we and noninertial frames This shown is Fig. in In the accelerated frame, the 12-3(b). object appears to have not only a constant downward component of acceleration due to gravity, but also a constant horizontal component of acceleration a nonvertical particle to follow simple example which causes the straight-line path. [A similar monkey-shooting drama described the is — x direction in the The outcome becomes almost Chapter 3 (p. 104). we choose to describe the events in the rest in self-evident if frame of the falling monkey. In this frame the bullet just follows a straight-line path directly toward the monkey, while the ground accelerates upward at 9.8 m/sec 2 .] There no mystery about the unfamiliar motion repre- is sented by Fig. 12-3(b). It is of describing the normal free-fall We itself accelerated. a direct kinematic consequence motion from frame that a is could perfectly well use this path, described by measurements made entirely within S', to discover the acceleration of this frame, provided that the direction of the true vertical were already known. However, a greater interest attaches to learning about the acceleration through dynamic methods. That is the concern of the next section. ACCELERATED FRAMES AND INERTIAL FORCES From what has bcen said, very special status. that it is AU it is inertial clear that inertial frames have a frames are eguivalent in the cover their motions motions are comes what There in significant. is called the any absolute sense— only Out of Newtonian this their relative dynamical equivalence principle of relativity: no dynamical observation that leads us to prefer one is frame to another. Hence, no dynamical experiment us whether we have a constant velocity through space. inertial tell sense impossible by means of dynamical experiments to dis- As we have two frames just seen, however, a relative acceleration is dynamically detectable. As observed frames, objects have unexpected accelerations. It will between in accelerating follows at once, Newton's law establishes a link between force and acceleration, that we have a quantitative basis for calculating the magnitude of the inertial force associated with a measured since Conversely, and more importantly, we have a dynamical basis for inferring the magnitude of an acceleration from the inertial force associated with it. This is the underlying acceleration. 497 Accelerated frames and inertial forees principle of all the instruments known They as accelerometers. function because of the inertial property of some physical mass. To makc the analysis explicit, consider the motion of a particle P with respect to two reference frames like those considered in the last section and S and an frame x = X y = The 1 shown accelcrated frame S'. in Fig. 12-2: We then an inertial have, once again, + x, y' velocity components of P as measured in the two frames are thus given by where vs = dx s any particular we can put d s = v /dt at acccleration a, constant acceleration is not at all If instant. + S' has a constant but the condition of at, necessary to our analysis. Taking the time derivatives of the instantaneous velocity components, we then get az Oy where = = a'x + a, a'y a, is the instantaneous acceleration of the frame S'. Although we have chosen to introduce the calculation of Cartesian components, it is relates the acceleration a of P, as tion a' as measured in terms clear that a single vector statement measured in S' together in S, to its accelera- with the acceleration a, of S' itself: a = (12-1) a' Multiplying Eq. (12-1) throughout by m, hand we recognize the side as giving the real (net) foree, F, that particle, since this defines the true measured in an inertial frame. That cause of is, in the its S is left- aeting on the acceleration as frame, F = «n (12-2) but, using Eq. (12-1), this gives us F = »ra' + We now come (12-3) ma, to the crucial question: How do we interpret Eq. (12-3) from the standpoint of observations made within the 498 Inertial forees and noninertial frames accelerated frame S' itself? — that the net force on an object is the Newton's viewpoint (F nct = ma)- is so deeply ingrained accelerated motion cause of in we our thinking that relationship at all are strongly motivated to preserve this When we observe an times. object accelerating, we interpret this as due to the action of a net force on the object. Can we achieve a mathematical format that has the appearance °f Fnct = wa Yes. reference? and f° r tne present case of an accelerated frame of By transferring all terms but resultant F', which is of the the observed acceleration left a' = F - ma, = ma' F' The ma' to the on m, and have a correct magnitude to produce just treating these terms as forces that act (12-4) net force in the S' frame force equal to —ma frame of reference which has s, itself Fx which case the is particle, as result expressed matical trick. From — ma to , and a parts: a "fictitious" origin in the fact that the An important F is observed in S', moves under alone. s by Eq. (12-4) is not mercly a mathe- the standpoint of an observer in the accelerating frame, the inertial force took steps Fy that in which the "real" force the action of the inertial force The its and has the acceleration +a„. special case of Eq. (12-4) zero, in made up of two thus is "real" force, F, with components actually present. is keep an object "at rest" in S', by tying If it one down with springs, these springs would be observed to elongate or contract in such a way as to provide a counteracting force to balance the inertial force. is therefore To describe such a force as "fictitious" somewhat misleading. One would like to have some convenient label that distinguishes inertial forces from forces that arise from truc physical interactions, and the term "pseudo- force" is oftcn used. Even this, however, does not do justice to such forces as experienced by someone accelerating frame. name, who Probably the original, "inertial force," which is free is actually in the strictly technical of any questionable over- tones, remains the best description. As an illustration of the way in which the same dynamical may be described from the different standpoints of an situation inertial frame, on the one hand, and an accelerated frame, on pendulum suspended from the roof the other, consider a simple of a car. The mass of the bob is m. In applying F = standpoint of a frame of reference 499 S ma from the attached to the earth Accelerated l'rames and inertial forces 12-4 Fig. acting on mass Forces a suspended in (a) a slationary car, (b) a car moving al consianl velocity, and (c) a car imdergoing a posiliue acceleration. (assumed nonrotating), one can draw isolation diagrams possible motions of the car as there are just two shown in Fig. 12-4. on the bob: (real) forces acting of gravity, and T, the tension in the string. Cases not involve acceleration and the application of F In (c), the bob undcrgoes string hangs at rto 7",). The isolation diagram F = ma as follows: Vertical component : F„, the foree (a) and = ma (b) do is trivial. acceleration toward the right and the an angle with some increase Horizontal component: for the In each case, in its tension (from of Fig. 12-5 (a) leads us to apply 7"i Ty cos 6 sin0 = ma — mg = In the S' frame, however, because of the acceleration of the frame, there will be an additional foree of magnitude ma direetion opposite to the acceleration of the frame. in the Figure 12-5(b) shows an isolation diagram for the bob as seen in S'. The bob is (because a' Fig. 12-5 in equilibrium. = Here, application of F' 0): Forces on an object thai resi relaliue lo is al an acceleraled car (a) as judged in an inerlial frame, and judged (b) as in the acceleraled frame. 500 Inerlial forces and noninertial frames = ma' gives Ti sin S 7"i cos 8 - ma = — mg = O Thus the equilibrium inclination of the pendulum is defined by the condition tan 9 = - (12-5) g ACCELEROMETERS The for result expressed by Eq. (12-5) provides the a simple accelerometer. vertical direction, If we have representing e = 0, first theoretical basis established the true the observation of the angle of inclination of a pendulum at any subsequent time tells us the value of a through the equation a = g tan For example, chain, /2-5 (a) Tie Ffe. hanging if a passenger in an airplane lets his freely the beginning of the Quantitatwe accelerometer based on measuring the eguilibrium angle of a simple (c) Car- penter's level (in Ihis case a pivoted marker immersed in liquid of greater densily) can be used as an acceleromeler. (d) A bubble irapped in a curved tube of liguid gices direct readings acceleralion. his fingers almost constant [Fig. 12-6(a)]. accelerated cehicle. plumbline. from tie, of Tliis form of accelerometer was deuised by W. V. Walton (Edueation Research Center, M.I.T.). Figure 12-7 shows an example of its use. 501 Accelerometer* or a key- during the takeoff run, he can make a rough estimate of the acceleration, which in equilibrium within an (h) hang d is usually If he also records the time from run to the instant of takeoff, he can obtain 10 20 30 40 50 70 60 /, 80 90 100 110 120 130 140 sec 12-7 (a) Record obtaimd with the accelerometer ofFig. 12-6(rf) before and afler takeoffofa commercial jet aircraft. The accelerometer was held seas to record the horizontal component of acceleration only. Not e the sharp decrease in a at takeoff. (b) Graph ofvelocity versus time, obtained by graphical integration of(a). (c) Graph of distance versus time, obtained by Fig. graphical integration of(b). good estimate of the length of the run and the takeoff speed. If he is more ambitious, he can go armed with a card, as a fairly marked out as a goniometer (= angle in Fig. 12-6(b), already measurer) or even directly calibrated in terms of acceleration measured in convenient units (e.g., mph per second). ' Another simple accelerometer carpenter's level immersed is made obtainable readymade in the form of a of a small pivoted float that in a liquid [Fig. 12-6(c)]. AH is completely these devices make use of the fact that the natural direction of a plumbline in an accelerated frame is defined by the combination of the gravitational acceleration vector g the frame and the negative of the acceleration a of itself. A quite sensitive accelerometer of this same basic type, with the further advantages of a quick response and a quick attain- ment of equilibrium (without much overshoot or can be made by curving a piece of arc and filling it oscillation) plastic tubing into a circular with water or acetone until only a small bubble remains [see Fig. 12-6(a)]. Figure 12-7 (a) shows the record of acceleration versus time as obtained with such an accelerometer during the takeoff of a jet aircraft. Figures 12-7(b) and (c) show the results of numerically integrating this record so as to obtain and the the speed total distance traveled. Accelerometers of a vastly more sophisticated kind can be made by using very sensitive strain gauges, with electrical measuring techniques, to record in minute detail the deformations of elastic systems to which a mass shows is attached. Figure 12-8 schematic form the design of such an instrument. If in the object on which the accelerometer is mounted undergoes an by the pendulum bob acceleration, the inertial force experienced begins to deflect trical two of which it. This, however, unbalances slightly an elec- capacitance bridge in which the pendulum forms part of An error signal is obtained used both to provide a measure of the acceleration and the capacitors, as shown. is to drive a coil that applies a restoring force to the pendulum. Such an accelerometer unit may have a useful range from about -5 "g" 10 to more than 10 g. 'A book entitled Science (Ballantine, New for the Alrplane Passenger by Elizabeth A. Wood York, 1969) has such a goniometer on its back cover and a discussion of its use in the text. in The book also describes a host of other ways which airplane passengers can discover or apply scientific principles during their travels. 503 Accelerometers Integrating circuit Amplifier Signal generator Bridge Fig. 12-8 Electro- unbalance mechanical acceler- signal omeler system. ACCELERATING FRAMES AND GRAVITY In our discussions of accelerated frames, we have assumed all — know "which way is up" i.e., they know the and magnitude of the force of gravity and treat it that the observers direction we have done) (as mass of the as a real force, whose source is the gravitating But suppose our frame of reference to be a earth. room with no access to the external surWhat can one then deduce about gravity and inertial through dynamical experiments wholly within the room? completely enclosed roundings. forces We shall suppose once again that there frame, S, attachcd to the earth. he is able to verify that the dropped from rest and hence surface is is downward along a is an observer in a is not isolated; This observer acceleration of a particle line perpendicular to the earth's directed toward the center of the earth. ' He is able to draw the orthodox conclusion that this acceleration is due to the gravitational attraction from the Iarge mass of the earth. Our second observer is shut up in a room that defines the frame We are S'. Initially it is known that the fioor of his 504 We is ignoring the rolation of the earth, which causes this statement to A falling object does nol fail exactly parallel to a plumbshall come back to this when we diseuss rotating reference frames. slill be not quite correct. line. room Incrlial 101 •ccs and non inertial frames' . horizontal and that its In subsequent mea- walls are vertical. suremcnts, however, the observer in S' finds that a plumbline hangs at an angle and vertical, what he had previously taken to be the to dropped from that objects rest travel parallel to The observers in S and S' report their findings to one another by radio. The observer in S' then concludes that he has three alternative ways of accounting for the component his plumbline. of force, parallel to the floor, that is now exerted on all particles as observed in his frame: In addition to the gravitational force, there 1 force in the the — x direction is due to the acceleration of an his inertial frame in +x direction. 2. His frame has been set is down not accelerating, but a large massive object in the —x direction outside his closed thus excrting an additional gravitational force on all room, masses in his frame. 3. His room has been tilted through an angle e and an extra mass has been placed beneath the room gravitational force. (This to increase the net close to being just a variant of is alternative 2.) In supposing that all three hypotheses work equally well to we must assume that the additional postulated in alternatives 2 and 3, produccs an explain what happens in S', massive object, uniform gravitational effectively field throughout the room. From dynamical experiments made entirely within the closed room, there is no way to distinguish among these hypotheses. The acceleration of the frame of reference produces effects that are identical to those of gravitational attraction. Inertial gravitational forces are both proportional to the mass of the The procedures object under examination. measuring them are scribablc i field) that An for detecting and and Moreover, they are both de- identical. n tcrms of the properties of a field (an acceleration has a certain strength and direction at any given point. object placed in this field experiences a certain force without benefit of any contaet with interesting parallel, or does its it surroundings. Is all this just an have a deeper significance? Einstein, after pondering these questions, concluded that there was indecd somcthing fundamental here. In particular, the completely exact proportionality (as far as could be determincd) between gravitational force and that 505 no inertial mass suggested to him physical distinetion could be drawn, at least within a Accelerating frames and eravity Fig. 12-9 (a) Apple falling inside a (b) Indistinguishable Ihe earth. is inside on rests an acceleraled box i n ouler space. limited region, (a) box t ha I molion when Ihe apple between a gravitational field and a general acceleration of the reference frame (see Fig. 12-9). He announced 1 this— his famous principle of equivalence— in 191 1. The proportionality of gravitational force to inertial mass now becomes an exact necessity, not a n empirical result. It is and inevitably approximate also implied that anything traversing a gravitational must follow a curved path, because such a curvature would appear on purely kinematic and geometrical grounds if we replaced the gravitational field by the equivalent acceleration of our own reference frame. I n particular, this should happen with field (b) rays of light (see Fig. 12-10). With the help of these ideas Einstein proceeded to construct his general theory of (as we pointed out in Chaptcr relativity, which 8) is primarily a geometrical theory of gravitation. Fig. 12-10 Successive stages in ihe path of a horizontally Iraveling object as observed within acceleraling certically upward. eauicalence ofgraviiy an enclosure This illustrates ihe and a general acceleration of the reference frame. 'A. Einstein, Ann. Phys. (4) 35, 898 (1911), reprinted in translated form in The Principle of Relaticiiy (W. Perrett and G. B. London, 1923 and Dover, New York, 1958. 506 Inertial forces Jeffery, translators), and noninertial frames Methuen, CENTRIFUGAL FORCE We now shall consider a particular kind of inertial force that always appears if the motion of a particle described and is analyzed from the standpoint of a rotating reference frame. This force — the centrifugal force — familiar to us as the force with is if we whirl we shall consider which, for example, an object appears to puli on us it around at the end of a string. ' To introduce it, a situation of just this kind. Suppose that a "tether ball" is being whirled around in horizontal circular motion with constant speed (Fig. We shall analyze the motion of the points: a stationary frame S, ball as seen venience, we S will that rotates be designated w The and z' axes rotational speed of S' relative Figure 12-11 shows the (in rad/sec). analysis with respect to these For con- ball. align the coordinate systems with their z (as well as origins) coincident. to from two view- and a rotating frame S' with the same (constant) rotational speed as the 12-11). two frames. The essential conclusions are these: 1. From the standpoint of the stationary (inertial) frame, the ball has an acceleration The force, F r ing cord, and (In S) 2. , 2 toward the axis of rotation. r) to cause this acceleration is supplied by the tether- we must have F, From (— = —mo>-r the standpoint of a frame that rotates so as to kcep exact pace with the ball, the acceleration of the ball can maintain the validity of Newton's law if, in addition to the force force Fi, equal F r, is zero. in the rotating We frame the ball experiences an inertial and opposite to F r, and so direeted radially outward ^{«-A + ft-O = [Fi The m<i> 2 r force Fi is then what we call the centrifugal force. The magnitude of the centrifugal force can be cstablishcd experimentally by an observer in the rotating frame S'. hold a mass •The name fugere, to 507 m Let him stationary (as secn in his rotating frame) by "centrifugal" ffee. Centrifugal force comes from ihe Latin: cenlrum, the center, and Viewed from stationary frame S Procedure —A<&-*—\^ *\gj l r —a - . h Pictorial sketch of Viewed from rotating frame i" ' -S problem is observed to move with speed u in a circle of radius (angular speed u<) Bali r Boy turns around at the same angular speed o> as the ball; from his point of view, the is "Isolate fhe body." \ Draw forces that act on the ball all , ball at rest T \>1 nng T= mg = F, tension in cord force of gravity = inertial force due to viewing the problem from a rotating frame T cos e T cos e For ease of calculation we resolve forces into components T r sin $ F, 9 sin in mutually perpendicular directions \ 'mg 1 1 Vertical Direction Vertical Direction We now F Because there analyze the problem in celeration, terms of = ma is no vertical ac- we conclude that the net vertical force must be zero; Because there The object is t' aveling in a circle. therefore acc elerating: the net force (i.e.. th B sum of all three forces) is hor zontal toward the center of the circle, and must be equal in n lagnitude to mv*lr; The object the sum is of "at rest," therefore the forces on it all the analysis in the it is given by left column. . —— = m<u'r = mv 2 ,. i e- and is di is directed outward. rected radially We call F, the centrifugal force Molion of a suspended ball, which is traveling in a horizontal as analyzed from the earth's reference frame and from a frame rotating Fig. 12-11 with the ball. 508 = mg must be zero; hence F, is equal in magnitude to T sin A From t circle, n Horizontal Direction hence This force inward no vertical we conclude T cos - mg Horizontal Direction sin is that the net vertical force must be acceleration, zero; hence; hence T cos T mg Inertial forces and noninerlial frames Spring balance fastened to the axis of rotation, which is perpendicular to the plane of the paper- Fig. 12-12 W Measurement of the \^1>1)00000 t>~- A-. force needed to hold an object at rest in a rotating reference frame. attaching it to a spring balance (Fig. 12-12). any location except on will show tional to that it is m and mass is at exerting on the mass an inward force proporIf the r. If the the axis of rotation, the spring balance observer in S' is informed that frame his is rotating at the rate of w rad/sec, he can confirm that this force is equal to mwV. The observer explains the extension of the spring by saying that it is counteracting the outward centrifugal force on m which is present in the rotating frame. Furthermore, if the spring breaks, then the net force fugal force and the object acceleration we can apply its is just the centri- have an outward response to this so-called "fictitious" in Once again the force. as u2r of on the mass will at that instant inertial force is "there" by every eriterion (except our inability to find another physical system source). The magnitude of seen, by the equation the centrifugal force given, as is we have A Fccntrifugai A nice = m— = mw 2 r (radially example of our almost conditions in which there is As a first sitting on step we We it, is under provided have been washing and we want to get dry on the it a inside. get rid of the larger drops of water that are the inside walls. And we do tube longitudinally, but by whirling 12-13(a)]. (12-6) intuitive use of this force, nothing to balance by situations such as the following: piece of straight tubing, outward) it The analysis of what happens this, in as not by shaking the a circular arc we [Fig. begin this rotation gives a particularly elear picture of the difference between the deseriptions of the process in stationary and rotating frames. also provides us with a different the centrifugal force way of itself. Suppose that a drop, of mass m, of the tube at a point axis of rotation. 509 A Assume Centrifugal force It deriving the formula for [Fig. is sitting on the inner wali 12-13(b)], a distance r from the that the tube is very smooth, so that Fig. 12-13 (a) Shak- ing a drop ofwater oul ofa tube. (A) Analysis of initial motion in lerms of centrifugal forces. no resistance if it moves along the tube. The drop must, however, be carried along in any transverse movement of the tube resulting from the rotation. Then if the the drop encountcrs tube suddenly is set into motion and rotated through a small angle A0, the drop, receiving an impulse normal to the wali of moves along the straight line AC. This, however, now further from the axis of rotation than if it the tube and had traveled along the circular been fixcd to had arc AB. We have, in fact, the tube at A, means that BC = it is r sec A0 - r Now sec AS = (cosA0)-' « [1 « 1 - 2 1 KAfl) ]" KA<0 2 + Therefore, BC ~ We £r(A0) 2 can, however, exprcss A0 = u A/. and the time Al: A0 in terms of the angular velocity w Thus we have BC = iw 2 r(Af) 2 This in is then recognizable as the radial displacement that occurs time Ar under an acceleration "ccntrifiiciil — °> = "'W u 2 r. Hence we can r and so fcentrifuRal 510 inertial forces 2 /" and noninertial frames put ' what Notice, then, that ment is, a small transverse displace- in fact, no in a straight line, with real force in the radial direction, appears in the frame of the tube as a small, purely radial dis- The physical moved outward along the tube is readily terms of either description. (We should add, how- placement under an unbalanced centrifugal force. drop fact that the understood in i s ever, that our analysis as it stands does only apply to the initial Once the drop has acquired an appreciable things become more complicated.) step of the motion. radial velocity, The term "centrifugal force" is frequently used incorrectly. For example, one may read such statements as "The satellite does not fail down as it moves around the earth because the centrifugal force just counteracts the force of gravity there no is Newton's net force to law first straight line. . . . make it on For the satellite if The only frame it. move at all. it is must and hence such statement flouts it travels in a described as moving in also have an unbalanced which the centrifugal force does in balance the gravitational force appears not to Any — A body with no net force on a curved path around the earth, force fail." the frame in which the satellite is One can, of course, consider the description of such motions with respect to a reference frame rotating at object some arbitrary itself. rate different from that of the orbiting In this case, however, the centrifugal contribution to the inertial forces represents only a part of the story, and the simple balancing of "real" and centrifugal forces does not apply. In particular, let reference there is us reemphasize that in a nonrotating frame of no such thing The as centrifugal force. long- standing confusion that leads people to use the term "centrifugal force" incorrectly has driven at least one author to extreme vexation. In an otherwise sober and quite formal text the author writes: good "There citizens. is no answer to these people. They vote Some of them the ticket of the party that is are responsible for the prosperity of the country; they belong to the only true church; they subscribe to the Red Cross drive no place in the Temple of Science; they profane — but they have it." CENTRIFUGES The laboratory centrifuge represents an immensely important and direct application of the dynamical principle of centrifugal force. The basic arrangement of a simple type of centrifuge is 'W. F. Osgood, Mechanics, Macmillan, 511 C?cntrifuacs New York, 1937. Fig. 12-14 lical section (a) Ver- through a simple centrifuge. (b) Analysis of radial sedimenlation in terms of cenirifugal forces. shown Carefully balanced tubes of liquid are 12-14(a). in Fig. When suspended on smooth pivots from a rotor. made to spin at high speed, the tubes swing into almost horizontal positions the rotor is upward and outward and may be maintained orientation for many hours on end. At any point P in in this one of the tubes [Fig. 12— 14(b)], distance r from the axis of rotation, there is an made be very much (co 2 about 4000m/sec or 400 g. = w 2 r, For example, greater than g. and the rotor spins at 25 rps is magnitude effective gravitational field of 50* sec -1 ), Small particles which may if r = 15 the value of in cm w 2r suspension in the liquid will be driven toward the outward (bottom) end of the tube much more quickly than they would ever be under the action of gravity alone. The basis for calculating the drift speed the resistive force to motion through a fluid at we met first and speed in Chapter v, this force is 5. For a spherical is the formula for low speeds, which particle of radius r proportional to the product ro. If the is water, the approximate magnitude of the force R(v) « 0.02rv R is medium is given by where value of v in is newtons, r in meters, and v in m/sec. attained when A steady this force just balances the driving force associated with the effective gravitational field strength, g'. In calculating this driving force buoyancy the particle force is — effects is i.e., it is important to allow for Archimedes' principle. p p and the density of the liquid given by F=y(p„-p,)rV 512 Inertial forces and noninertial (rames If the density of is p ( the driving , This can be more simply expressed mass m To (= of the particle if 3 4irp p r /3), in we which case we can put take a specific example, suppose that suspension of bacterial particles of radius of about 5 water. If F« We X w6 2 10 -15 10~ 12 1 we have an aqueous p., each with a mass kg and a density about take for g' the value 400 X introduce the true g 1.1 times that of we calculated earlier, find N thus obtain a drift speed given by X a ^ 2X10-^10-» 10~ 12 2 ]n _4 m/S6C '° , This represents a settling rate of several centimeters per hour, which makes for effective separation in reasonable times, whereas under the normal gravity force alone one would have only a millimeter or two per day. The above example represents what one may regard as a more or less routine type of centrifugation, but in 1925 the Swedish chemist T. Svedberg opened up a whole new field of research when, by achieving centrifugal fields thousands of times stronger than g, he succccded in measuring the molecular weights The type of of proteins by studying their radial sedimentation. machine he developed the ultracenlrifuge, purpose was appropriately named for this and Svedberg succeeded fugal fields as high as about 50,000 g. The in producing centri- physicist has taken the technique even further through J. W. Beams development his of magnetic suspensions, in vacuum, that dispense with me- The chanical bearings altogether. it at a constant vertical level against the normal puli of gravity. By such methods Beams has produced centrifugal to about 10° g in a usable centrifuge and fields fields as tion is set mum by the bursting speed of the rotor; value of u The field g' is speed sets an upper limit to ur, value of g' varies as it equal to may u 2 r, # limita- this defines a proportional to l/r (see Chapter 7, Since the centrifugal p. maxi208). and the limiting be seen that the attainable l/r. The techniquc of Centrifuees equivalcnt high as 10 9 in tiny objects (e.g., spheres of 0.001-in. diameter). 513 empty rotor simply spins in space, with carefully controlled magnetic fields to hold ultracentrifuge methods has been brought to an extraordinary pitch of refinement. It has become possible to determine molecular weights to a precision of bettcr than over a range from about 10 8 (virus particles) about The 50. possibitity of weights by this method pointed out that in amu and analysis normal measuring the very low molecular a solution of sucrose in water, the calculatcd earlier) (A gravity. Beams has particularly impressive. radius about 5 A, we gave 1% low as to as an individual sucrose molecule, of mass about rate of descent of 340 is down be would (according to less than 1 mm rate as slow as this the kind of 100 years under in becomes mean- in fact Beams points out, it would be completely swamped by random thermal motions.) If a field of 10 5 g is ingless because, as howevcr, the time constant of the sedimentation available, process is reduced to the order of day or 1 measurement well within the range of which brings the less, possibility. ! This whole subject of centrifuges and centrifugation particularly good application of the concept of because the phenomena are so appropriately described of static or quasistatic is a inertial force, in terms equilibrium in the rotating frame. CORIOLIS FORCES We 2 have seen how the centrifugal on a mass m in a frame rotating at a given angular w, depends only on the distance r of the particle from force, mo> r, exerted particle of a givcn velocity the axis of rotation. In general, however, another inertial force appears in a rotating frame. This is the Coriolis force, depends only on the velocity of the particle (not on We shall introduce this force in a simple way for its 2 and it position). some specific situations. Later, by introducing vector expressions for rotational motion, we centrifugal shall develop a succinct notation that gives both the and Coriolis forces in a form valid in three dimensions using any type of coordinate system. The need comparing the to introduce the Coriolis force straight-line motion of a frame S with the motion of the same is easily shown by particle in an inertial particle as seen in a rotating frame S'. For further reading on this extremely interesting subject, see T. Svedberg and K. O. Pedersen, The Ulrracentrifuge, Oxford University Press, New York, 1960, and J. W. Beams, "High Centrifugal Fields," Physics Teacher 1, 103 (1963). 2 514 G. Coriolis, J. de VEcole Polytechnique, Cahier 24, 142 (1835). Inertial forces and nonincrtial irames Suppose that S' a coordinate system attached to a hori- is zontal circular table that rotates with constant angular speed w. Let the vertical axis of rotation define the z' axis that the table surface (in the x'y' plane) has no fastened to the origin holds a partiele is at rest in equilibrium forees of the tension in the string vertical foree of gravity A string y' coordinate axis from the axis of rotation. Thus, at a radial distance r'a frame, the partiele on the and suppose frietion. in the S' under the combined and the centrifugal foree. (The and the normal foree of the table surface always add to zero and need not concern us further.) The same partiele coincides with S' at travels with (= radius r f is = viewed from an uniform speed r' ) in the string. inertial frame S which In this stationary frame, the partiele 0. ve = in ov" a cirele of constant under the single unbalanced foree of the tension There is, of course, no centrifugal foree in this S the partiele then travels inertial frame. At / = Fig. 12-15 an object the string breaks. Two different deseriptions of the molion of that is initially tethered begins molion under no forees at on a rotating disk and t = 0. As seen 0, In the string breaks. The dots The in partiele the rotating S' frame is initially at rest. After the move indicate the positions of the partiele string breaks, successive equal time intervals At the same time instants 0, 1, 2, 3 the rotating y' axis has the positions outward, but as soon as it acquires some speed, it veers toward the right of the at shownby(O), 515 (1), (2), (3),.... Coriolis forees y' axis. it begins to radially speed v in a straight line with constant To 12-15(a). motion find the frame the positions of the y' in S', = u>r Q shown as we compare in the in Fig. stationary and the corresponding locations axis We discover moves radially of the particle at successive equally-spaced times. that the particle, as observed in S', not only outward, but also moves farther and farther to the right of the / formed by the rotating single radial line To plotted in Fig. 12— 15(b). in the rotating frame, explain this motion as observed we cussion, determine shall its magnitude and show that always acts at right anglcs to any velocity We can find the This deflecting In the course of the following dis- the Coriolis force. is v' in motion in these it the S' frame. magnitude of the Coriolis force by vestigating another simple is necessary to postulate, in addition to it is the centrifugal force, a sideways deflecting force. force This result axis. in- two frames. Suppose we make a that, instead of the situation just described, particle follow a radially outward path in the rotating frame at constant velocity In this frame there must be no net force on the v'r . Hence we particle. shall have to supply some real (inward) force to counteraet the varying (outward) centrifugal force as the particle We moves. shall not concern ourselves with these radial components but will concentrate our attention only on the transverse-foree components. In this way we can remove from consideration the distortion of the trajectory by the centrifugal force, which How purely radial. is does the motion appear in the two frames? Figure 12— I6(b) shows the straight-line path of the object in the rotating frame. But the path of the object curved line velocity vg AB (= as ur) shown is greater at distance from the axis is in the stationary in Fig. 12-16(a). B In S frame is a the transverse than at A, because the radial greater at B. Hence there must be a real transverse force to produce this increase of velocity seen in the stationary frame. This real force might be provided, for example, by a spring balance. What does this motion look like in the rotating frame? In moves outward with constant speed and hence has S' the object no acceleration [see Fig. 12-16(b)]. This means, as we have said, that there can be no net force on the object in the rotating frame. But since an observer in S' sees the spring balance exerting a real sideways force on the object in the infers that there tion to balance 516 it. Inertial forees +6 direetion, he a counteraeting inertial force in the is This is the Coriolis force. and noninertial frames — diree- : _____^_____^_____________^__ Fig. 12-16 (a) Lab- oratory view of the As seen path of a particle that moves as it As seen the stationary s frame in S' the rotating frame radially out- The mass ward on a rotaling table. in (b) The mass point moves point foilows the curved path The motion A frorrr radially outward with constant speed. to 8. appears in the rotaling frame itself. To determine its magnitude, be successive positions of the by Let A/. OC perpendicular to Av e = [co(r let same OA and OB in Fig. 12-17 radial line at times separated The be the bisector of the angle A0. velocity OC changes by the amount Avg during At, + Ar) cos(A0/2) + where v T sin(A0/2)] - [tar cos(A0/2) For small angles we can put the cosine equal to 1 v, sin(A0/2)] and the sine equal to the angle, which leads to the following very simple expression for Ao$ Avb The a» coAr + vr A6 transverse acceleration ao a6 = ai(Ar/Al) + is thus given by u r (A0/Al) But Ar/Al = cr and A6/At = o> Hence at co(r + Ft 8r) Fig. a = 2uv, = 2mwo r 12-17 particle Basis of calculaling the Coriolis force for mocing radially at constant speed with spect to a rotating tahle. 517 Coriolis forces re- This gives us the real force needed to cause the real acceleration But as observed as judged in S. is no no acceleration and — 2mcov' Coriolis force, equal to in the rotating T, is frame S', there Hence the existence of the net force. (Note that v T inferred. — Vr.) n the negative B' direction, opposite to the This inertial force is i spring force, and is at right angles to the direction of motion of the particle Fs'(Coriolis) An is = -2mwo'r important feature, which you should verify for yourself, that if we had considered a we would have then (12-7) acting in the positive radially inward motion (v'r negative), inferred the existence of a Coriolis force In both cases, therefore, the B' direction. Coriolis force acts to deflect the object in the same way with respect to the direction of the velocity v' itself— to the right the frame S' or to the prove shall is left if rotating counterclockwise, as S' rotates clockwise. later, that It if we have assumed, we turns out, in fact, as even in the case of motion in an arbitrary direction the Coriolis force is always a deflecting force, exerted at right angles to the direction of motion as observed in the rotating frame. The Coriolis force is very real from the viewpoint of the want toconvince yourself of rotating frame of reference. If you the reality of this "fictitious" force, ride a rotating merry-go- round and try walking a radial — the cautiously that it is Coriolis force line is outward or inward. (Proceed so unexpected and surprising easy to lose one's balance!) DYNAMICS ON A MERRY-GO-ROUND As we have just mentioned, the behavior of objects in motion within a rotating reference frame can run strongly counter to one's intuitions. It is not too hard to get used to the existence of the centrifugal force acting on an object at rest with respect to the rotating frame, but the combination of centrifugal Coriolis effects that appear when the object is set in and motion can be quitc bewildering, and somctimes entertaining. Suppose, for example, that a man stands at point A on & merry-go-round [Fig. 12-18(a)] and tries to throw a ball to someone at B (or perhaps aims for the bull's-eye of a dart board placed there). Then the thrown object mystcriously veers to the right and misses its target every time. One can blame part of this, of course, on the centrifugal force 518 Inertial forccs itself. Howevcr, it is and nonincrlial framcs to be noted that Fig. 12-18 (a) Trajectories ofobjects as lliey appear to observers on a rotating lable. (b) An object projected on africtionless rotating table return to its can starting point. since the magnitude of the centrifugal force the Coriolis force portional to v'fur. is is mu 2 r and 2mwv', the ratio of these two forces Thus if v' is made much that of pro- is greater than the actual peripheral speed of the merry-go-round, the peculiarities of the motion are governed almost entirely by the Coriolis effects. this condition holds, the net deflection of a moving object always be to the right with respect to v' rotating counterclockwise. Thus if If will on a merry-go-round the positions A and B in by two people trying to throw a ball have to aim to the left in order to make Fig. 12-18(a) are occupied back and forth, each will a good throw. An extrcme case of this kind of behavior can cause an object to follow a continuously curved path that brings to its starting point, although forces at all. it is it back not subjected to any real This phcnomenon has been demonstrated in the highly entertaining and instructive film, Frames of Reference. A dry-ice puck, launched at point [Fig. 12-18(b)], A on 1 a tabletop of plate glass can be caused by a skilled operator to follow a trajectory of the kind indicated. GENERAL ECjUATION OF MOTION The goal of IN A ROTATING FRAME 2 this discussion will be to relate the time derivatives of the displacement of a moving object as observed in a sta- '" Frames of Reference," by J. N. P. Humc and D. G. Ivey, Education Development Center, Newton, Mass., 1960. 2This section may be omitted by a reader who is willing to take on trust its final results — that the total inertial force in a rotating frame is the combination of the centrifugal force with a Coriolis force corresponding to a generalized formof Eq. 519 (12-7). General equation of motion in a rotating frame Fig. 12-19 Use of angular velocity as a vector to define the linear velocity parlicle lable: v ofa on a roiating = to X r. tionary frame S' and in a rotating frame 5'. shall introduce the idca that angular velocity To set the stage, we may be represented as a vector. Consider It P on a point first a rotating disk [Fig. 12-19(a)]. has a purely tangential velocity, magnitude and direction, same convention That We OP. angles to the radius that if we we w is a direction at right is Thus with hand are curled around in the as shown in represented as a vector, of length propor- x direction pointing along the positive z direction, one toward the positive y the disk is in this The of to velocity of = to X direction. P from the The rotation case counterclockwise as seen from above. P is now with the radius vector v given by the vector (cross) produet r: (12-8 r This veetor-produet exprcssion if which the thumb defining a rotation that carries each point such as positive of to 4. thumb extended tional to the angular speed, in the direction in points. both this velocity, in introduced for torque in Chapter sense of rotation, keeping the the figure, then in define a vector according to the the fingcrs of the right is, if v$, can describe the position vector r of P is is valid in three dimensions also, measured from any point on the shown in Fig. 12-19(b). The radius of the which P moves is R = r sin e. Thus we have v = vg = axis of rotation, as cirele in a direction perpendicular to the plane defined by to what Eq. (12-8) gives us. Next, we consider how the change of any vector during a cor sin 6, in and 520 r. That is precisely Inertial forees and nonincrtial frames Fig. 12-20 (a) Change of a vector, analyzed in terms ofits change as measured on a rotaling lable, togel/ier with the change I due lo rolalion of the lable iiself. (6) Similar analysisfor an arbitrary veclor referred to any origin on the axis ofrotation. sum of two small time interval At can be exprcssed as the vector contributions: 1. The change that would occur of constant length embedded 2. The if it were simply a vector in the rotating further change described by its frame S'. change of length and direction as observed in S'. we show this analysis for motion confined The vector A at time / is represented by the line CD. In Fig. 12-20(a) to a plane. remains fixed with respect to a rotating table, If it at time its t + At is given by the line CE, where A8 its direction = w At. Thus change due to the rotation alone would be represented by DE, where DE = A Ad = Au At. frame S' change would not be observed. There might, however, this From be a change represented by the — A the standpoint of EF; we line shall denote this as AA S sum of DE and EF, e., the line DF, then represents the true of A as observed in 5. We therefore denote this as AX S change the change of . 12-20(b) we show the corresponding analysis for three dimensions. direction Since A<p The vector i. In Fig. its as observed in S'. is The length of DE is now equal to A perpendicular to the plane defined by = u At, we sin 6 Aip; w and A. can put vector displacement DE = (u X A) A/ The displacement A A s- may be in any direction with respect to DE, but the two again combine to give a net displacement DF which is to be identified with A As. Thus we have 521 General eqnalion of motion in a rotating frame AA S = AA S + (w ' We A)Af can at once proceed from of change of A \drjs This we X this to a relation between the ratcs S and S', respectively: as observed in +«X \ di J s- A (12-9) A a very powerful relation because is can be any vector please. we First, (d\/dt)s is choose shall A the true velocity, the apparent velocity, to be the position vector v, as observed in S'. v', Then r. as observed in S, and (dA/dt)g> is Thus we immediately have = v v' +« X (12-10) r Next, we shall choose Now (dv/dt)s in to be the velocity v: the true acceleration, a, as observed in S. is quantity (d\/dt) s change A however, a sort of hybrid is, - S' of the velocity as observed sense of this if we substitute for v in S. — We it is The the rate of can make more from Eq. (12-10); we then have ®, -(& + •*$„ The two terms on the right of this equation are nizable; (d\'/dl)s(dr/dt)s> is (£), just = is v'. the acceleration, = now quite recog- as observed in S', and Thus we have •*- X v' Substituting this in Eq. (12-11) a a', we thus get «Xv a'+«Xv' + We do not need to have both v and v' on the right-hand and we shall again substitute for v from Eq. (12-10). This side, gives us finally a A 522 = a' + remark + 2u X v' is in order regarding the Inerlial fbrees u X (w X (12-12) r) last term, which involves and noninertial Frames According to the rules of the cross product of three vectors. vector algebra, the cross product inside the parentheses taken answer is will result for all cases other than 0° or 180°. would result in zero for all between these vectors. m Multiplying Eq. (12-12) throughout by the mass we recognize the r Performing the cross products in the cases, regardless of the angle the nonzero where the angle formed by u and reverse (incorrect) order, however, object, to be is A then the other cross product performed. first, of the side as the net external force on left mass as seen in the stationary system. ma = F nct = m»' + 2m(w X v') + m[u X In the rotating frame of reference, the object tion a'. We may this accelerated (<o X r)] m has the accelera- preserve the format of Newton's second law in frame of reference by rearranging the above equation, so as to be able to write Fi* = (12-13a) roa' where FL = Fnet - 2m(« X v') - m[u force X (to X (12-13b) r)] centrifugal force Coriolis force "real" inertial forces The mathematical form of Eq. (12-13b) shows Coriolis force and the centrifugal force are in right angles to the axis of rotation defined by u. force, in particular, is is that both the a direction The at centrifugal always radially outward from the axis, as clear if one considers the geometrical relationships of the vectors involved in the product Fig. 12-21. The equation also — u X (w X r), as shown in shows that the Coriolis force es uxr -» Fig. 12-21 x (u x Relalion of the vectors involved informing the centrifugal acceleration 523 — c» X (w X r). General equation of motion in a rotating frame r) would reverse if thc direction of u were reversed, but the direction of the centrifugal force would remain unchanged. The on made specification of F' in Eq. (12-13) can be and the basis of measurements of position, velocity, tion as observed within the rotating fratne term, involving the vector we could just as well put the two frames do agree on itself. centrifugal r, might seem to contradict r' instead of r, To summarize, we have established but this, because observers in the vector position of a at a given instant, granted that they use the that the dynamics of The entirely accelera- moving object same choice of origin. by the above calculation motion as observed uniformly rotating in a frame of reference may be analyzed in terms of the following three categories of forces: This is sum the of all the "real" forces on the object such as forces of contaet, tensions "Real": the force of gravity, electrical in strings, Fnnt forces, magnetic forces, and so Only on. these forces are seen in a stationary frame of reference. The Coriolis force is a deflecting force always mass at right angles to the velocity v' of the m. no the object has If velocity in the Coriolis: -2m(oi X frame rotating of reference, v') Coriolis force. in a centrifugal force depends < inertial is is no Note on position It is an always radially outward. force not seen in of reference. (as there inertial force nol seen sign. only and I an stationary frame of reference. minus The It is — m[o> X We (<•> X a stationary frame could equally well write r')]. Note the minus it sign. THE EARTH AS A ROTATING REFERENCE FRAME In this seetion we shall consider a few examples of the way in which the earth's rotation affects the dynamical processes occurring on it. The local value of g If a partiele P is at rest at latitude X near the earth's surface, then as judged in the earth's frame 524 Inertial forces it is subjected to the gravita- and noninertial frames Fig. 12-22 on an (a) Forces objecl at rest Ott the earth, as interpreted in a reference frame that roiates wilh (b) eart/i. An l/ie objecl falling from rest relalive to Ihe earth undergoes an eastward displacement. from a frame ing motion of (b), as seen (c) The fall- thal does nol roiale wilh ihe earth. and the tional force F„ centrifugal force The magnitude of the 12-22(a). F ccnt shown latter is given, in Fig. according to Eq. (12-13b), by the equation 2 = mu R cosX ^cent = rnu> R is the earth's radius. where R sin 9 Chapter 8 the way local in which We have already discussed this centrifugal magnitude of g and also modifies the in term reduces the Iocal direction of the by a plumbline. The analysis is in fact much simpler and clearer from the standpoint of our natural reference frame as defined by the earth itself. We have, as Fig. 12-22(a) vertical as dcfined shows, the following relations: Fr = F, - F$ = Fcnt sin X Fccnt COS X Fa — moi Rcos X (12-14) = mu R sin X cos X Deviation offreelyfalling objects If a particle is releascd 12-22(a), it net force F' 525 The from rest at a point begins to accelerate such as downward under P in Fig. the action of a whose components are given by Eq. (12-14). earth as a rotating reference frame As soon as has any appreciable velocity, however, it = -2mn X Fcorioiu Now the velocity axis. The v' is i Coriolis force (12-15) v' n the plane PON containing the earth's must be perpendicular to eastward. Thus by the we if up a set local plumbline vertical local coordinate and the this plane, u and a consideration of the actual directions of it is also expe- it by the equation riences a Coriolis force given v' system defined local easterly direction, as in Fig. 12-22(b), the falling object deviates eastward AB plumbline and ground hits the and shows that at a point C. The from a effect is very small but has been detected and measured in careful experiments (see Problem 12-24). To calculate what the deflection should be for an object from a given height falling h, to be inserted in Eq. (12-15) we is use the fact that the value of v' extremely well approximated by the simple equation of free vertical = v' where gt v' is measured as positive downward. Thus eastward direction as m— fail r-r- = x', = we label the (2mu> cos X)gt Integrating this twice with respect to x' if we have t, we have $gwt a cos X (12-16) For a total distance of vertical U2 (2h/g) which thus gives us fail equal to h, we have t = , ,_2V|«C<|X X 3,2 (121?) A = 2xday -1 Inserting approximate numcrical values (w 10 -5 sec x' -1 ~ ), 2 one X 10- 5A 3/2 cos X Thus, for example, with h x" « 5 It (x' = 50 and h m in m) at latitude 45°, one has mm, is or about ? in. perhaps worth reminding oneself that the inertial forces 526 ~7X finds can always be calculated, Inertial forces if effects of one wishes, from the and noninertial frames standpoint of an inertial frame in which these forces simply do not exist. In the present case, one can begin by recognizing that a particle held at a distance h above the ground has a higher eastward velocity than a point on the ground below. plicity, let how us consider this operates at the For sim- equator (X = 0). Figure I2-22(c) shows the trajectory of the falling object as The seen in a nonrotating frame. DO, = + o>(R After a time *» horizontal has traveled a horizontal distance x given, very t it With the object now gravitational force acting on Fx initial h) nearly, by coRt. the negative object has an by velocity given x We direction. it at P (see the figure) the has a very small component in have, in fact, — jf( « —mgmt Hence A we have Integrating once, dx j ~ "o*- is°» , 2 t = w(R Substituting the value v Qx + h), this gives, as a very good approximation, d f= co(« + - iga 2 /o t we have Integrating a second time, x - «(* + h)i - However, the point \gut 3 O at the ground at C, the point earth's surface is also moving, with Thus, when the falling object hits the a constant speed of coR. O has reached O', where 00' = uRl. Hence we have x' If 527 we The = O'C w oj/>/ substitute h = - i*cor 3 \gt 2 , we at once obtain the result given by cari h as a rotating reference frame Low Fig. J 2-23 Format ion ofa cyclone In the northern hemisphere, under the aclion of Coriolis forces on the Eq. (12-16) for X V2h/g and moving = arrive at air masses. [Or, of course, 0. we can substitute t = Eq. (12-17)]. Patterns of atmospheric circulation Because of the Coriolis effect, air masses being driven radially inward toward a low-pressure region, or outward away from a high-pressure region, are also subject to deflecting forces. This causes most cyclones to be in a counterclockwise direction in the northern hemisphere and clockwise in the southern hemisphere. The origin of these preferred rotational directions may be seen shows the motions of air in the northern moving toward hemisphere a region of low pressure. The hori12-23, which in Fig. zontal components of the Coriolis force deflect these motions i 31 ^f^m BJty.\~ ^fl ^^^-*" *•*-»• Fig. 12-24 Tiros satellite photograph of a cyclone. (Courtesy of Charles W. C. Rogers and N.A.S.A.) 528 Inertial forces ^^^B^^ |§£$ «.'— RJ— V' -— . - C^K2l^ fl_ ^frfp-1 and noninerlial frames I toward the Thus, as the right. air masses converge on the center of the low-pressure region, they produce a net counterclockwise rotation. For air the Coriolis force surface. If moving north or south over the earth's surface is due east or due west, parallel to the earth's we consider a mph) 10 m/sec (about 22 mass of air at a wind velocity of 45° north latitude, a direct applica- 1-kg at tion of Eq. (12-15) gives us Fcorioii, , - 2mui/ sin X « (2)(1)(2tt X lO" 5) X (10)(0.707) w 10" 3 N we had considered air flowing in from east or west, the Coriolis would not be parallel to the earth's surface, but their components parallel to the surface would be given by the same If forces equation as that used above. (Verify The approximate may be this.) radius of curvature of the resultant motion obtained from 2 R or R = m — 2 « ^ 1 rCorioli. As air X -r^r-r. 10 = 5 10' m (about 60 miles) •» masses move over hundreds of miles on the earth's surface, they often in the Tiros form huge weather satellite vortices —as photograph is dramatically shown in Fig. 12-24. Occasionally one reads that water draining out of a basin also circulates in a preferred direction because of the Coriolis force. In negligible however, most cases, the Coriolis force on the flowing water is compared with other larger forces which are present; if formed, the extremely precise and careful expcriments are pereffect cari be demonstrated. 1 The Foucault pendulum No account of Coriolis forces would be complete without some mention of the famous pendulum experiment named after the French physicist J. B. L. Foucault, who first demonstrated in 1851 how the slow rotation of the plane of vibration of a pen- 'See, for example, the film "Balhtub Vortex," an exccrpt from "Vorticity," by A. H. Shapiro, National Council on Fluid Mechanics, 1962. 529 The earth as a rotating reference frame Fig. 12-25 (a) A pendulum swinging along a north-south line at lalilude X. (b) Palh of pendulum bob, as seenfrom above. (The change ofdireclion per swing is, however, grossly exaggerated.) dulum could be used as evidence of the earth's own rotation. It is easy, but rather too glib, to say that of course we are simply seeing the effect of the earth turning beneath the pendulum. This description might properly be used for a pendulum suspended at the north or south pole. One can even press things a 12-25(a)] little further and say that at a given latitude, X [see Fig. w sin X along component the earth's angular velocity vector has a the local vertical. This that the plane of the would indeed lead to the correct result— pendulum rotates at a rate corresponding to one complete rotation in a time 7"(X) given HX) = lir 24 esc X by hours (12-18) >sinX But the pendulum is, after all, connected to the earth via its suspending wire, and both the tension in the wire and the gravitathe tional force on the bob lie in the vertical plane in which pendulum is first set swinging. (So, too, is the air resistance, if that can be this needs to be considered.) It is the Coriolis force invoked to give a more cxplicit basis for the rotation. For a pendulum swinging in the northern hemisphere, the Coriolis force acts always to curve the path of the swinging bob to the As right, as indicatcd in exaggerated form in Fig. 12-25(b). with the Coriolis force on moving air, the effect does not depend on the direction of swing— contrary to the intuition most of us probably have that the rotation is likely to be more marked when the pendulum swings along a north-south line than whcn it 530 swings east-west. Inertial forces and noninertial frames THE TIDES As everyone knows, the production of ocean tides the consequence of the gravitational action of the to a lesser extent, the sun. is basically moon — and, Thus we could have discussed this as an example of universal gravitation in Chapter 8. The analysis of the phenomenon is, however, considerably helped by introducing the concept of inertial forces as developed in the present chapter. one The feature that probably causes the first learns about the tides places on the earth's than just one. is two high surface, most puzzlement when the fact that there are, at most day rather tides every This corresponds to the fact that, at any instant, the general distribution of ocean levels around the earth has two bulges. On the simple would be highest and farthest performs its model that we shall discuss, these bulges on the earth's surface nearest to at the places from the moon [Fig. 12-26(a)]. While the earth rotation during 24 hr, the positions of the bulges would remain almost stationary, being defined constant position of the moon. Thus, if by the almost one could imagine the earth completely girdled by water, the depth of the water as measured from a point pass through two fixed to the earth's solid surface maxima and two minima A better approximation to the observed facts would in each revolution. is obtained by considering the bulges to be dragged eastward by friction from the land and the ocean floor, so that their equilibrium positions with respect to the To moon are more nearly as indicated in Fig. 12-26(b). conclude these preliminary remarks, we may point out that the bulges are, in fact, also being carried slowly eastward all Fig. 12-26 (a) tidal bulge would be as the time by the Double motion it if the earlh's rotation did not displace it. the bulge The is size of enormously exaggerated. (b) Approximate true orientation ofthe tidal bulges, carried eastwardby the earth's rotation. 531 The moon's own motion around the earth. This (one complete orbit relative to the fixed stars every tides it takes more than 24 hr for a given point on the earth to make successive passages past a 27.3 days) has the consequence that particular tidal bulge. Specifically, this causes the theoretical time interval between successive high tides at a given place to be close to 12 hr 25 min instead of precisely For example, a high day, its if tide is 12 hr (see Problem 2-15). observed to occur at 4 p.m. one counterpart next day would be expected to occur at about 4:50 p.m. Now The first manner in which the earth as a whole is being accelerated toward the moon by virtue of the gravitational attraction between them. With respect to the CM of the earth-moon system (inside the earth, at about 3000 miles from the earth's center), the earth's center of mass has an acceleration us consider the dynamical situation. let point to appreciate is the of magnitude a c given by Newton's laws: MrOc = GMeM„, rm 2 i.e., ac where = GMm Mm and (12-19) r„, are the moon's mass and not be immediately apparent receives this Flg. 12-27 bital motion ofthe is distance. that every point in the earth same acceleration from the moon's The or- earth about the moon does not by itselfinvolce any rotation of the earth; the line AiBi is carried into the parallel configuration A2B2. 532 Inertial forces What may and noninertial frames attraction. If one draws a sketch, as shown which the of time, one is of the arcs along in Fig. 12-27, and the earth's center moon travel in a certain span tempted to think of the earth-moon system as a kind of rigid dumbbell that rotates as a unit about the center of mass, O. it It is moon, for its part, does move so that same face toward the earth, but with the true that the presents always the earth itself its axis, things are different. every point on it and direction to the arc size center. A 2B2 . The The If the C C2 AiBy would be line earth were not rotating on would follow a X translated into the parallel line earth's intrinsic rotation superposed on this circular arc identical in traced out by the earth's about general displacement its axis is simply and the associated acceleration. This is where noninertial frames come into the The dynamical consequences the picture. of the earth's orbital motion around CM of the earth-moon system can be correctly described terms of an mass m inertial force, wherever then added to it all may —mac, be, in or in experienced by a particle of on the the other forces that This force earth. may is be acting on the particle. In the model that we are using — corresponding to what — the water around the is called the equilibrium theory of the tides earth simply moves until it attains an equilibrium configuration that remains stationary the moon. Now from the viewpoint of an observer on we know that for a particle at the earth's center, the centrifugal force and the moon's gravitational attraction are equal and opposite. If, however, we consider a particle on the earth's surface at the nearest point to the Fig. 12-28(a)], the gravitational force centrifugal force by Fig. 12-28 (a) Dif- ference belween centrifugal force and the earth's gravity at the points nearest to and farthest from the moon. (b) Tide- producing force at an arbitrary point P, showing existence of a transverse component. 533 The tides an amount that we on moon it is shall [point A in greater than the call/ : GMm m « rm (R E RE Since GMm m _ « r m /60), we can approximate this expression as follows: GMm m = /o [(-r-] i.e., /„«H™^*, By an (12_20) exactly similar calculation, force on a particle of mass B [point we find that the tide-producing m at the farthest point from the moon —f in Fig. 12-28(a)] is equal to ; the tendency for the water to be pulled or midplane drawn By going C, The we can just a little further P tidal force now Consider recognize a on in the it = R E cos with x (x, y), x get a much better in- a particle of water at an Relative to the earth's center, [Fig. 12-28(b)]. has coordinates it we through the earth's center (see the figure). sight into the problem. arbitrary point hence pushed away from direction is 6, y = RE sin 6. given by a calculation just like those above: m IGM^n IGM^n x _ ^ (]2_ 21) cqs g This yiclds the results already obtained for the points if we put d = or line joining the centers of the earth and the moon there however, a transverse force, because makes a small angle, a, with the x the line center tional force, A and B In addition to this force parallel to the ir. GMm m/r' 2 , is also, from P to the moon's axis, and the net gravita- has a small component perpendicular to x, given by /„ GMj— mm sin a . = Now we a wc can 534 ... = , r*) have tana = Since , (with r is y a very small angle [<tan safely (/?/.;//•„,), which approximate the above exprcssion: Inertial forces and noninerlial frames is about 1°] Fig. 12-29 Pattern of tide-producing forces around the earlh. The circular dashed line shows where the undisturbed waler surface would be. y_ _ R E sin d r- rm The component/„ of the fv » GMm5— m tidal force is GMm5— m Re sin y = We see that this transverse force point it is then given by equal to half the is (12-22) greatest at d maximum = value (/ ) ir/2, at which of fx Using . we can develop an over-all picture of the tide-producing forces, as shown in Fig. 12-29. This shows much more convincingly how the forces act in such Eqs. (12-21) and (12-22) together, directions as to cause the water to flow the manner already 1 high ought the equilibrium tidal bulge to be? familiar with actual tidal variations 'This section goes well added 535 redistribute itself in qualitatively described. TIDAL HEIGHTS; EFFECT OF THE SUN How and If you are you may be surprised at the beyond the scope of the chapter as a whole but for the interest that it may have. Tidal heights; eflect of the sial is The equilibrium result. 2 We ft. by the tential would be a rise and fail of less than can calculate this by considering that the work done moving a tidal force in 12-29) (Fig. tide is particle of water from D to A equivalent to the increase of gravitational po- energy needed to raise the water through a height h against the earth's normal gravitational puli. the difference of water levels between Eqs. (12-21) and (12-22) dW = f = + dx z The D. distance h Now, is using we have f„dy GMm m ,_ ir~ (2xdx - ydy) , W . r RE GMm m r Wda = r° 2x dx Uo rm3 3GMm m — ydy / Jr b 2 amount of work equal energy, mgh, we have Setting this potential ' A and to the gain of gravitational 2 _ 3GMm R B (12 _23) 2grma The numerical values of the G = Mm = = RE = g = rm 10~ n 7.34 X X 3.84 X 10 8 m 6.37 X 10° m 6.67 10 22 9.80 m/sec relevant quantities are as follows: m 3 /kg-sec 2 kg 2 Substituting these in Eq. (12-23) A The « 0.54 m« we find 21 in. great excess over this calculated value in factors of 10 or even many places (by more) can only be explained by considering the problem in detailed dynamical terms, in which the accumula- and resonance effects, can phenomenon. The value that completely alter the scale of the we have calculated should bc approximated in the open sea. The last point that we shall consider here is the effect of the tion of water in sun. Its narrow estuaries, mass and distance are as follows: Technically, this condition corrcsponds to the water surface being an energy equipotential. 536 Incrlial forces and noninertial irames M, = r„ If we 1.99 10 30 kg X = 1.49X10" m directly sun and the compare the moon on a gravitational forces exerted by the particle on the earth, we discover that the sun wins by a large factor: MJr? F, Mm/rm2~ F„ M.(rm \\ Mm ° \r.) What matters, however, for tide production is the amount by which these forces change from point to point over the earth. This is expressed in terms of the gradient of the gravitational force: • m = 2p /=AF=-Âr Putting ±/ M = Mm , r = rm , (,2-24) and Ar = ±RE , we obtain the forces corresponding to Eq. (12-20). We now to the sun see that the comparative tide-producing forces due and the moon are given, according to Eq. (12-24), by the following ratio: Substituting the numerical values, one finds fy Jm « 0.465 This means that the tide-raising ability of the of the sun by a factor of about 2.15. The combine linearly— and, of course, relative angular positions of the moon effects vectorially, moon and exceeds that of the two depending on the the sun. When they on the same line through the earth (whether on the same side or on opposite sides) there should be a maximum tide equal to 1.465 times that due to the moon alone. This should happen are once every 2 weeks, approximately, when the full. tions should fail to moon. The about 537 moon is new or At intermediate times (half-moon) when the angular posiof sun and moon are separated by 90°, the tidal amplitude a minimum value equal to 0.535 times that of the of maximum to minimum values is thus ratio 2.7. Tidal heights; efl'cct of the sun THE SEARCH FOR A FUNDAMENTAL INERTIAL FRAME The phenomena that we have discussed in this chapter seem to leave us in no doubt that the acceleration of one's frame of They suggest reference can be dctected by dynamical means. that a very special status does indeed attach to inertial frames. But how can we be sure that we have identified a true frame in which Galileo's law of inertia holds exactly? We saw at the very beginning of inertial our discussion of dynamics good approximation to such a frame for many purposes, especially for dynamical phenomena whose scale in distance and time is small. But we have now seen abundant evidence that a laboratory on the earth's surface is that the earth itself represents a * If the accelerated. each point laboratory in it is accelerating is at latitude X (see Fig. 12-30), toward the earth's axis of rotation with an acceleration given by a\ = 2 oi R cos X with = R = 01 27r/86,400sec- 1 X 6.4 10 6 m This gives a\ = 3.4 X 10- 2 cosXm/sec 2 This acceleration of a frame of reference as we tied to the earth know, not the simplest case of an accelerated frame. linearly accelerated frames with much more readily analyzed. which we began It this chapter are was, however, the associated with rotating frames that led Newton is, The phenomena to his belief in absolute space and in the absolute character of accelerations. Near the beginning of the Principia he describes a celebrated 538 Fig. 12-30 axis by Acceleration toward ihe earth's virtue ofits rotation. Inertial Corces and noninertiai frames Bucket rotating water stationary Main features Fig. 12-31 ofllie experiment that Newton guolecl as evidence oflhe absolute character of rotation and the associated acceleration. made experiment that he with a bucket of water. It is ari experiment that anyone may readily repeat for himself. The bucket is hung on a strongly twisted rope and is then released. There are three key observations, depicted in Fig. 12-31: Bucket and water rotating together At 1. almost at before the viscous forces have had time to set The water surface rotating. was the bucket spins rapidly, but the water remain's first rest, is flat, just as it it was before the bucket released. The water and 2. the bucket are rotating together; the water surface has become concave (see Problem 12-18). The bucket 3. is suddenly stopped, but the body of water continues to rotate, and Bucket stationary water rotating and it motion of the bucket and It must be the absolute rotation of the water in attendant acceleration, that its the phenomenon. for relative not the factor that determines the curvature of the is water surface. space, surface remains curved. Newton, the Clearly, said the water its And with the help of F= is at the bottom of ma, we can account quantitatively. Newton's argument is a powerful one. He could point to further evidence in support of his views in the bulging of the earth by virtue of itself of the earth in 300. It is is the equator and is, The equatorial diameter by about 1 part rotation. seems almost obvious, even without detailed calculation, that this culation its greater than the polar diameter closely tied to the fact that a^/g is is about g£t5 at zero at the poles (although the detailed cal- in fact, a bit messy). Newton did not stop here, of course, He held the key of Even a nonrotating earth would not be frame, because the whole earth is accelerating toward universal gravitation. an inertial the sun. For this system a> a2 = R = 2 = o) /? = Newton represents we have 2ir/(3.l6 X lO^sec- 1 1.49 X 10 n m 5.9 X 10- 3 m/sec 2 does not suggest that he actually performed this third step, but a natural completion of the experiment as one might perform for oneself. 539 The search for a fundamental inertial frame it it If we could conceive of an object that was immune to the gravita- would not obey the law of inertia as observed from a reference frame attached to the earth. From tional attraction of the sun, it Newton's standpoint the acceleration is real and absolute and is linked to the existence of a well-defined gravitational force provided by the sun. That was about the end of the road as far as Newton was concerned. For him the system of the stars provided the arena in which the motions that he so A analyzed took place. brilliantly reference frame attached to these fixed stars could be taken to constitute a true inertial system, even though might not coin- it cide with the absolute space in which he believed. Today, thanks to the work of astronomers, we know a good deal about the motions of some of those "fixed" stars. We have come to be aware of our involvement in a general rotation of our Galaxy. The sun would appear to be making a complete circuit of the Galaxy about 2.5 X about in 10 4 2.5 X 10 8 years at a radial distance of from the light-years For center. this motion we would have co~ 2tt/(8 «« a3 It « 2.4 X X 10 10 20 15 )sec-' m 10- 10 m/sec 2 looks as though this acceleration can be reasonably accounted for by means of Newton's law of universal gravitation, if we regard the solar system as having a centripetal acceleration under the attraction of all the stars lying within dynamical experiments that we do on But no its orbit. earth require us to take — into account this extremely minute effect or, even, for most purposes, the revolution of the earth about the sun. (The rotation of the earth on sideration — its own axis is, however, an important con- and indeed an important aid gyroscopic navigation.) rotating frames in which Figure we in such matters as 12-32 schematizes the three find ourselves (we ignore here the acceleration caused by the moon). But we still have not found an unaccelerated object to which In fact, we could we can attach our inertial frame of reference. extend this tantalizing dence search even further. There is some that galaxies themselves tend to cluster together in containing a few galaxies to perhaps thousands. consists of about 10 galaxies. Our evi- groups local group Although individual galaxies could have rather complex motions with respect to each other, 540 Inertial force* and non inertial frames Fig. 12-32 Accelera| ofany laboratory reference frame attions tached to the earth's surface. group this is believed to have a more or less common motion through space. So where are the "fixed" stars or other astronomical we can attach our inertial frame of reference? objects to which pears that referring to the "fued stars" is ap- It not a solution and contains an uncomfortable element of metaphysics (although we frequently usc this phrasc as a shorthand designation for the establishment of an inertial frame). This does not mean that the astronomical search for an inertial frame has been without value. For, at lcast up to the galactic level, it would seem that apparent departures from the law of inertia can be traced to identifiable accelerations of the reference are observed. However, the quest likely to remain. 541 Tho is frame in which motions incomplete, and so Ultimatcly, therefore, we rely it seems on an operational search for a fundamental inertial frame upon definition based We tion. an define local dynamical experiments inertial frame to be one mentally, Galileo's law of inertia holds. and observa- in which, experi- The very existence of the inertial property remains, however, a deep and fascinating problem, and we shall end the chapter with a few remarks about most fundamental feature of dynamics. this SPECULATIONS ON THE ORIGIN OF INERTIA Not everyone accepted Newton's view was perhaps the phenomena demonstrated the absolute objects The philosopher-bishop, George of acceleration. character Berkeley, rotating with associated that the first person to argue 1 that all motions, including rotational ones, only have meaning as motions relative The to other objects. two spheres around circling of their center of mass could not, he said, be imagined in a space that was otherwise empty. Only when we introduce the background do we have a basis for recognizing the represented by the stars existence of such motion. About 150 Ernst He Mach German philosopher much more cogent form. years later (in 1872) the presented the same idea in wrote: For mc, only regard, viously motions exist . . . distinction between rotation no it relative does not matter can and I and translation. see, in this Ob- we think of the earth as turning if round on its axis, or at rest while the fixed stars revolve around of the earth at rest and the fixed stars it . But if we think revolving around it, there is no flattening of the earth, no . . on— at least according to our usual conception of the law of inertia. Now one can solve the Foucault's experiment and so difflculty in of inertia two ways. Either is The law of all motion wrongly expressed ... inertia I is absolute, or our law prefer the second way. must be so conceived that exactly the thing results from the second supposition as from the this it will be evident that in its to the masses of the universe. expression, regard same By first. must be paid 2 In his iraet De Moiu, written in 1717, 30 years after the publication of Newton's Principia. 2 E. Mach, Hislory and Rool of the Principle of the Consercation of Energy, (2nd ed.), Barth, Leipzig (1909). English translation of the 2nd edition by P. Jourdain, Open Court Publishing Co., London, 1911. Actually the first sentence of the quolation is taken from Mach's classic book, The Science of Mechanics, 542 first published in 1883. Inertial forees and nonincrtial frames the profound Thus was born become famous — subsequently to — that the property and novel idea as Mach's principle inertial of any given object depends upon the presence and the distribu- took and Einstein himself accepted this idea tion of other masses. as a central principle of cosmology. it If one admits the validity of this point of view, then one whole basis of dynamics sees that the method the we described that in the ratio of the inertial masses of is Chapter 9 two For consider involved. (p. 319) for finding This ratio objects. given is as the negative inverse ratio of the accelerations that they produce by their mutual interaction: /Ml _ (22 W2 Oi This looks very simple and straightforward, but our clear that is it ability to attach specific values to the individual accelerations, as distinct from the depends total relative acceleration, completely on our having identified a reference frame For these accelerations can be measured. physical background provided by other objects this which in purpose the is essential. In looking critically at the phenomena of rotational motion, Mach some attacked intuitive notions that are seated than any that He motion. we have deep- considered the evidence provided by Newton's rotating-bucket experiment which It is much more in connection with straight-line we discussed in the last section. quite clear that the curvature of the water surface is related overwhelmingly to the cxistence of rotation relative to the vast When amount of distant matter of the universe. rotation stopped, the water surface becomes is bucket rotates and the water remains fixed stars), the out if may bc competent "No only a matter of degree. to say how one," would in fact the equivalent of centrifugal forces on the water insidc this This mass were ultimately several leagues thick." His own belief that this rotation of a monster bucket though the the experiment would turn the sides of the vessel increased in thickness and until they was "is When (both relative to the shape of the water surface remains unaffected. But, said Mach, that he wrote, still that relative fiat. is water had no rotational motion in the a startling idea indced. Let us prcsent dirTerent context. We know generate it, even accepted sense. it in a slightly that the act of giving an object an acceleration a, with respcct to the inertial frame defined by the — /wa, that expresses the resistance of the object to being accelerated. In fixed stars, calls into play 543 an inertial force, equal to Spcculations on the origin of inertia — Mach's view we are equally accept a description of the attached to the object has the acceleration a' phenomenon a frame always in In this frame the rest of the universe itself. the object experiences (indeed, compelled) to entitled (= —a) and the inertial force ma' that must be ascribable to the acceleration of the other masses. This then brings us to the quantitative question: M, at distance what contribution does object, ma and relativity that this point analogy we is know it q t a mass must more to the total inertial force we know that the force is the grounds of symmetry be proportional to M A speculative realm. But at also. very suggestive provided by electromagnetic interactions. that the static _ make Since we can argue on entcr a electric charges, two If and q it are separated by a distance force exerted by q x on q 2 is given by r, we fr?l<?2 where k units. is it that the object experiences? proportional to m, If given the acceleration a relative to a given is r, is a constant that depends on the particular choice of is given the acceleration a there however, the charge q x If, an additional force that comes into play, directly proportional to a and , ri2 where inversely proportional to the distance: = kq x q 2 a 5 c2r c is the speed of light. Since this force with distance than the static interaction, it falls off more slowly can survive in appreciable magnitude at distances at which the static l/r has become negligible. magnetic radiation This field is, in fact, 2 force the basis of the electro- by which signals can be transmitted over large distances. Suppose now that we assume an analogous situation for The basic static law of force is known gravitational interactions. to be GMm The on be given by force , Fl2 544 m associated with an acceleration of = GMma . n ( ~&T Inertial forces M would then and noninerlial framcs ,,., } — On this basis we can estimate the relative magnitudes of the contributions from various masses of interest sun, our do to is own Galaxy, and — the the rest of the universe. M/r to calculate the values of earth, the Ali we have The for these objects. shown in Table 12-1, using numbers to the nearest power of 10 only. (The value of for the universe as a whole is the somewhat speculative value quoted in Chapter 1.) We results are M TABLE 12-1: RELATIVE CONTRIBUTIONS TO INERTIA M, kg Source Earth Sun Our Galaxy Universe r, m M/ kg/m M/r 'r, (relative) 10 25 10 3O 10 7 10 18 10- 8 10 11 10 19 10* 10« 10 20 io- 7 10-° 10 2G 10 2e l 52 ] 1 see that, according to this theory, the effect of a nearby object, even one as massive as the earth pared to the The effect itself, of the universe would be negligible com- at large. total inertial force called into existence if everything in the universe acquires an acceleration a with respect to a given object would be obtained by (12-26) over _ «"inertial all summing masses other than = ma m ^GM — £_, c*r This, however, should be identical with magnitude of the the forces F'l2 of Eq. itself: what we know by the value of ma. Thus the theory would require the following identity to hold: Ef-1 universe u 02-27) ' from Table 12-1 that even such a large our Galaxy represents only a minor contribution It is clear invoived with is If we regard it p and radius univerao ' local ; mass as what we are a summation over the approximately uniform distribution of matter represented 545 to be the inertial force as directly given as a sphere, centered R v (« «'O 10 10 by the universe as a whole. on light-years mean density m), we would have ourselves, of = 10 26 » Speculations on the oriain of inertia ' The mass total Mu = however given by is — pRu Thus we have, on (based on Euclidean simple picture this geometry) M=-— Mu « — E ** Using the values we ,-20 1 / 10 kg/m 3 universe G m 10 -10 N-m 2 /kg 2 and c 2 m 10 17 m 2 /sec 2 , would then have E^-io— 1 f " universe c&r Taking into account the uncertainties in our knowledge of the distribution of matter throughout space, many would say that the factor of about 10 that separates the above empirical value from the theoretical value (unity) called for by Eq. (12-27) is not significant. The result is intriguing, to say the least, and many cosmologists have accepted as fundamentally correct this development from the primary ideas espoused by Mach and Einstein. PROBLEMS 12-1 A single-engine airplane flies horizontally at a constant speed v. In the frame of the aircraft, each tip of the propeller sweeps out a of radius R at the rate of n revolutions per second. Obtain an equation for the path of a tip of the propeller as viewed from the earth. circle 12-2 A person observes the position of a post from the origin of a reference frame (S') rigidly attached to the rim of a merry-go-round, as shown in the figure. The merry-go-round (of radius R) is rotating with angular velocity u, the distance of the post from the axis of the merry-go-round is D, and at / = 0, the coordinates of P in 5' are x' = D - R,y' = (equivalently, Find the coordinates corresponding *'(/) and y'(t). (a) r' = D - r'(i), 6'(t) R, d' = 0). of the post; also give the •For further reading on this fascinating topic, see, for example, R. H. Dicke, "The Many Faces of Mach," in Gracitation and Relalicily (ed. H.-Y. Chin and W. F. Hoffmann, eds.), W. A. Benjamin, New York, 1964; N. R. Hanson, "Newton's First Law," and P. Morrison, "The Physics of the Large," both in Beyond the Edge of Cenainly (R. G. Colodny, ed.), Prentice-Hall, Engle- wood New Cliffs, N.J., York, Doubleday, 546 1965; D. New W. Sciama, The Unity ofihe Unicerse, Doubleday, and The Physical Foundaiions of General 1961, York, 1969. Inertial forccs and noninertial frames Relalicily, By (b) differentiating the results of (a), obtain the velocity acceleration of the post in both Cartesian Make (c) and and polar coordinates. a plot of the path of the post in S'. 12-3 A boy is riding on a railroad flatcar, on level ground, that has an acceleration a in the direction of its motion. At what angle with the vertical should he toss a ball so that he can catch ing his position 12-4 A on railroad train traveling 20 m/sec begins A m. = sliding friction n without shift- on a straight track at a speed of slow down uniformly as it enters a station and comes to to a stop in 100 it the car? suitcase of mass 10 kg having a 0.15 with the train's floor slides coefficient of down the aisle diiring this deccleration period. (a) What the acceleration of the suitcase (with respect to the is ground) during this time? (b) What is the velocity of the suitcase just as the train comes toahalt? (c) The suitcase continues sliding for a period after the train When has stopped. original position 12-5 comes it on to rest, how far is it displaced A man weighs himself on a spring balance calibrated in which indicates weight as his from its the floor of the train? mg = 700 N. What will repeats the observation while riding an elevator from the manner? and third floors the elevator newtons he read first if he to the twelfth floors in the following (a) Belween the the rate of 2 m/sec (b) first accelerates at 2 . Between the third and tenth floors the elevator travels with the constant velocity of 7 m/sec. (c) Between the tenth and twelfth at the rate of 2 (d) He (e) If m/sec 2 then makes a similar trip on another say of his motion? 12-6 truck 547 what Problems is trip the Which way If the cocHicient is /x, the floors the elevator decelerates . is down again. balance reads 500 N, what can you he moving? of friction between a box and the bed of a maximum acceleration with which the truck can climb a hill, slipping on 12-7 A making an angle 6 with the horizontal, without the box's the truck bed ? block of mass 2 kg rests on a frictionless platform. attached to a horizontal spring of spring constant 8 Initially the in the figure. platform begins to 2 m/sec. 2 As move whole system is N/m, stationary, but at as t It is shown = to the right with a constant acceleration the of a result the block begins to oscillate horizontally relative to the platform. 2kg 8 N/m 2 m/sec 2 immfmr- TJ What is the amplitude of the oscillation? At t = 2v/3 sec, by what amount is the spring (a) (b) it was in T? its initial longer than unstretched condition? -1 plane surface inclined 37° (sin f) from the horizontal is accelerated horizontally to the left (see the figure). The magnitude of 12-8 A the acceleration is gradually increased until a block of mass m, orig- up the plane. inally at rest with respect to the plane, just starts to slip The static friction force at the Draw (a) just before it block-plane surface characterized by is a diagram showing the forces acting on the block, slips, in an inertial frame fbced to the floor. (b) Find the acceleration at which the block begins to (c) Repeat part slip. frame moving along with (a) in the noninertial the block. A 12-9 and nervous passenger in an airplane at takeoff removes his lets it hang loosely from his fingers. takeoff run, which lasts 30 sec, the What the vertical. runway 12-10 is A is 8 N/m 2 = 7500 kg/m 3 rod (density steel m X tion of length by a constant force applied to its away from is ) of length the center of allowable acceleration if 1 is is ultimate tensile one end and directed What mass of the rod. the rod , accelerated along the direc- not to break? is the maximum If this acceleration exceeded, where will the rod break? 12-11 (a) A train liquid level of horizontal ? from the A slowed with deceleration hit the a. What angle would the a bowl of soup in the dining car have made with the child dropped an apple from a height h and a distance d front wali of the dining car. observed by the child ? 548 makes an angle of 15° with and how much runway is level. tie strength 5 10 tie observes that during the the speed of the plane at takeoff, needed ? Assume that the uniform He Under what What path did the apple take as conditions would the apple have ground? The front wali ? Tncrtial forces and noninertial frames As (b) making the above observations, the parents a reward for bought the child a helium-filled balloon at the next stop. they asked him what would happen to the balloon station with acceleration An 12-12 */3 For fun, the train left the Subsequently, they were surprised to a'. find his predictions corrcct. if What did the precocious child answer? downward acceleration equal to g/3. Inside mounted a pulley, of negligible friction and inertia, over which passes a string carrying two objects, of masses m and 3»i, elevator has a the elevator is respectively (see the figure). tt the acceleration of the object of mass (a) Calculate 3m relatice to the elevator. on (b) Calculate the force exerted 3^ joins it the pulley by the rod that to the roof of the elevator. How (c) could an observer, completely isolated inside the m in terms of forces that elevator, explain the acceleration of he him- could measure with the help of a spring balance? self 12-13 In each of the following cases, find the equilibrium position as pendulum of length L: well as the period of small oscillations of a (1) In a train moving with acceleration a on level tracks. In a train free-wheeling on tracks making an angle 6 with the (2) horizontal. (3) In an elevator falling with acceleration a. 12-14 The world record for the 16-lb hammer throw Assuming about 2 m before being 12-15 (a) A is man rides in is about 70 m. whirled around in a circle of radius let fly, must be able that the thrower He hammer that the estimate the magnitude of the puli to withstand. an elevator with vertical acceleration a. swings a bucket of water in a vertical circle of radius R. With must he swing the bucket so that no water spills? (b) With what angular frequency must the bucket be swung if the man is on a train with horizontal acceleration a? (The plane of what angular velocity the circle again vertical and contains the direetion of the train's is acceleration.) 12-16 Consider a thin rod of material of density stant angular velocity (a) Show that a> i f p rotating with con- about an axis perpendicular to the rod is to have a constant force per unit area of cross seetion) along its its length. stress 5 (tensile length, the cross-sectional area must deercase exponentialIy with the square of the distance from the axis: A = Aue— frr a where k = 2 po>~/2S [Consider a small segment of the rod between r and r -f Ar, having a im = pA(r) Ar, and notice that the differcnce in tensions at its mass ends 549 is AT = Problem s A(SA).] What (b) 5m ax, and is maximum ihe angular velocity o) max in terms of p, A:? The (c) ultimate tcnsilc strength of steel maximum Estimate the number of rpm of possible = which the "taper constant" k about 10 is m~ 2 100 (p = 9 N/m 2 . a steel rotor for 7500 kg/m 3 ). A spherically shaped influenza virus particle, of mass 6 X -5 and diameter 10 cm, is in a water suspension in an ultrag centrifuge. It is 4 cm from the vertical axis of rotation, and the speed 12-17 10 -16 of rotation 10 3 rps. is The density of the virus particle is 1.1 times that of water. From (a) (b) what the standpoint of a rcference frame rotating with the what is the effective value of 'V ? Again from the standpoint of the rotating reference frame, centrifuge, the net centrifugal force acting is outward at given by on the virus particle? Because of this centrifugal force, the particle moves radially (c) a small Frcs = speed Jnrr\vd, The motion o. where d is is resisted by a viscous force the diameter of the particle the viscosity of water, equal to 10~- cgs units (g/cm/sec). (d) Describe the situation and What from the standpoint of an t\ is is cl inertial frame attached to the laboratory. [In (b) and account must be taken of buoyancy (c), effects. Think of the ordinary hydrostatics problem of a body completely immersed in a fluid of different density.] 12-18 Show (a) that the equilibrium form of the surface of a rotating body of liquid is parabolic (or, strictly, a paraboloid of revolution). This problem is most simply considered from the standpoint of the rotating frame, given that a liquid cannot withstand forees tangential to its surface disappear. and will tend toward a configuration in which such forees It is instructive to consider the situation from the standpoint of an inertial frame also. (b) It has been proposed that a parabolic mirror for an astronomical teleseope might be formed from a rotating pool of mercury. What rate of rotation (rpm) would make a 12-19 To a first More proportional to m? approximation, an object released anywhere within an orbiting spacecraft spacecraft. mirror of focal length 20 its remain will in the accurately, however, same it place relative to the experiences a net force distance from the center of mass of the spacecraft. This force, as measured in the noninertial frame of the craft, arises from the small variations in both the gravitational force and the centrifugal force due to the change of distance from the earth's center. Obtain an expression for this force as a funetion of the mass, m, of the object, its distance AR from the center of the spacecraft, the radius R of the spacecraft's orbit around the earth, and the gravitational acceleration 550 gR at the distance Inertial forees R from the earth's center. and noninertial frames 12-20 A velocity w = t 0.2 rad/sec. = velocity v' = m circular platform of radius 5 1 A man rotates with an angular of mass 100 kg walks with constant At time jumps off the edge m/sec along a diameter of the platform. he crosses the center and time at = t 5 sec he of the platform. Draw a graph of (a) the centrifugal force function of time in the interval Draw a (b) = / to t = felt by the man as a 5 sec. similar graph of the Coriolis force. For both diagrams, give the correct vertical scale (in newtons). Show on a (c) sketch the direction of these forces, assuming the platform to rotate in a clockwise direction as seen from above. On 12-21 a long-playing record (33 rpm, 12 Assume crawl toward the rim. and the record its legs Does is 0.1. an in.) insect starts to that the coefficient of friction between it reach the edge by crawling or otherwise ? A child sits on the ground near a rotating merry-go-round. With respect to a reference frame attached to the earth the child has no acceleration (accept this as being approximately true) and experiences no force. With respect to polar coordinates fixed to the 12-22 merry-go-round, with origin at (b) What What (c) Account for (a) its center: is the motion of the child? is his acceleration? this acceleration, as acting on measured in the rotating Coriolis forces judged to be and frame, in terms of the centrifugal the child. 12-23 The text (p. 516) derives the Coriolis force in the transverse (8) motion of an object along a radial direction by considering the line one considers an object in the that is moving transversely rotating frame, one can obtain the Coriolis and net radial force due to centrifugal effects. Consider a Correspondingly, in the rotating frame. on a particle The particle from the if frictionless turntable rotating with angular velocity w. at rest relative to the turntable, at a distance r is initially axis of rotation. up a fixed coordinate system S with axes x transversely and with its origin O at the position of the particle at (see the figure). Set up another coordinate system S', with O' and axes x' and y', which rotates with the turntable and (a) Set and y t = origin radially which coincides with S at / = 0. Show that at a later time ordinates of a given point as measured in S' and following equations, where 8 x' = x cos 8 + y sin 8 + y' = (b) 551 = y cos 8 Suppose Problcms — S / the co- are related by the wt: r sin 8 x sin 8 — r(l that, at / = 0, the particle is given a velocity v' — cos 8) , relative to O' in the x' direction. the x subsequent motion Its — direction at the constant velocity o' this to obtain its coordinates x' (c) Making for small values of / and at a later time one can put y' ~ %a'r i 2 , where be along Use to O. /. the approximations for the case wt t will tar relative a'T « 1, = co show 2r — that Tmvf This corresponds to the required combination of centrifugal and Coriolis accelerations. (d) If you are feeling ambitious, apply the for an initial velocity in 12-24 In an an article entitled 16, 246 (1903)], same kind of analysis arbitrary direction. "Do Objects Edwin Hall reported fail South?" [Phys. Rev., the results of nearly 1000 trials in which he allowed an object to fail through a vertical distance of 23 m at Cambridge, 42° N). Mass. eastward deflection of 0.149 (a) Comparc (lat. cm and a He found, on the average, an southerly deflection of 0.0045 cm. the easterly deflection with what would be expected from Eq. (12-17). (b) Consider the fact that the development of an eastward component of motion relative to the earth would indeed lead in turn to a southerly component of Coriolis force. Without attempting any detailed analysis, estimate the order of resulting southerly deflection to the Do you think flection that magnitudc of the ratio of the predominant easterly an explanation of HalPs results deflection. on southerly de- can be achieved in these terms? 12-25 Calculate the Coriolis acceleration of a satellite in a circular polar orbit as observed by someone on the rotating earth. Obtain the direction of this acceleration throughout the orbit, thereby explaining why the satellite always passes through the poles even though subjected to the Coriolis force. in a similar force on it is a satellite an equatorial orbit? 12-26 Imagine that a and of radius R, 552 Is there frictionless horizontal table, circular in is fitted Inertial forces shape with a perfectly elastic rim, and that a dry-ice and nonincrtial frames 2 puck is . The launchcd from a point on the rim toward the center. puck bounccs back and forth across the table at constant speed because of the Coriolis force path along a diameter. it does not quite follow a straight-line Consider the rate at which the path of the puck gradually turns with respect to the table, and compare the with that for a Foucault pendulum at the same latitude, X. 12-27 In the text (p. but v, 536) the height of the equilibrium tide result is cal- carrying a particle work done by the tide-producing force in of water from point D to point A (see the figure). By work from culated by considering the considering the D to an intermediate point P, one can obtain a general expression for the elevation or depression h{6) of the water at an arbitrary point, relative to what the water level would be in the absence of the tide-producing force. two The calculation involves parts, as follows: work P(x = (a) Evaluate the D(x = y = Re) water of mass m. 0, to (b) Re cos 0, y = Re sin d) Equating potential energies, mg(iip ference hp — integral of the tide-producing force — of gravitational this to the difference an expression ho), one gets from for a particle of for the dif- ho- The total volume of water is a constant. Hence, if ho represents the water depth in the absence of the tide-producing force, we must havc r/ L 2TR K -[h(6) - ho]sinddd = Putting the results of (a) and (b) together, you should be able to verify that the deviation of the water level is 553 proportional to 3 cos 2 Problem s — 1 from its undisturbed state What makes planet s go around the sun ? At the time of Kepler some people answered this problem by saying that there were angels behind them beating their wings pushing the planet s around an answer is is orbit. not very far front the truth. that the angels sit in a As you and will see, the The only difference different direction and their wings push inwards. R. P. feynman, The Character of Physical Law (1965) 13 Motion under central forces we have already seen, especially in Chapter 8, how the motion of objects under the action of forces directed toward some welldefined center is one of the richest areas of study in mechanics. Twice in the history of physics the analysis of such motions has been linked with fundamental advances in our understanding of nature through the explanation of planetary motions, on the — macroscopic scale, particle scattering, Up subatomic world. studies of alpha- and through Rutherford's which gave man his first until this point to the study of circular orbits, and clear view of the we have it is limited ourselves remarkable how much can be learned on that basis. But now we shall begin a more general analysis of motion under the action of central forces. BASIC FEATURES OF THE PROBLEM As we saw in Chapter 1 1 (p. 444), a central force field that conservative must be spherically symmetric, and important fields in is also some of the most nature (notably electrical and gravitational) are precisely of this type. The frequent occurrence of symmetric models to describe physical the basic assumption that space is reality isotropic is and spherically closely linked to is the intuitively natural starting point in building theoretical models of various kinds of dynamical systems. We shall begin with the specific single particle of mass w in problem of the motion of a a spherically symmetric central field 555 Fig. 13-1 (a) Unit veciors associated with radial and transverse directions in a plane polar coordinale system. (b) Radial and transverse components ofan elementary vector displacement Ar. of force. we Initially, at least, sponsible for this central ficld assume that the object shall so massive that is it re- can be regarded as a fixed center that defines a convenient origin of coordinates for the analysis of the motion. The first thing to notice This plane particle and the defined by the is initial plane of and v r To . is the motion it, vector v no component of initial velocity must remain confined to this analyze the motion we must appropriate coordinate system. Because the force of the scalar distance r only and (positively or negatively), of the of the particle with re- r Since the force acting on the particle and sincc thcrc in this plane, perpendicular to initial velocity vector position spect to the center of force. is moving par- a fixed plane that passes through the center of ticle will lie in force. that the path of the is it is with the plane polar coordinates is F first is pick an a function along the line of the vector r clearly (r, 0), most convenient to work as indicated in Fig. 13-l(a). This means that we shall be making use of the acceleration vector expressed in these coordinates. lated (r this = vector for the constant). Now we In Chapter 3 (p. 108) particular shall unit vectors e r calcu- case of circular motion develop the more general expression that embraces changes of both r Using the we and and 0. e» as indicated in Fig. 13-1 (a) we have r = V= (13-1) re, dr d, = dr dt C' + This equation for v dd r is (13-2) e° di by recognizing that a Ar, is obtained by com- readily constructed general infinitesimal change in position, bining a radial displacement of length Ar (at constant 0) and a 556 Motion under cenlral forccs : : transverse displacement of length r A6 (at constant r), as indicated Alternatively, one can just differentiate both in Fig. 13-1 (b). of Eq. (13-1) with respect to time, remembering that sides = d(e r)/dt {dO/dtytt [see Eq. (3-17a)]. We now proceed to differentiate both sides of Eq. respect to = a (13-2) with / 2 r dv d dt = dT* d e + + dr d d, dl ' t <f$ + r dT2 e , ee d-t ' dd d + , Substituting d(e r)/dt dr dO , , er r = dtdt* (d6/dt)ee , and d(e$)/dt = - (dd/dt)c T , the expression for a can be rewritten as follows: ir. er dfi + r ^1+ 2— — dfi Cfl (13-3) di dt be convenient to extract from this the separate radial and It will transverse a' components of the = d\ dfid (de\ r d + , dfi 2 (13-4) \jt ) 2 at=r-rT total acceleration , dr dB 2-r-r (13-5) dt dt The statement of Newton's law in plane polar coordinates can then be made n terms of these separate acceleration components: i FT = m d\ F» = m (13-6) 1 dfi d^£ r dfi + dr dd (13-7) dt dt The above two equations provide a problem of motion coordinates. We basis for the solution of in a plane, referred to shall, any an origin of polar however, consider their application to central forccs in particular. THE LAW OF EQUAL AREAS In the case of any kind of conservative central force, FT = F(r) simply, implies that as 557 = and Fe = 0. The law of equal 0. The second of we have these immediately Substituting the specific expression for as areas 13-2 Fig. Illustrating the basis of calculating areal velocity (area swept oul per unit lime by ihe radius vector). from Eq. (13-5), we have r£+2$£-0 dt 2 03-8) dt dt This equation contains a somewhat veiled statement of a simple geometrical result by — that the vector r sweeps out area at a constant One way of seeing rate. 2 2d ' The 6 dfi + lr / may we 2d2 r di) dt2 2 r dd/dt: drdd + ' dt dt we integrate this expression, r Now then be recognized as the derivative with of the product rf/s«\ dt\ in Eq. (13-8) throughout drdd = Q 7iJi left-hand side respect to If this is to multiply r: 2 therefore have = const. ? dt in Fig. 1 (13-9) 3-2 we a short time Al. show It the area AA the triangle is (shaded) swept out by r POQ indistinguishable from a straight line if it is (we take PQ to be short enough) and we have + Ar) sin A6 AA - iKr The rate at limit of Ar/r for At and Ad —» A= d dt which area AA/At \ 2 r sin being swept out, instantaneously, —* > 0. A0, Since, as we we approach is the this limit, arrive at the result ^l (,3-10a) dt Thus we recognize 558 — is the constant on the right-hand side of Eq. Motion undcr ccntral f'orces s A por- Fig. 13-3 tion of Newton' (a) manuscript, (a) De Motu, showing the basis of his proofofthe law of equal areas for a cenlral force. (6) En- larged copy of Newton' s diagram. J. Herivel, (.Front The Back- ground to Newton's Principia, Oxford University Press, London, 1965.) (13-9) as twice the rate (a constanl rate) at which the radius vector r sweeps out area, and (Any The Kepler It was therefore have —=s central force) result expressed in his we dA 1 dt 2 r 2 dd — = (13-10b) const. t/r by Eq. (13-10b) was first discovered by analysis of planetary motions (of which stated by him in what is known more later). as his second law (although Newton understood it on the samc dynamical grounds as we have discussed above, it was actually the i.e., chronologically). as a feature of the motion of an object acted kind of force that 559 first The law of cqunl is directed always to the areas same on by any point. He visualized the action of such a force as a succession of small kicks or impulses, which in the limit would go over into a continuously He applied infiuence. out this view of the process in a tract set written in 1684 (about 2 years before the Principia). 13-3(a) 1 Figure a reproduction of a small fragment of the work, is indicating Newton's approach to the problem. With the help of an enlarged version of his sketch [Figure 13-3(b)] we can more readily follow Newton's argument, which as usual was geometrical. Newton imagines an object traveling along receiving an irripulse directed toward the point S. now carry and to D, E, it BC along the line travels instead of Bc. To make F. visualizes the displaccmcnt AB and then As a things quantitative, BC as being, in effect, result it Similar impulses Newton the combination of the displacement Bc, equal to AB, that the object would have undergone if it had continued for an equal length of time with original velocity, together with the displacement BS the line along which the impulse was applied. yields the Iaw of areas by a simple argument: The and SBc are equal, having equal bases altitude. The triangles and (SB) base between the This at once triangles equal, having a same its parallel to {AB and Bc) and SBc and SBC are lying cC parallels. SAB the same common Hence ASAB = ASBC. THE CONSERVATION OF ANGULAR MOMENTUM We give a more modern and more fundamental it in angular momentum. a particle at If slant to the law terms of the conservation of orbital of areas by expressing P [Fig. 13-4(a)] is acted on by a force F, we have F = /na = & m-7dt Let us now form the vcetor (cross) produet of the position veetor r with both sides of this equation: rXF The = rXmât left-hand side is (13-11) the torque M due to F about O. De Motn (Concerning the Motion of Bodies), contains of the important results that were later incorporated in the Principia. 'This tract, called many 560 Motion under central forees Fig. 13-4 (a) Vector relationship ofposilion, linear momentum, and orbital momentum, (b) Analysis of the velocity into radial and transverse components. We now introduce the moment of the momentum, of the I, particle with respect to O'. l Thus the = rX/wv = 1 rXp (13-12) bears the same relation to the linear moment of the force, M, does to F. It is momentum, in Fig. 13-4(a), perpendicular to the plane defined If we calculate the rate of change of with respect to d\ dt = ji Jt — we have 1 — i. e., p, as a vector, as shown by r and v. the derivative of 1 / w w dv x m y+rxm, \Xmv + rXm dt dt = However, the value of v X mv product of two parallel vectors. of the above equation zero, because is the cross The second term on the right is it by Eq. (13-1 1), equal to the torque or moment of the applied force about O. Thus we have the simple is, result M=rXF=^dt (13-13) moment of F to the rate of change of the momentum, or orbital angular momentum, of the about the origin O. If, now, we take F to be a central This, then, relates the moment particle of force direeted radially is zero. toward or away from O, the value of we have In this case, therefore, ?=° dt 561 The conservation of angular momentum M and = 1 X r This, then, = p the statement of the conservation of orbital angular is momentum for under zero force. motion under a central force we look If (13-14) const. I magnitude = of course, the situation in the plane of the motion at [Fig. 13-4(b)] the vector scalar — or, / is 1 points upward from the page, and — = mr2 de rmo» its given by (13-15a) dt This means that the constant rate of sweeping out area, as expressed by Eqs. (13-10a) and (13-10b), dA _\_ dt ~ The d8 2 T 2 dt (13-15b) As we have by Eqs. (13-14) and (13-15) if we some process pointed out before, — these bccome powerful phenomena. Angular momentum sort which problems. is is a valu- find quantities — that remain constant throughout "conserved" given quantitatively by = ~ J_ 2m result expressed able one. is i.e., they are tools in the analysis of a conserved quantity of this particularly valuable in the analysis of central field is It should be noted that the conservation of angular momentum depends only on the absence of a torque and is independent of the conservation or nonconservation of mechanical energy. A nice example of this last feature an object that whirls around compared to at the r itself, the velocity v at each instant, is tension T one revolution Fig. is 13-5 = 562 mwi Small portion of the path of an object that O at the end of a string the being whirled around length of which is heing steadily shorlened. Moiion under is small is very nearly in the string that pulls the object inward exerts no torque about O, and so we have a change of r from r\ to r 2 we thus have mv\r\ up of almost perpendicular to r and the angular momentum (rmog ) The equal to mur. the speeding end of a gradually shortening If the decrease of r in string (Fig. 13-5). is central forees / = const. In or V2 Thus, r\ — 01 = than n, there if r 2 is less K2 The work Kt is JW (^ - = a gain of kinetic energy given by l) equivalent to this gain of kinetic energy must be provided by the tension in the string. In an equilibrium orbit we would have — 2 T= r Substituting v .2 T= = I/mr, this becomes mvi /wr 3 2 2 r\ r* The work done by 7" in a change from r to r T acts radially inward). is given by w It may 2 f 2 be seen that Thus dr . the total + dr —T dr (since is work between 22/1 The problem has a good of the earth satellite spiraling servation of orbital angular deal in K2 — ENERGY CONSERVATION If we IN r2 ATi already common with that inward, although in that casc con- momentum does not apply, because the air resistance represents a transverse force providing tive torque (see discussion and l\ equals the value of this calculated. rx a nega- on pp. 470-473). CENTRAL FORCE MOTIONS are dealing with a conservative central force (which the case in the example analyzed above) we can was not write a statement of the conservation of total mechanical energy in the following form: a I [Or + V,*}+ In addition, 563 we have f/(r) = E the law of conservation of angular Fnergy conservation in central force motions (13-16) momentum as given by Eq. (13-15a): mrvt = l The quantities U(r) is E and / are thus "constants of the motion." the potential energy of the particle m in The orbital-momentum equation allows us problem of a to the form of a particle the central field. to reduce Eq. (13-16) moving in one dimension only undcr the action of a conservative force. This is the key to From Eq. the method of handling we take the value of u$ and use this value in Eq. (13-16). central field problems. (13— 15a) There follows ¥ h ™ + + - B (13-17) which describes the radial part of the motion. This ^- + U'(r) is of the form = E (13-18) where 2 u'(r) = / + V(r) Imr-i The quantity V '(r) plays the rolc of an equioalent potential energy for the one-dimcnsional radial problem. l 2 /2mr 2 takes complete account, The additional term as far as the radial motion is concerned, of the fact that the position vector r in the actual motion is continuously changing direction. its But it is to be noted that nothing that depends explicitly on the angular coordinate d or its time derivatives appears The term 2 l /2mr 2 is in Eq. (13-18). often referred to as the "centrifugal*' potential energy, because the force represented by the negative gradient of this potential energy rwntrifugnl Putting / = T = mr* -\ 2 is identical with the centrifugal force mui r in a rotating at an angular velocity co Molion under frame cqual to the instantaneous value of dO Idi. 564 given by z mr' (de/dl), this becomes ftmttrtfugal which dr \2mr2/ is central forees must be remembered, of course, that the "centrifugal It potential energy" is particle: that part due in fact a portion of the kinetic energy of the to the componcnt of its motion transverse The circum- to the instantaneous direction of the radius vector. stance that this kinetic-encrgy contribution can be expressed as a function of radial position alone enables us to treat the radial motion as an independent one-dimensional problem. The potential-encrgy function U(r) in Eq. (13-17) can be any In developing this discussion function of radius. we shall first consider the general properties of motion in any central field for which U(r) approaches zero as Subsequently we shall r increases. take up the important special case of an inverse-square foree field in which the potential energy of a particle portional to is inversely pro- r. USE OF THE EFFECTIVE POTENTIAL-ENERGY CURVES The general properties of the motion of a particle in a central field can be most readily obtained by using the energy-diagram method of Chapter 10 applied to Eq. (13-17) or Eq. (13-18). There are, however, two significant differences between the use of the energy method for one-dimensional motion and for one-dimensional motions, as in the case of central in use its any two-dimensional motion which can be reduced to two fields. First, one dimension, the energy alone determines the general charaeter of the motion in a conservative angular momentum / In the central-field field. case, however, specification of the energy is The not enough. of the particle must also be specified, and on both parameters E and /, instead of on E alone. This shows up clearly in the energy diagram, because the 2 2 plot of U' = U + (l /2mr ) will be different for different values the motion depends of /. tial We have, then, a whole family of curves of equivalent poten- energy, corresponding to different values of the angular momentum, and must consider what happens to the motion with a given energy for the different values of /. In addition, to translate the results obtained for the radial motion by this seheme into the actual motion of the must remember that while the radial coordinate r with time, so is 6. The changes in r are taneous rotation of the vector particle 565 depends on both. Use of the The r, particle, is we changing accompanied by a simul- and the actual orbit of the rotation of the vector r elTcctive poiemial-energy curves is not uniform except in the special case of circular motion, for which the length of the position vector r does not change with time. The angular momentum / and the energy E define the basic dynamical conditions and are related to the position r velocity v E= = I Since we a total pletely. each) Fig. at an arbitrary time —=- + LH.ro) (13-19) are dealing with a two-dimensional problem, bf four A initial statement of the vectors r and v purpose. (a) True potential energy U(r), ferent values ofl. we need conditions to specify the situation com- and several "centrifugal poleiitial energy" curves belonging to dif(b) Resultan! effective potential energy U' (r) for different orbital momenta. 566 and by mro(c t )o fulfills this 13-6 / Motion undcr ccntral forccs (two components Now let us look at a typical energy diagram. In Fig. 13-6(a) are plotted the curves of values of the angular t 2 /2mr 2 against momentum, and r for several different U(r) against r for an at- from tractive potential that rises with increasing distance U(r) — » energy —> oo curves U '(r) as r r U(r) with each of the separate centrifugal potential curves. U how consider 0. obtained by combining the single function obtain curves with minima as shown, variation of = Figure 13-6(b) shows the effective potential- . with r to be is it rapid than l/r less 2 Let us .) draw valuable to use such curves to (To necessary for the now qualitative conclusions about the possible motioijs. In Figure 13-7 drawn we show an for a particular value of value of /, effective potential -energy curve We /. no physically meaningful then see that, given this situation can exist for Em value of the total energy less minimum value of U'(r). At no motion can occur; the motion must be radial than the energy this minimum any equal to the possible value of E, circular with a For a range of larger values of E, between Em and zero, the radial motion will be periodic (e.g., E = Ea or Eb ). With the assumed form of U(r), all motions with a positive value radius equal to r of E (e.g., E c) but no upper . are unbounded; there limit. If given, but the orbital seen [most readily momentum from permissible value of /; is a least possible value of r one chooses to regard the value of as a variable, then Fig. 13-6(b)] that there is a any value from this down and the maximum value corresponds to a Fig. 13-7 to zero Effective diagram from wlrich the character ofthe radial motion for different values total ofthe energy can be inferred. 567 Use ofthe is effective potential-energy curves as allowed, circular orbit. polential-energy E may be maximum it BOUNDED ORBITS With the hclp of the above preliminary analysis of the radial component of the motion, one can then proceed to develop some ideas about the appearance of the actual orbits in space. Suppose, we had the situation corresponding to the The radial motion is bounded between for example, that Eb energy in Fig. 13-7. minimum and maximum values certain period T,. Thus we know periodic with certain of a that the partiele r. It is moves always within the area between two circles as shown in Fig. 13-8. Furthermore, the radial distances r min and rmax represent turning points of the radial motion. The orbit must be such that it is tangent to both these circles, because at these points the radial velocity is zero but the tangential velocity cannot be zero, given that the partiele has angular after its has reached point it and continuing radial period Tr . changing eloekwise through 2w OA until On its or line on the takes for this part of the motion the other hand, the radius veetor direetion, always in the counterclockwise) O A' represents the veetor ratio of the radial which the and angular periods differ. effect is that The of an approximately elliptic orbit that keeps turning {precessing) in own its plane. 568 the con- same sense (i.e., will have turned In Fig. 13-8 the line some instant, position at the time Tg later. two periods, T, and Plan view in and is is the charaeter of the orbit depends strongly of an orbital motion for a case C again becomes tangent to the outer cirele it after a charaeteristie period T$. It is clear that 13-8 cirele at point represents the veetor position of the partiele at and the Fig. it The time at point B. either momentum. Consider the partiele moves in as indicated, of the figure. It becoming tangent to the inner trajectory tinually A Motion under central forees Tg, of the doubly periodic motion. If the periods are commensurable be expressed as the ratio of two will ultimately (after Tr and 7» their ratio (i.e., if a time equal to the lowest can moving particle common multiple integers), the same position as initially and the orbit will thus have been closed. If the two periods happen to be exactly equal, this closure will happen after only of find itself in exactly the one radial period and one increment of this sees that this is in it is precisely In Fig. 13-8 6. a unique and (on the face of improbable situation, yet is lir A and B would would mean that the points coincide. One a thoroughly it) what one has if the force one of attraction proportional to the inverse square of Thus we of this remarkable character, as trostatic) yield orbits show r. the most important forces in nature (gravitational and elccshall shortly. and angular periods are comparable but defone has just the kind of situation shown in If the radial initely different, then Fig. 13-8. This corresponds to a case in which T, greater than 7"«, somewhat so that the radius vector rotates through rather more than 2x before r completes and back again. In a situation to being a closed curve but its variation is, from rmax to where the path like this, precession orbital is rmin near in effect, also turning (either forward or backward), one says that the orbit The study of is is is precessing. important in astronomical systems. If the radial and angular periods are incommensurable, the orbit will never close and will eventually fiil up the whole region between rmi „ and rm „. UNBOUNDED ORBITS We have already pointed out that a positive total energy leads to a lower limit of r but to no upper limit in the effective potential of Fig. 13-7. This is, in fact, a rather any potential that tcnds to zero at r = general result, applying to oo, because it corresponds to the fact that the particle has a positive kinetic energy at infinity. If U(r) is a repulsive potential, ing that it everywhere positive, then (assum- decreases monotonically with increasing r) there are no bounded motions at all. Figure 13-9 compares the situations obtained by taking a given centrifugal potential and combining it with attractive and repulsive potential-energy curves that differ only in sign. Figure 13-10 shows what this means in terms of the trajectories that a particle of a given energy 569 Unbonnded orbits would follow in these Fig. 13-9 (a) Centrifugal potential-energy curve, two potential-energy curves, differing only and in sign, that might arise from electrical interactions oflike and unlike charges. (b) Effectice potential-energy curves corresponding to tlie two cases shown different distances ofclosest in (a), indicating approach for a gicen posiliue total energy. At large distances from the center of force, such 2 2 that the magnitudes of U(r) and l /2mr are both negligible, the two situations. travels particle (2£/w) 1/2 in a straight equal to with a speed v linc This line of motion . offsct is from a parallel line through the center of force by a certain distance b that related to the angular / = mv momentum we is directly have, in fact, ; (13-20) b Thus an assumption of given values of expressed by using the values of Po ar| d E and / in Fig. 13-9 is b " n F'g- 13-10(a), which corresponds to an attractivc potcntial, and in Fig. 13— 10(b), which corresponds to a repulsivc potential. called the impact parameter, and it is The distance b is a very useful quantity for characterizing situations in which a particle, in an unbounded a large distance away. approachcs a center of force from For a given value of u a the value of b completely defines the orbit, , orbital angular It is momentum, clear that thesc via Eq. (13-20). unbounded trajectories represent single is no bounded encounters of the particle with the center of force; there possibility of successive rcturns as 570 Motion tituler central forees we have with the 13-10 Fig. (a) Plan view oftrajectory ofa parlicle moving around a center of attraction. The angular momentum is defined by Ihe "impact parameter" (b) b. Corresponding trajectory with the same impact parameter, but with a repulsive center of force. orbits of the kind shown in Fig. 13-8. The situations shown in Fig. 13-10 thus represent individual collisions or scattering processes, We of the kind so important in atomic and nuclear physics. shall return to them Certain potentials Fig. 13-11 struction Con- of effective potential-energy diagram for an attractive potential that varies more strongly than l/r". 571 Unbounded orbits later. may lead to the possibility of having both bounded and unbounded motions at the same energy. This any attractive potentiai curve U(r) that falls possibility exists for 2 more rapidly than l/r with increasing r. An example is shown in Fig. 13-11. As long as / is not zero, the combination off of U(r) with the centrifugal potentiai gives an effective potential- much energy curve that looks very of Fig. 13-7. maximum of For a positive £/'(/), there are total like an upside-down version energy now two E that is than the less distinct regions of possible motion: ri This is < < r < r < w bounded r\ orbits unbounded orbits not just an academic curiosity. The effective potential- energy curve representing the interaetion between, us say, a let proton and an atomic nuclcus resembles Fig. 13-11 and suggests that the partiele may be either trapped within r it or frec outside r2> with no possibility of going from one state to the other. It is, we first mentioned in Chapter 10, one of the faseinating results of quantum mechanies that a transition through the classically as forbidden region can in faet oecur with a certain probability a probability that is far too small to be significant in most cases but can become very important CIRCULAR ORBITS As IN in atomic and nuclear systems. AN INVERSE-SQUARE FORCE FIELD a quantitative example of the general approach diseussed in the preceding seetions, cireular orbits under we shall take the case of circular or almost 2 an attractive central foree varying as l/r . This means that we shall mainly be diseussing, from a different point of view, a situation that detail in earlier chapters. we have There is already analyzed in merit in this, because it some enables one to see more readily the relationship between the different approaches. To be specific, let us consider the motion of a satellite of mass m in the gravitational field of a vastly more massive planet. For this case, the potentiai energy in the field of a planet of mass takes the special form [see Eq. (11-31)] M u(r) where G = is -9Mm the universal gravitational constant of the attracting object. 572 (i3-2i) Motion under and M the mass Equation (13-17) becomes for central forees this case 2 GMm ,2 2 2mr2 = E (13-22) In Fig. 13-12 are plotted the equivalent potential-energy curves U'(r) GMm f = 2mr2 for two valucs of the angular momentum A circular orbit corresponds, as minimum energy equal to the Now dr mr* Putting dU'/dr f The = GMm total /. U '(r), taking / as fixed, we have = r* = 0, we thus find that const. orbital radius orbital angular a seen, to GMm = _ _T_ ' we have value of U'(r) for a given value of from the above equation for di/ /. is (13-23) therefore proportional to the square of the momentum. This is indicated qualitatively in Fig. 13-12. Let us orbit is now consider the energy of the motion. A v r always equal to zero. , Thus in Eq. (13-22) we can put GMm E= (Circular orbit) 2/W/-2 From Fig. 13-12 Eq. (13-23), however, we have Energy diagrams for orbils of different orbital angular momentum around a given center offorce. 573 circular characterized by having the radial velocity component, Circular orbils in an inverse-square force field GMm 2 l mr* r Hence the energy can be expressed E= - (Circular orbit) The second form shows portiond to that \E\ in either of the following -^-r = (13-24) that the orbital radius \E\; the first, is inversely pro- taken together with Eq. (13-23), shows inversely proportional to is ^^ - ways: l 2 . These properties, too, are qualitatively indicated in Fig. 13-12. The period of the orbit can be obtained, in this approach, with the help of the law of areas. Earlier in this chapter we established the following result [Eq. (13-15b)]: dA _ J 2ra dt For a is irr in circular orbit, the total area Thus we have dA/dt simply mvr. 2 /(ur/2) more putting — hardly 2irr/v interesting terms is = *r 2 and the value of vr/2 and hence 7» a surprising result! we can make To / = express Tg use of Eq. (13-23), = mvr: / (mvr) mr = A = GMm giving 1/2 °-m Using this explicit expression for uasa function of r, we then arrive once again at the equation that expresses Kepler's third law: T9 _ . 2t - (G/uy* 312 (13-25) r SMALL PERTURBATION OF A CIRCULAR ORBIT we limitcd ourselves to redeveloping some now let us do something new, which exploits In the previous scction familiar results. But the insights provided by the effcctive potential-energy method. Imagine that a energy 574 E and radius satellitc r is describing a circular orbit of around a sphcrical planet whose center Mo

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