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Chapter 12 Homework
Due: 9:00am on Wednesday, November 11, 2009
Note: To understand how points are awarded, read your instructor's Grading Policy.
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Ladybugs on a Rotating Disk
Two ladybugs sit on a rotating disk, as shown in the figure (the ladybugs are at rest with respect to the surface of the disk and do not slip). Ladybug 1 is
halfway between ladybug 2 and the axis of rotation.
Part A
What is the angular speed of ladybug 1?
ANSWER:
one-half the angular speed of ladybug 2
the same as the angular speed of ladybug 2
twice the angular speed of ladybug 2
one-quarter the angular speed of ladybug 2
Correct
Part B
What is the ratio of the linear speed of ladybug 2 to that of ladybug 1?
Hint B.1
Relation between linear and angular speeds
The relation between the linear speed
and angular speed
of an object is given by
,
where
is the distance between the object and the axis of rotation.
Answer numerically.
ANSWER:
2
Correct
Part C
What is the ratio of the magnitude of the radial acceleration of ladybug 2 to that of ladybug 1?
Hint C.1
Radial (centripetal) acceleration of an object moving on a circle
Hint not displayed
Answer numerically.
ANSWER:
=2
Correct
Although the trajectory of ladybug 2 has twice the radius as that of ladybug 1, ladybug 2 also has twice the linear velocity of ladybug 1. Thus, according to the formula
, where
is centripetal
acceleration, ladybug 2 has twice the centripetal acceleration of ladybug 1.
Part D
What is the direction of the vector representing the angular velocity of ladybug 2? See the figure for the directions of the coordinate axes.
Hint D.1
Direction of the angular velocity vector
Hint not displayed
ANSWER:
Correct
Part E
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Now assume that at the moment pictured in the figure, the disk is rotating but slowing down. Each ladybug remains "stuck" in its position on the disk. What is the direction of the tangential component of the
acceleration (i.e., acceleration tangent to the trajectory) of ladybug 2?
ANSWER:
Correct
A Spinning Electric Fan
An electric fan is turned off, and its angular velocity decreases uniformly from 550
to 210
in a time interval of length 4.05 .
Part A
Find the angular acceleration
Hint A.1
in revolutions per second per second.
Average acceleration
Hint not displayed
ANSWER:
= -1.40
Correct
Part B
Find the number of revolutions made by the fan blades during the time that they are slowing down in Part A.
Hint B.1
Determine the correct kinematic equation
Which of the following kinematic equations is best suited to this problem? Here
and
are the initial and final angular velocities,
is the elapsed time,
is the constant angular acceleration, and
and
are the
initial and final angular displacements.
Hint B.1.1
How to chose the right equation
Hint not displayed
ANSWER:
Correct
ANSWER:
25.7
Correct
Part C
How many more seconds are required for the fan to come to rest if the angular acceleration remains constant at the value calculated in Part A?
Hint C.1
Finding the total time for spin down
To find the total time for spin down, just calculate when the velocity will equal zero. This is accomplished by setting the initial velocity plus the acceleration multipled by the time equal to zero and then solving for
the time. One can then just subtract the time it took to reach 210
from the total time. Be careful of your signs when you set up the equation.
ANSWER:
2.50
Correct
Torques on a Ruler
Acentimeter ruler, balanced at its center point, has two coins placed on it, as shown in the figure. One coin, of mass
the other, of unknown mass
, is placed at the 4.7
mark. The center of the ruler is at the 3.0
, is placed at the zero mark;
mark. The ruler is in equilibrium; it is perfectly
balanced.
Part A
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Does the pivot point (i.e., the triangle in the diagram upon which the ruler balances) exert a force on the ruler? Does it exert a nonzero torque about the pivot?
Choose the correct series of answers to these two questions.
ANSWER:
Correct
Note that the weight of the ruler itself also does not exert a torque with respect to the pivot point since the rule is uniform and is pivoted at its midpoint.
Part B
Although the pivot exerts a force on the ruler it does not exert a torque with respect to the pivot point. Why not?
ANSWER:
Because the exerted force is vertical.
Because the distance from the exerted force to the pivot is zero.
Because the exerted force is along the ruler.
Correct
Part C
Find the mass
Hint C.1
.
Sum of torques
Hint not displayed
Hint C.2
Determine the moment arm of
Hint not displayed
Express your answer for the unknown mass numerically, in grams, to two significant digits.
ANSWER:
= 18
g
Correct
Precarious Lunch
A uniform steel beam of length
exerts an unknown force,
and mass
is attached via a hinge to the side of a building. The beam is supported by a steel cable attached to the end of the beam at an angle , as shown. Through the hinge, the wall
, on the beam. A workman of mass
sits eating lunch a distance
from the building.
Part A
Find
, the tension in the cable. Remember to account for all the forces in the problem.
Hint A.1
Pick the best origin
Hint not displayed
Hint A.2
Calculate the sum torques
Hint not displayed
Express your answer in terms of
,
,
, , , and , the magnitude of the acceleration due to gravity.
ANSWER:
=
Correct
Part B
Find
, the -component of the force exerted by the wall on the beam (
Hint B.1
), using the axis shown. Remember to pay attention to the direction that the wall exerts the force.
Find the sign of the force
The beam is not accelerating in the -direction, so the sum of the forces in the -direction is zero. Using the given coordinate system, is
Express your answer in terms of
ANSWER:
going to have to be positive or negative?
and other given quantities.
=
Correct
Part C
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Find
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, the y-component of force that the wall exerts on the beam (
Express your answer in terms of
ANSWER:
, ,
,
), using the axis shown. Remember to pay attention to the direction that the wall exerts the force.
, and .
=
Correct
If you use your result from part (A) in your expression for part (C), you'll notice that the result simplifies somewhat. The simplified result should show that the further the luncher moves out on the beam, the lower
the magnitude of the upward force the wall exerts on the beam. Does this agree with your intuition?
Acceleration of a Pulley
A string is wrapped around a uniform solid cylinder of radius , as shown in the figure . The cylinder can rotate freely about its axis. The loose end of the
string is attached to a block. The block and cylinder each have mass . Note that the positive y direction is downward and counterclockwise torques are
positive.
Part A
Find the magnitude
Hint A.1
of the angular acceleration of the cylinder as the block descends.
How to approach the problem
Hint not displayed
Hint A.2
Find the net force on the block
The block has two forces acting on it: the tension of the string and its own weight. What is the net force
Express your answer in terms of
,
(the magnitude of the acceleration due to gravity), and
acting on the block? Use the coordinate system shown in the figure.
(the tension in the string).
ANSWER:
Correct
Hint A.3
The tension
Find the net torque on the pulley
in the string produces a torque that acts on the pulley. What is the torque?
Hint A.3.1
Formula for torque
Hint not displayed
Express your answer in terms of the cylinder's radius
and the tension
in the string.
ANSWER:
Correct
The moment of inertia of a uniform cylinder about its axis is equal to
. Substituting this into the above equation gives
.
Hint A.4
Relate linear and angular acceleration
The string does not stretch. Therefore, there is a geometric constraint between the linear acceleration
of the block?
Express your answer in terms of
and the angular acceleration
. What is the cylinder's angular acceleration
in terms of the linear acceleration
and . Be careful with your signs.
ANSWER:
=
Correct
From this equation,
Hint A.5
. Substitute for
in the force equation for the block.
Putting it together
Solve the system of equations to eliminate
and obtain an expression for
Express your answer in terms of the cylinder's radius
.
and the magnitude of the acceleration due to gravity .
ANSWER:
=
Correct
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Note that the magnitude of the linear acceleration of the block is
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, which does not depend on the value of .
Problem 12.72
The 3.50
, 33.0-cm-diameter disk in the figure is spinning at 310
.
Part A
How much friction force must the brake apply to the rim to bring the disk to a halt in 2.50 ?
ANSWER:
3.75
N
Correct
Problem 12.76
A 5.0 kg, 60-cm-diameter disk rotates on an axle passing through one edge. The axle is parallel to the floor. The cylinder is held with the center of mass at the same height as the axle, then released.
Part A
What is the cylinder's initial angular acceleration?
ANSWER:
21.8
Correct
Part B
What is the cylinder's angular velocity when it is directly below the axle?
ANSWER:
6.60
rad/s
Correct
Problem 12.37
An 7.00-cm-diameter, 330
sphere is released from rest at the top of a 2.00-m-long, 20.0 incline. It rolls, without slipping, to the bottom.
Part A
What is the sphere's angular velocity at the bottom of the incline?
ANSWER:
88.4
rad/s
Correct
Part B
What fraction of its kinetic energy is rotational?
ANSWER:
0.286
Correct
Twirling a Baton
A majorette in a parade is performing some acrobatic twirlings of her baton. Assume that the baton is a uniform rod of mass 0.120
and length 80.0
.
Part A
Initially, the baton is spinning about a line through its center at angular velocity 3.00
. What is its angular momentum?
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Hint A.1
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Angular momentum for a rigid body rotating about an axis of symmetry
Hint not displayed
Hint A.2
Moment of inertia
Hint not displayed
Express your answer in kilogram meters squared per second.
ANSWER:
1.92×10−2
Correct
Part B
With a skillful move, the majorette changes the rotation of her baton so that now it is spinning about an axis passing through its end at the same angular velocity 3.00
as before. What is the new angular momentum
of the rod?
Hint B.1
How to approach the problem
Hint not displayed
Hint B.2
Moment of inertia
Hint not displayed
Express your answer in kilogram meters squared per second.
ANSWER:
7.68×10−2
Correct
Here is another way to solve this problem. There is a theorem that relates the angular momentum
center of mass
of an object about an arbitrary axis to the angular momentum of the object about the axis passing through its
:
,
where
is the mass of the object,
is the length of the position vector of the center of mass with respect to the point chosen, and
is the velocity of the center of mass with respect to the point chosen.
Substituting for the values on the right-hand side would yield the same angular momentum that you calculated.
Problem 12.84
A 10 g bullet traveling at 400 m/s strikes a 10 kg, 1.0-m-wide door at the edge opposite the hinge. The bullet embeds itself in the door, causing the door to swing open.
Part A
What is the angular velocity of the door just after impact?
ANSWER:
1.20
rad/s
Correct
Problem 12.45
Force
is exerted on a particle at
.
Part A
What is the magnitude of the torque on the particle about the origin?
ANSWER:
50.0
Nm
Correct
Part B
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What is the direction of the torque on the particle about the origin?
ANSWER:
along the
vector
along the
vector
along the
vector
Correct
Record and Turntable
Learning Goal: To understand how to use conservation of angular momentum to solve problems involving collisions of rotating bodies.
Consider a turntable to be a circular disk of moment of inertia
rotating at a constant angular velocity
around an axis through the center and perpendicular
to the plane of the disk (the disk's "primary axis of symmetry"). The axis of the disk is vertical and the disk is supported by frictionless bearings. The motor of
the turntable is off, so there is no external torque being applied to the axis.
Another disk (a record) is dropped onto the first such that it lands coaxially (the axes coincide). The moment of inertia of the record is . The initial angular
velocity of the second disk is zero.
There is friction between the two disks.
After this "rotational collision," the disks will eventually rotate with the same angular velocity.
Part A
What is the final angular velocity,
Hint A.1
, of the two disks?
How to approach the problem
Hint not displayed
Hint A.2
Initial angular momentum
Hint not displayed
Hint A.3
Final angular momentum
Hint not displayed
Express
in terms of
,
, and
.
ANSWER:
=
Correct
Part B
Because of friction, rotational kinetic energy is not conserved while the disks' surfaces slip over each other. What is the final rotational kinetic energy,
Hint B.1
, of the two spinning disks?
Initial rotational kinetic energy
Hint not displayed
Hint B.2
Final rotational kinetic energy
Hint not displayed
Hint B.3
Putting it all together
Hint not displayed
Express the final kinetic energy in terms of
,
, and the initial kinetic energy
of the two-disk system. No angular velocities should appear in your answer.
ANSWER:
=
Correct
Some of the energy was converted into heat and sound as the frictional force, torque acted, stopping relative motion.
Part C
Assume that the turntable deccelerated during time
before reaching the final angular velocity (
the same angular velocity). What was the average torque,
Hint C.1
is the time interval between the moment when the top disk is dropped and the time that the disks begin to spin at
, acting on the bottom disk due to friction with the record?
Average angular acceleration
Hint not displayed
Hint C.2
Formula for torque
Hint not displayed
Express the torque in terms of
,
,
, and
.
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ANSWER:
=
Correct
Score Summary:
Your score on this assignment is 121.4%.
You received 83.01 out of a possible total of 85 points, plus 20.13 points of extra credit.
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