Version 020 – midterm3 – shih – (58505)
This print-out should have 16 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
001 10.0 points
A wheel starts from rest at t = 0 and rotates
with a constant angular acceleration about a
fixed axis. It completes the first revolution in
3.2 s.
How long after t = 0 will the wheel complete the second revolution?
1. 7.63675
2. 3.39411
3. 8.6267
4. 11.5966
5. 4.52548
6. 8.20244
7. 8.06102
8. 4.24264
9. 11.8794
10. 7.77817
Correct answer: 4.52548 s.
Explanation:
Let :
t1 = 3.2 s ,
θ1 = 1 rev ,
θ2 = 2 rev .
and
The wheel starts from rest, so
θ = θ0 +
1
1
α t2 = α t2 ∝ t2 .
2
2
Thus
θ2
t2
= 22
θ1
t1
r
θ2
t1
θ
r 1
2 rev
=
(3.2 s)
1 rev
t2 =
= 4.52548 s .
002
10.0 points
1
The figure below shows a rigid 3-mass system
which can rotate about an axis perpendicular
to the system. The mass of each connecting
rod is negligible. Treat the masses as particles.
The x-axis is along the horizontal direction
with the origin at the left-most mass 5 kg.
5 kg
7 kg
4m
x
5 kg
4m
The masses are separated by rods of length
4 m, so that the entire length is 2 (4 m).
Determine the x-coordinate of the center of
mass for the three-mass system with respect
to the origin.
1. 5.06667
2. 4.0625
3. 6.0
4. 5.53846
5. 2.14286
6. 1.33333
7. 4.36364
8. 4.0
9. 2.2
10. 3.55556
Correct answer: 4 m.
Explanation:
First find the center of mass. Define the
origin to coincide with the far left mass M .
P
mi xi
XCM = P
mi
(5 kg)(0 )
=
(5 kg) + (7 kg) + (5 kg)
(7 kg)(1 (4 m)
+
(5 kg) + (7 kg) + (5 kg)
(5 kg)(2 (4 m)
+
(5 kg) + (7 kg) + (5 kg)
17 kg
=
(4 m)
17 kg
= 4 m.
3.1 m
Version 020 – midterm3 – shih – (58505)
2
r
003 10.0 points
10
(9.8 m/s2 ) (3.1 m)
=
7
A solid sphere of radius 18 cm is positioned
= 6.58787 m/s .
at the top of an incline that makes 28◦ angle
with the horizontal. This initial position of
the sphere is a vertical distance 3.1 m above
keywords:
its position when at the bottom of the incline.
The sphere is released and moves down the
004 10.0 points
incline.
18 cm
Two disks of identical mass but different radii
(r and 2 r) are spinning on frictionless bearM
ings at the same angular speed ω0 but in opposite directions. The two disks are brought
ℓ
slowly together. The resulting frictional force
between the surfaces eventually brings them
µ
28◦
to a common angular velocity.
Calculate the speed of the sphere when it
reaches the bottom of the incline if it rolls
without slipping. The acceleration of gravity
is 9.8 m/s2 . The moment of inertia of a sphere
with respect to an axis through its center is
2
M R2 .
5
1. 7.93725
2. 4.58258
3. 5.91608
4. 7.29383
5. 6.58787
6. 5.6745
7. 7.38918
8. 6.37181
9. 7.0
10. 7.57628
r
ω0
What is the magnitude of that final angular
velocity in terms of ω0 ?
1. ωf =
2. ωf =
Correct answer: 6.58787 m/s.
3. ωf =
Explanation:
From conservation of energy we have
4. ωf =
5. ωf =
1
M v2 +
2
1
= M v2 +
2
7
M v2
=
10
r
10
gh
v1 =
7
M gh =
1
I ω2
2 2
v
1 2
2
MR
2 5
R2
ω0
2r
6. ωf =
7. ωf =
8. ωf =
9. ωf =
1
5
1
4
2
3
3
5
2
5
3
4
4
5
1
2
1
3
ω0
ω0
ω0
ω0 correct
ω0
ω0
ω0
ω0
ω0
Version 020 – midterm3 – shih – (58505)
Explanation:
Note: Since the disks are spinning in opposite directions, let ω1 = ω0 and ω2 = −ω0 .
The inertia of the larger disk is
I1 =
1
m (2 r)2 = 2 m r 2 ,
2
and of the smaller disk
I2 =
1
m r2 .
2
Using conservation of angular momentum,
I i ωi = I f ωf
I1 ω0 − I2 ω0 = (I1 + I2 ) ωf
I1 − I2
ω0
ωf =
I1 + I2
1
2 m r2 − m r2
2
=
ω0
1
2
2
2mr + mr
2
3
=
ω0 .
5
005 10.0 points
A system of two wheels fixed to each other is
free to rotate about a frictionless axis through
the common center of the wheels and perpendicular to the page. Four forces are exerted tangentially to the rims of the wheels,
as shown below.
2F
3
4. τ = 2 F R correct
5. τ = 5 F R
Explanation:
The three forces F apply counter-clockwise
torques while the other force 2 F applies a
clockwise torque, so
X
τ=
Fi Ri
= (−2 F ) (3 R) + F (3 R) + F (3 R)
+ F (2 R)
= 2F R.
006 (part 1 of 2) 10.0 points
Consider a spool with inner radius b and outer
radius c. A string wraps around the inner
stem in a counterclockwise manner. Starting
from rest, the spool rolls when the string is
pulled horizontally along AB as shown below.
The force F is applied in such way that there
is no slippage.
b
F
c
A
B
Determine the direction of rotation of the
spool.
1. Cannot be determined
3R
2. counterclockwise
2R
F
F
3. clockwise correct
Explanation:
F
What is the magnitude of the net torque on
the system about the axis?
1. τ = 14 F R
2. τ = 0
3. τ = F R
Fnet = m a ,
and
(1)
a
τnet = I α = I ,
(2)
r
Under the action of force F , the spool tends
to accelerate linearly to the right, and to rotate counterclockwise. Therefore the contact
point (presently at P , see the picture below),
will have a tendency to move to the right
Version 020 – midterm3 – shih – (58505)
X
a
against the surface. In turn, the friction force
τ : f c+F b= I
c
must point to the left.
Explanation:
Using equations (1) and (2),
4
b
F
c
A
f
B
P
Choose either A or P as the pivoting point.
Choose A: the torque about A is τ = f (c − b)
clockwise and the spool will rotate clockwise.
Choose P : the torque about P is τ = F (c − b)
clockwise and spool will still rotate clockwise.
The agreement in the direction between these
two different points of view actually tells us
that our choice of the direction of f is correct.
007 (part 2 of 2) 10.0 points
Choose the correct set of equations that govern the motion of the spool. Here f is the
magnitude of the frictional force between the
spool and the horizontal surface, a is the linear acceleration of the center of mass, and I is
the moment of inertia about the center of
mass of the spool. The torque will be evaluated with respect to the center of the spool.
Be careful about the relative signs.
X
1.
Fx : F − f = ma
X
τ : fc−F b=Ia
X
2.
Fx : F − f = ma
X
a
τ : fc−F b=I
b
X
3.
Fx : F + f = ma
X
a
τ : fc−F b=I
b
X
4.
Fx : F + f = ma
X
τ : fc−F b=Ia
X
5.
Fx : F − f = ma
X
a
τ : f c − F b = I correct
c
X
6.
Fx : F − f = ma
Fnet = F − f = m a ,
and for clockwise torque about the center of
mass
a
τnet = f c − F b = I α = I ,
c
a
for rolling
where we use the relation α =
c
without slipping. So the correct pair of equations is
X
Fx : F − f = m a
X
τ:
fc−F b=
Ia
.
c
keywords:
008 10.0 points
A ball of mass 0.4 kg is initially at rest on the
ground. It is kicked and leaves the kicker’s
foot with a speed of 5.0 m/s in a direction 60◦
above the horizontal.
The magnitude of the impulse k~Ik imparted
by the ball to the foot is most nearly
1. k~Ik = 1 N · s.
√
2. k~Ik = 3 N · s.
2
3. k~Ik = √ N · s.
3
4. k~Ik = 2 N · s. correct
5. k~Ik = 4 N · s.
Explanation:
Using Newton’s third law, the impulse imparted by the ball on the foot has the same
magnitude as the impulse imparted by the
foot on the ball:
k~Ik = = k∆~pk = m v
= (0.4 kg)(5.0 m/s)
= 2.0 kg · m/s
= 2.0 N · s .
Version 020 – midterm3 – shih – (58505)
009 10.0 points
Consider a lever rod of length 6.03 m, weight
65 N and uniform density. The lever rod is
pivoted on one end and is supported by a
cable attached at a point 1.62 m from the
other end:
5
W = 65 N ,
b = 1.62 m ,
α = 76◦ , and
β = 68◦ .
For equilibrium,
X
~ = 0 and
F
X
~τ = 0 .
Consider the free-body diagram of the lever
rod:
T
β
1. 62
m
68◦
76
◦
F
6. 03
α
W
m
where T is the tension of the cable and F is
the force on the lever rod at the pivot point.
Let the pivot point be the reference point
of all torques, and let the positive direction be
clockwise. Then
The lever rod is in equilibrium at angle of 76◦
from the vertical wall. The cable makes angle
of 68◦ with the rod.
What is the tension of the supporting cable?
1. 44.1568
2. 13.8436
3. 31.5466
4. 46.5051
5. 57.7092
6. 21.2462
7. 27.7156
8. 28.0963
9. 11.5311
10. 28.8077
Correct answer: 46.5051 N.
Explanation:
Let :
L = 6.03 m ,
τpivot = Fpivot (0) = 0 ,
L
τgrav = +W
sin α ,
2
τcable = −T (L − b) sin β ,
and hence in equilibrium
0 = τnet = τpivot + τgrav + τcable
L
sin α
=0+W
2
− T (L − b) sin β .
Solving this equation for the cable tension T ,
we find
L
sin α
T =W 2
L − b sin β
6.03 m
sin 76◦
2
= (65 N)
6.03 m − 1.62 m sin 68◦
= 46.5051 N .
Version 020 – midterm3 – shih – (58505)
010 (part 1 of 2) 10.0 points
Four particles with masses 4 kg, 6 kg, 4 kg,
and 5 kg are connected by rigid rods of negligible mass as shown. Assume the system
rotates in the yz plane about the x axis with
an angular speed of 7 rad/s.
y
4 kg
5 kg
7 rad/s
O
6 kg
mi ri2 = r 2
X
mi
= 232.75 kg · m2 .
011 (part 2 of 2) 10.0 points
Now assume the system rotates in the xy
plane about the z axis (origin, O) with an
angular speed of 7 rad/s.
y
4 kg
5 kg
x
7m
O
Correct answer: 232.75 kg · m2 .
Explanation:
top left
bottom left
bottom right
top right and
Each mass is a distance r = 3.5 m from the
x-axis and
X
mi = 4 kg+6 kg+4 kg+5 kg = 19 kg , so
x
7 rad/s
4 kg
Find the moment of inertia of the system
about the x axis.
1. 120.0
2. 294.0
3. 143.75
4. 117.0
5. 207.0
6. 171.5
7. 384.0
8. 232.75
9. 272.0
10. 193.75
Let : m1 = 4 kg ,
m2 = 6 kg ,
m3 = 4 kg ,
m4 = 5 kg ,
s = 7 m.
X
= (3.5 m)2 (19 kg)
7m
7m
7 rad/s
Ix =
6
6 kg
4 kg
7m
Find the moment of inertia of the system
about the z axis.
1. 208.0
2. 465.5
3. 350.0
4. 224.0
5. 200.0
6. 378.0
7. 234.0
8. 784.0
9. 144.0
10. 392.0
Correct answer: 465.5 kg · m2 .
Explanation:
Each mass is a distance
p
√
d = r2 + r2 = 2 r2
from the axis of rotation, so
Iz =
X
i
mi ri2 = d2
X
i
mi = 2 r 2
X
i
mi
Version 020 – midterm3 – shih – (58505)
7
= 2 (3.5 m)2 (19 kg)
8. III, IV and V only
= 465.5 kg · m2 .
9. I only
012
10. All of these
10.0 points
A uniform rod has a mass 2 m and a length
ℓ, and it can spin freely in a horizontal plane
about a pivot point O at the center of the rod.
A piece of clay with mass m and velocity v
hits one end of the rod, and causes the rodclay system to spin. Viewed from above the
scheme is as follows:
v
m
ℓ
0
ω
ω
2m
(a) before
(b) during
(c) after
After the collisions the rod and clay system
has an angular velocity ω about the pivot.
Which quantity/quantities
I) total mechanical energy
II) total linear momentum
III) total angular momentum with respect to
pivot point O
IV) total gravitational potential energy
V) total kinetic energy
is/are conserved in this process?
1. II and III only
2. II, III, IV and V only
3. None of these
4. III and IV only correct
5. I, II and III only
6. II and V only
7. II, III and IV only
Explanation:
V) Total KE is not conserved. One can
Kf
Ii
verify that
=
< 1. The inequality
Ki
If
here is due to the fact that Ii is the moment
of inertia of the clay and If is the moment of
inertia of the clay+rod.
IV) Since the motion is in a horizontal plane
there is no change in the potential energy.
IV) and V) together imply that the total
mechanical energy (KE + PE) is not conserved.
II) Linear momentum is not conserved by
inspection. Apparently the linear momentum is altered by the presence of the pivot.
The pivot provides an external force to the
rod+clay system.
III) Angular momentum is conserved, since
there is no external torque applied to the
rod+clay system.
013
10.0 points
A spring is compressed between two cars
on a frictionless airtrack. Car A has four
times the mass of car B, MA = 4 MB , while
the spring’s mass is negligible. Both cars are
initially at rest. When the spring is released,
it pushes them away from each other.
Which of the following statements correctly
describes the velocities, the momenta, and the
kinetic energies of the two cars after the spring
is released? Note: Velocities and momenta are
given below as vectors.
1
1. ~vA = + ~vB
5
4
~pA = + ~pB
5
4
KA =
KB
25
2. ~vA = −2 ~vB
~pA = −8~pB
Version 020 – midterm3 – shih – (58505)
KA = 16 KB
The velocities and the kinetic energies follow from the momenta:
3. ~vA = −4 ~vB
~pA = −16~pB
KA = 64 KB
4. ~vA = −~vB
~pA = −~pB
KA = KB
1
5. ~vA = − ~vB
3
2
~pA = − ~pB
3
4
KA = KB
3
1
6. ~vA = − ~vB
4
~pA = −~pB
1
KA = KB correct
4
1
7. ~vA = + ~vB
4
~pA = +~pB
KA = 4 KB
1
8. ~vA = − ~vB
4
~pA = −~pB
KA = 4 KB
1
9. ~vA = − ~vB
2
~pA = −2~pB
KA = KB
10. ~vA = −~vB
~pA = −4~pB
KA = 16 KB
Explanation:
Let : MA = 4 MB .
There are no external forces acting on the
cars, so their net momentum ~pA + ~pB is conserved. The initial net momentum is obviously zero, hence after the spring is released,
~pA + ~pB = 0; i.e., ~pA = −~pB .
8
~v =
~p
.
M
So given MA = 4 MB , it follows that
1
~vA = − ~vB .
4
Likewise,
~p2
1
2
.
K = M ~v =
2
2M
Since ~pA = −~pB ,
KA =
1
KB .
4
014 10.0 points
A rock has a mass of 1 kg and hangs from the
0 cm end of a meter stick.
What is the mass of the measuring stick
if it is balanced by a support force at the
one-quarter mark?
1. 4 kg
2. 3 kg
3. 0.25 kg
4. 0.5 kg
5. 1 kg correct
6. 2 kg
7. 0.75 kg
Explanation:
The mass of the meter stick is concentrated
at its center, so equal torque arms means
equal masses.
015 10.0 points
Two identical balls (labelled A and B) move
on a frictionless horizontal tabletop. Initially,
ball A moves at speed vA,0 = 10 m/s while
ball B is at rest (vB,0 = 0).
Version 020 – midterm3 – shih – (58505)
The two balls collide off-center, and after
the collision ball A moves at speed vA = 6 m/s
in the direction θA = 53 ◦ from its original
velocity vector:
9
after
b
8.
53◦
8 m/s
6 m/s
A
53
after
b
◦
10 m/s
0 m/s
b
A before B
9.
B
b
after
Which of the following diagrams best represents the motion of ball B after the collision?
1.
4 m/s
after
b
B 37◦
6 m/s
after
3.
10.
B
b
6 m/s
Explanation:
Because we are given both the speed and
the direction of ball A after the collision, the
problem can be solved in terms of momentum
conservation only.
Indeed, there are no (horizontal) external
forces acting on the two balls, so their net
(horizontal) momentum vector is conserved
during the collision:
after
b
2.
37◦
4 m/s
after
B
B
~ net = m ~vA,0 + m ~vB,0
P
= m ~vA + m ~vB
B
b
8 m/s
[before]
[after]
and since ~vB,0 = ~0 (ball B is initially at rest),
after
~vA + ~vB = ~vA,0 .
b
4.
37◦
correct
Pictorially, this means
B
8 m/s
~vA
after
θA
~vA,0
b
5.
4 m/s
B 53◦
~vB
θB
after
b
and in components,
6.
53
◦
6 m/s
7.
B
b
0 m/s
B
after
vA × cos θA + vB × cos θB = vA,0 ,
vA × sin θA − vB × sin θB = 0.
Solving for the two components of ball B’s
velocity, we find
vB,x ≡ vB × cos θB
Version 020 – midterm3 – shih – (58505)
=
=
=
≡
=
=
=
vB,y
vA,0 − vA cos θA
(1)
◦
(10 m/s) − (6 m/s) cos 53
6.4 m/s,
vB sin θB
vA sin θA
(2)
◦
(6 m/s) sin 53
4.8 m/s,
Comparing this formula to eq. (3), we immediately see that
2
2 vA
− 2 vA,0 vA cos θA = 0
and therefore
vA = vA,0 × cos θA .
and hence
= 8 m/s,
vB,y
vB,x
4.8 m/s
= arctan
6.4 m/s
◦
= 37 .
θB = arctan
vB = vA,0 × sin θA
vB,x
vB
1 − cos2 θA
=
sin θA
= sin θA
cos θB ≡
+
2
2
2
2
m vA
m vB
=
+
2
2
and therefore
[before]
[after]
2
2
2
vA
+ vB
= vA,0
.
(3)
At the same time, eqs. (1) and (2) above imply
2
2
2
vB
= vB,x
+ vB,y
2
2
= vA,0 − vA cos θA
+ vA sin θA
2
2
= vA,0
− 2 vA,0 × vA × cos θA + vA
(the cosine theorem), and hence
2
vA
+
2
vB
2
− vA,0
=
2
= 2 vA − 2 vA,0
=⇒
Knet =
(6)
and hence
Solving an Elastic Collision:
If we know that the collision in question
is perfectly elastic — and it is — then we
do not need to be given both the speed vA
and the direction θA of the ball A after the
collision: Any one of these two parameters
would determine the other. The reason for
this is conservation of kinetic energy in an
elastic collision:
2
m vA,0
(5)
Note that the problem data indeed satisfy this
equation, which confirms that the collision in
question is perfectly elastic.
Given eq. (5), the energy conservation condition (3) implies
q
2
2
+ vB,y
vB,x
q
=
(6.4 m/s)2 + (4.8 m/s)2
vB =
2
m vA,0
10
θA + θB = 90◦ .
016 10.0 points
A solid sphere rolls along a horizontal, smooth
surface at a constant linear speed without
slipping.
What is the ratio between the rotational
kinetic energy about the center of the sphere
and the sphere’s total kinetic energy?
1.
2.
3.
4.
(4)
× vA × cos θA .
(7)
5.
2
5
3
7
2
correct
7
7
2
3
5
Version 020 – midterm3 – shih – (58505)
6.
5
3
7. None of these
Explanation:
Krot
1
=
2
2
v 2 1
2
mr
= m v2
5
r
5
and
Ktot = Ktrans + Krot
1
1
7
= m v2 + m v2 =
m v2 ,
2
5
10
1
m v2
10
2
1
Krot
5
=
=
=
7
Ktot
5
7
7
m v2
10
keywords:
so
11