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Exam 3 Practice Solutions
Multiple Choice
1. A thin hoop, a solid disk, and a solid sphere, each with the same mass and radius, are at rest
at the top of an inclined plane. If all three are released at the same instant, in which order
will they reach the bottom?
(a) disk, sphere, hoop
(b) sphere, disk, hoop
(c) hoop, disk, sphere
(d) hoop, sphere, disk
(e) sphere, hoop, disk
At the top of the ramp, each object has the same gravitational potential energy. The gravitational
potential energy is converted to translational and rotational kinetic energy. More of the energy will
show up as rotational kinetic eneryg for objects with higher moments of inertia. Therefore, the objects
with lower moments of inertia will reach the bottom first: sphere ( 52 M R2 ), disk ( 21 M R2 ), hoop (M R2 ).
2. A student is sitting on a heavy stool with their feet off of the ground. The stool’s seat is free
to rotate. The student holds a spinning wheel with a heavy rim, as shown below. As viewed
from the front, the student tilts the wheel to the right (the student tilts the wheel to their
left). What happens?
(a) the stool rotates clockwise as viewed from above
~ as shown
When the student tips the wheel to the right, their is an angular momentum change of ∆L
above. Since angular momentum is conserved, there must be a corresponding angular momentum
~ for the stool. This corresponds to a clockwise rotation when viewed from above.
change of −∆L
1
3. The vectors shown below represent forces of equal magnitude applied to boxes of equal mass
(shown in gray). All forces are either vertical, horizontal, or at a 45◦ angle. Which of the
boxes are in translational and rotational equilibrium?
(a) I only
(b) II only
(c) III only
(d) I and II
(e) I and III
(f) II and III
(g) I, II, and III
Box II will rotate clockwise. Boxes I and III are in translation equilibrium: for every force vector, there
is a force vector pointing in the opposite direction. Boxes I and III are also in rotational equilibrium:
for every torque, there is an equal and opposite torque.
4. A child stands of mass m at the edge of a rotating disk of mass M (like a merry-go-round).
The disk rotates with angular velocity ω0 . Suppose that she starts walking opposite the
direction of rotation so that she is at rest with respect to the ground. What is the final
angular velocity of the disk?
(a) ω0 1 − 2m
M
(b) ω0 M
m
2m (c) ω0 1 +
M
(d) ω0 M − 2m
M
m
(e) ω0 M
Angular momentum must be conserved. The angular momentum before and after she starts walking
are equal. Since she is walking so that she is motionless with respect to the ground, her angular
velocity must be zero. She no longer carries any angular momentum, so the angular velocity of the
2
disk must increase so that it carries more angular momentum.
L = L′
Idisk ω0 + Ichild ω0 = Idisk ω ′
1
1
M R2 ω0 + mR2 ω0 = M R2 ω ′
2
2
2 1
M + m ω0 = ω ′
M 2
2m 1+
ω0 = ω ′
M
5. The moment of inertia of a disk of mass m and radius R about an axis through the edge of
the disk and perpendicular to the disk’s surface is
(a)
(b)
(c)
1
2
2 mR
mR2
3
mR2
2
(d) 2mR2
(e) none of the above
Use the parallel axis theorem to find the new moment of inertia.
I = Icm + mh2
1
I = mR2 + mR2
2
3
I = mR2
2
6. A particle of mass m moves in the +î direction with velocity v as shown below. The ĵ
direction is up and the k̂ direction is out of the page.
The angular momentum of the particle about the point P is
(a) mrv k̂
(b) −mrv k̂
(c) mrv sin θ k̂
3
(d) −mrv sin θ k̂
(e) none of the above
~ = ~r × ~p. The vector ~r points from P to the particle, so the direction of
The angular momentum is L
~r × ~p is into the page, or −k̂. The magnitude of the cross product is rp sin θ = mrv sin θ.
7. A thin rod of mass M and length L spins clockwise with angular velocity ω about an axis
that passes through one end of the rod, perpendicular to its length. An object of mass m
moving to the left with velocity v strikes the rod at its midpoint and sticks to it.
The rod stops spinning. What was v?
(a)
(b)
(c)
(d)
M Lω
3m
M Lω
6m
M Lω
12m
2M Lω
3m
(e) there is not enough information to solve this problem
Angular momentum is conserved in the collision, so the angular momenta of the object and the rod
must be equal and opposite. The angular momentum of the object is mvL
2 , out of the page. The
angular momentum of the rod is Iω 2 = 31 M L2 ω.
mvL
1
= M L2 ω
2
3
2M Lω
v=
3m
4
8. A nonuniform rod of length L is suspended horizontally by two cords as shown below. The
left cord makes an angle of 30◦ with the vertical and the right cord makes an angle of 60◦
with the vertical. The rod is in equilibrium. Measuring from the left, where is the center of
mass of the rod?
(a)
(b)
(c)
(d)
L
4
2L
3
3L
4
7L
8
(e) there is not enough information to solve this problem
Since the rod is in equilibrium and shifted to the right, we expect that the center of mass will be closer
to the right. The tension in the right string is TR , in the left string is TL , and the location of the center
of mass is x. The rod is in translational equilibrium, so:
X
Fx = 0
−TL sin 60◦ + TR sin 30◦ = 0
−(.866)TL + (.5)TR = 0
TL = (.577)TR
and
◦
X
Fy = 0
−M g + TL cos 60 + TR cos 30◦ = 0
−M g + (.5)TL + (.866)TR = 0
−M g + (.5)(.577)TR + (.866)TR = 0
−M g + (1.155)TR = 0
TR = (.866)M g
The rod is also in rotational equilibrium. Setting the pivot point at the left end of the rod:
X
τ =0
M gx − (.866)TR L = 0
M gx − (.866)(.866)M gL = 0
x = .75L
5
9. A mass is attached to a cord which is wound about a pulley that is fixed in placed and free
to rotate. If the mass is allowed to drop, which set of graphs best describes the angular
displacement, angular velocity, and angular acceleration of the pulley? The horizontal axis
on each graph is time, while the vertical axes are θ, ω, or α, as labeled.
Angular acceleration is the derivative (with respect to time) of angular velocity, which is the derivative
of angular displacement. Furthermore, for a falling mass, we expect a constant acceleration, a linearly increasing velocity, and a parabolic displacement. The only choice that correctly displays these
characteristics is (b).
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Problems
1. A cart of mass m (cart A) slides with velocity v to the right on a frictionless surface. Another
cart (cart B), also of mass m is at rest a distance x1 ahead of it. A distance x2 beyond the
second cart, a spring of spring constant k is attached to a wall. Assume that all collisions are
perfectly elastic. What are the final velocities of carts A and B? Explain briefly.
Cart A is moving to the right with velocity v, so it will collide with cart B. Since both carts have the
same mass and the collision is elastic, they will simply exchange velocities. Cart A stops while cart
B moves to the right with velocity v. Cart B then collides with and compresses the spring. Since the
spring force is conservative, no energy is lost in the compression and subsequent expansion of the
spring. Cart B is sent back to the left with the same velocity v. It then collides again with cart A.
Once again, since the carts have the same mass and the collision is elastic, they simply exchange
velocities. The final velocity of cart A is v to the left and cart B is at rest.
7
2. A block of mass m=1 kg is released from rest at the top of a frictionless inclined plane. The
plane is h=5 m high at an angle of 45◦ . When the block reaches the end of the plane, it
transitions smoothly to a rough surface where friction cannot be neglected. The block slides
a distance d before coming to a stop.
(a) Sketch a graph that shows the block’s potential energy, kinetic energy, and total mechanical energy as a function of position x (see figure).
It is convenient to set the zero of gravitational potential energy to be the bottom of the ramp
(that definition will be used for the remainder of this problem). The total mechanical energy,
the potential energy, and the kinetic energy of the block are graphed below. The blocks initially
has only gravitation potential energy. As it slides down the ramp, it gains kinetic energy but
loses gravitational potential energy. When it reaches the bottom of the ramp, all of the gravitational potential energy has been converted into kinetic energy. While the block is on the ramp,
the total mechanical energy remains constant because only conservative forces (gravity) are
acting. When the block begins to slide horizontally, friction starts to do work. Friction is a nonconservative force, so the mechanical energy drops until the work done by friction has removed
all of the mechanical energy from the system and converted it to heat. Thus, the mechanical
energy drops to zero over some distance (and the kinetic energy, of course, which is equal to
the mechanical energy).
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(b) If the frictional force is a constant 5 N, what is d?
When the block stops, it has zero mechanical energy. The initial mechanical energy was entirely
gravitational potential energy. The mechanical energy was lost to work done by friction, or FF d,
where FF is the force of friction and d is the distance the block slides on the horizontal surface.
We can use conservation of total energy to determine how far the block slides:
mgh = FF d
mgh
d=
FF
(1)(9.8)(5)
d=
5
d = 9.8 m
(c) Take a look at your graph. Is mechanical energy conserved? Why or why not? What
about total energy?
No, mechanical energy is not conserved. Mechanical energy is lost to work done by friction.
Total energy is conserved, since the work done by friction is the same as the mechanical energy
lost. Total energy is always conserved; unfortunately it is most often converted into unusable
forms such as heat (via friction in this case).
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3. A block of mass m=.306 kg is dropped from height h=1.33 m. Directly underneath the
block is a spring in its equilibrium position with spring constant k=2 N/m. The spring has
a massless platform that is attached to its top (the top of the spring with platform is even
with the ground level). The block lands on and compresses the spring by a distance x.
This is not a homework problem - feel free to round your answers as you go.
Hint: The following relation may be useful: if ax2 + bx + c = 0, then x =
√
−b± b2 −4ac
.
2a
(a) What is the speed of the block just before it hits the spring?
Set the zero of gravitational potential energy as the ground level from which h is measured (and
keep this definition throughout the problem).
Mechanical energy is conserved in this system since we have only conservative forces (gravity
and spring forces). The block starts out at rest with only gravitational potential energy mgh.
When the block is about to hit the spring, the gravitational potential energy is entirely converted
to kinetic energy:
1
mv 2
2
v 2 = 2gh
p
p
v = 2gh = 2(9.8)(1.33)
mgh =
v = 5.1 m/s
(b) When the block returns to height h after bouncing off of the spring, what is its speed?
Explain.
Since mechanical energy is conserved, the block will always have the same mechanical energy.
The block starts with total mechanical energy mgh and speed v=0. When the block returns to
height h, it has gravitational potential energy mgh which is the same as the total mechanical
energy. Thus, the speed of the block must be zero again.
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(c) Write an expression for the conservation of energy when the spring is at maximum
compression (in terms of the givens m, g, h, x, or k).
At maximum compression, the spring is compressed a distance x by the block. Thus, the spring
will have potential energy 21 kx2 . The block stops moving at this point and has no kinetic energy.
However, the block is now located a distance x below the ground level where we set the zero of
gravitation potential energy. Thus, the block has gravitational potential energy −mgx. The block
started out with gravitational potential energy mgh. Conservation of energy says that
Einitial = Efinal
1
mgh = kx2 − mgx
2
(d) Find x, the maximum distance the spring is compressed.
To find x, we need to solve the equation we determined in (c). This is a quadratic equation:
1 2
kx − mgx − mgh
2
1
0 = (2)x2 − (.306)(9.8)x − (.306)(9.8)(1.33)
2
0 = x2 − 3x − 4
0=
Using the quadratic formula with a = 1, b = −3, and c = −4, we obtain
√
3 ± 9 + 16
x=
2
3±5
x=
2
x = 4 or − 1
Since x is positive based on my definition of the gravitational potential energy, the solution is
x=4 m.
11
4. Two identical spools of thread (solid cylinders with several windings of thread) are held the
same height above the floor. The thread from spool A is tied to a support, while spool B is
not connected to a support. The thread has negligible mass. Both spools are released from
rest at the same instant.
(a) Draw a free-body diagram for each spool corresponding to an instant after they are
released, but before they hit the ground. Be sure to draw each force exactly where it
acts. Compare the magnitude of all forces on your diagrams.
The tension T~ < F~g .
(b) Which spool will reach the floor first? Explain how your answer is consistent with your
free-body diagrams.
Spool B will reach the ground first, as it has only one force acting on it, F~g , in the downward
direction. Spool A has two forces acting on it, one of which is in the upward direction, the tension
T~ . Thus the downward acceleration of spool B will be larger.
12
(c) Will spool A strike the floor directly below the point where it was released, to the right
of this point, or to the left of this point? Explain.
Spool A will strike the floor directly below where it was released. There are no horizontal forces
on spool A (it is in equilibrium in the horizontal direction) so there is no acceleration in the
horizontal direction.
A
)?
(d) What is the ratio of the accelerations of the two spools ( aaB
The acceleration of spool B is simply aB = g, the acceleration due to gravity. For spool A, we
need to find the acceleration by taking the tension into account. From Newton’s second law for
translational motion, we have T~ + F~g = ma~A where m is the mass of the spool. From Newton’s
second law for rotational motion, we have ~τ = I~
α. Working on the rotational motion first, we can
relate the linear acceleration aA to the angular acceleration α by α = aRA where R is the radius
of the spool. This is because the string constrains the rotation of the spool.
τ = Iα
aA
1
T R = mR2
2
R
1
T = maA
2
Now we can substitute our expression for T into the equation for the translational motion.
T − mg = −maA
1
maA − mg = −maA
2
3
aA = g
2
2
aA = g
3
Where we were careful to explicitly note the directions of tension, acceleration, and acceleration
due to gravity in the first step. The ratio of the accelerations is thus
2
aA
2g
=
=
aB
3g
3
A
(e) What is the ratio of the velocities of the two spools just before each hits the floor ( vvB
)?
The easiest way to solve this problem is using conservation of energy. Using the kinematic
equations is challenging because each spool is in the air for a different amount of time. Both
spools start with the same gravitational potential energy, mgh, where h is the height of the spool
above the floor. This energy is shared between rotational and translational motion as the spool
falls. Since spool B will not rotate, it will have a higher velocity.
1
mv 2 = mgh
2 B p
vB = 2gh
13
Spool A will rotate:
1
1 2
2
mvA
+ IωA
2
2
11
1
2
2 vA 2
mv +
mR ( )
2 A 22
R
1
1
2
2 vA 2
mv + mR ( )
2 A 4
R
3 2
v
4 A
= mgh
= mgh
= mgh
= mgh
r
4gh
vA =
3
The ratio of the velocities is
vA
=
vB
s
14
4gh
=
(3)2gh
r
2
3