AP Physics C Multiple Choice Practice – Rotation

SECTION A – Torque and Statics

1. Torque is the rotational analogue of
(A) kinetic energy (B) linear momentum (C) acceleration (D) force (E) mass

2. A square piece of plywood on a horizontal tabletop is subjected to the two horizontal forces shown above.
Where should a third force of magnitude 5 newtons be applied to put the piece of plywood into equilibrium?

3. A uniform rigid bar of weight W is supported in a horizontal orientation as shown above by a rope that makes a
30° angle with the horizontal. The force exerted on the bar at point O, where it is pivoted, is best represented by
a vector whose direction is which of the following?

(A) (B) (C) (D) (E)

4. A rod of negligible mass is pivoted at a point that is off-center, so that length l 1 is different from length l 2 . The
figures above show two cases in which masses are suspended from the ends of the rod. In each case the
unknown mass m is balanced by a known mass, M 1 or M 2 , so that the rod remains horizontal. What is the value
of m in terms of the known masses?
(A) M l + M 2 (B) ½(M l + M 2 ) (C) M l M 2 (D) ½M 1 M 2 (E) M1M 2

131
5. 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 above. The magnitude of the net torque on the system about the axis is
(A) zero (B) FR (C) 2FR (D) 5FR (E) 14FR

6. For the wheel-and-axle system shown above, which of the following expresses the condition required for the
system to be in static equilibrium?
(A) m 1 = m 2 (B) am 1 = bm 2 (C) am 2 = bm 1 (D) a2m l = b2m2 (E) b2m 1 = a2m 2

132
SECTION B – Rotational Kinematics and Dynamics
Questions 1-2
A cylinder rotates with constant angular acceleration about a fixed axis. The cylinder’s moment of inertia about
the axis is 4 kg m2. At time t = 0 the cylinder is at rest. At time t = 2 seconds its angular velocity is 1 radian per
second.

1. What is the angular acceleration of the cylinder between t = 0 and t = 2 seconds?

(A) 0.5 radian/s² (B) 1 radian/s² (C) 2 radian/s² (D) 4 radian/s² (E) 5 radian/s²

2. What is the kinetic energy of the cylinder at time t = 2 seconds?

(A) 1 J (B) 2 J (C) 3 J (D) 4 J (E) cannot be determined without knowing the radius of the cylinder

3. A particle is moving in a circle of radius 2 meters according to the relation = 3t2 + 2t, where is measured in
radians and t in seconds. The speed of the particle at t = 4 seconds is
(A) 13 m/s (B) 16 m/s (C) 26 m/s (D) 52 m/s (E) 338 m/s

4. A uniform stick has length L. The moment of inertia about the center of the stick is I o . A particle of mass M is
attached to one end of the stick. The moment of inertia of the combined system about the center of the stick is
(A) I 0 + 1 ML2 (B) I 0 + 1 ML2 (C) I 0 + 3 ML2 (D) I 0 + ML2 (E) I 0 + 5 ML2
4 2 4 4

5. A light rigid rod with masses attached to its ends is pivoted about a horizontal axis as shown above. When
released from rest in a horizontal orientation, the rod begins to rotate with an angular acceleration of magnitude
g g g 5g g
(A) 7 l (B) 5l (C) 4 l (D) 7 l (E) l

6. In which of the following diagrams is the torque about point O equal in magnitude to the torque about point X
in the diagram above? (All forces lie in the plane of the paper.)

133
Questions 7-8

An ant of mass m clings to the rim of a flywheel of radius r, as shown above. The flywheel rotates clockwise on
a horizontal shaft S with constant angular velocity . As the wheel rotates, the ant revolves past the stationary
points I, II, III, and IV. The ant can adhere to the wheel with a force much greater than its own weight.

7. It will be most difficult for the ant to adhere to the wheel as it revolves past which of the four points?
(A) I (B) II (C) III (D) IV
(E) It will be equally difficult for the ant to adhere to the wheel at all points.

8. What is the magnitude of the minimum adhesion force necessary for the ant to stay on the flywheel at point III?
(A) mg (B) m 2r (C) m 2r2 + mg (D) m 2r - mg (E) m 2r + mg

9. A turntable that is initially at rest is set in motion with a constant angular acceleration . What is the angular
velocity of the turntable after it has made one complete revolution?
(A) 2 (B) 2 (C) 4 (D) 2 (E) 4

10. A 5-kilogram sphere is connected to a 10-kilogram sphere by a rigid rod of negligible mass, as shown above.
The sphere-rod combination can be pivoted about an axis that is perpendicular to the plane of the page and that
passes through one of the five lettered points. Through which point should the axis pass for the moment of
inertia of the sphere-rod combination about this axis to be greatest?
(A) A (B) B (C) C (D) D (E) E

Questions 11-12
A wheel with rotational inertia I is mounted on a fixed, frictionless axle. The angular speed of the wheel is
increased from zero to f in a time interval T.

11. What is the average net torque on the wheel during this time interval?
2
f f I f I f I f
(A) (B) 2
(C) (D) 2
(E)
T T T T T
12. What is the average power input to the wheel during this time interval?
2 2
I f I f I f I2 f I2 2
f
(A) (B) (C) (D) (E)
2T 2T 2T 2 2T 2 2T 2

134
13. Two blocks are joined by a light string that passes over the pulley shown above, which has radius R and
moment of inertia I about its center. T 1 and T 2 are the tensions in the string on either side of the pulley and is
the angular acceleration of the pulley. Which of the following equations best describes the pulley's rotational
motion during the time the blocks accelerate?
(A) m 2 (B) (T 1 + T 2 (C) T 2 (D) (T 2 - T 1 (E) (m 2 – m 1

14. a disk is free to rotate about an axis perpendicular to the disk through its center. If the disk starts from rest and
accelerates uniformly at the rate of 3 radians/s2 for 4 s, its angular displacement during this time is
(A) 6 radians (B) 12 radians (C) 18 radians (D) 24 radians (E) 48 radians

Questions 15-16

A solid cylinder of mass m and radius R has a string wound around it. A person holding the string pulls it
vertically upward, as shown above, such that the cylinder is suspended in midair for a brief time interval t and
its center of mass does not move. The tension in the string is T, and the rotational inertia of the cylinder about
its axis is

15. the net force on the cylinder during the time interval t is
(A) (Β) mg (C) T – mgR (D) mgR - ( ) zero

16. The linear acceleration of the person's hand during the time interval t is
(A) (B) 2g (C) (D) (E) zero

17. A block of mass m is placed against the inner wall of a hollow cylinder of radius R that rotates about a vertical
axis with a constant angular velocity , as shown above. In order for friction to prevent the mass from sliding
down the wall, the coefficient of static friction between the mass and the wall must satisfy which of the
following inequalities?

(A) (B) (C) (D) (E)

135
SECTION C – Rolling
1. A bowling ball of mass M and radius R. whose moment of inertia about its center is (2/5)MR2, rolls without
slipping along a level surface at speed v. The maximum vertical height to which it can roll if it ascends an
incline is
2 2 2 2 2
(A) v (B) 2v (C) v (D) 7v (E) v
5g 5g 2g 10 g g

2. A wheel of mass M and radius R rolls on a level surface without slipping. If the angular velocity of the wheel is
, what is its linear momentum ?
(A) M R (B) M 2R (C) M R2 (D) M 2R2/2 (E) Zero

Questions 3-4

A sphere of mass M, radius r, and rotational inertia I is released from rest at the top of an inclined plane of
height h as shown above.

3. If the plane is frictionless, what is the speed v cm , of the center of mass of the sphere at the bottom of the incline?
2Mgh 2Mghr 2 2Mghr 2 2Mghr 2
(A) 2gh (B) (C) (D) (E)
I I I I + Mr 2

4. If the plane has friction so that the sphere rolls without slipping, what is the speed v cm of the center of mass at
the bottom of the incline?
2Mgh 2Mghr 2 2Mghr 2 2Mghr 2
(A) 2gh (B) (C) (D) (E)
I I I I + Mr 2

5. A wheel of 0.5 m radius rolls without slipping on a horizontal surface. The axle of the wheel advances at
constant velocity, moving a distance of 20 m in 5 s. The angular speed of the wheel about its point of contact on
the surface is
(A) 2 radians • s-1 (B) 4 radians • s-1 (C) 8 radians • s-1 (D) 16 radians • s-1 (E) 32 radians • s-1

6. A car travels forward with constant velocity. It goes over a small stone, which gets stuck in the groove of a tire.
The initial acceleration of the stone, as it leaves the surface of the road, is
(A) vertically upward (B) horizontally forward (C) horizontally backward
(D) zero (E) upward and forward, at approximately 45° to the horizontal

136
SECTION D – Angular Momentum
1. An ice skater is spinning about a vertical axis with arms fully extended. If the arms are pulled in closer to the
body, in which of the following ways are the angular momentum and kinetic energy of the skater affected?
Angular Momentum Kinetic Energy
(A) Increases Increases
(B) Increases Remains Constant
(C) Remains Constant Increases
(D) Remains Constant Remains Constant
(E) Decreases Remains Constant

2. A cylinder rotates with constant angular acceleration about a fixed axis. The cylinder’s moment of inertia about
the axis is 4 kg m2. At time t = 0 the cylinder is at rest. At time t = 2 seconds its angular velocity is 1 radian per
second. What is the angular momentum of the cylinder at time t = 2 seconds?
(A) 1 kg m²/s (B) 2 kg m²/s (C) 3 kg m²/s (D) 4 kg m²/s
(E) It cannot be determined without knowing the radius of the cylinder.

3. A figure skater is spinning on frictionless ice with her arms fully extended horizontally. She then drops her arms
to her sides. Which of the following correctly describes her rotational kinetic energy and angular momentum as
her arms fall?
Rotational Kinetic Angular
Energy Momentum
(A) Remains constant Remains constant
(B) Decreases Increases
(C) Decreases Decreases
(D) Increases Decreases
(E) Increases Remains constant

4. A particle of mass m moves with a constant speed v along the dashed line y = a. When the x-coordinate of the
particle is x o , the magnitude of the angular momentum of the particle with respect to the origin of the system is
(A) zero (B) mva (C) mvx o (D) mv x0 2 + a 2 (E) mva
x0 2 + a 2

5. The rigid body shown in the diagram above consists of a vertical support post and two horizontal crossbars with
spheres attached. The masses of the spheres and the lengths of the crossbars are indicated in the diagram. The
body rotates about a vertical axis along the support post with constant angular speed . If the masses of the
support post and the crossbars are negligible, what is the ratio of the angular momentum of the two upper
spheres to that of the two lower spheres?
(A) 2/1 (B) 1/1 (C) 1/2 (D) 1/4 (E) 1/8

137
6. A long board is free to slide on a sheet of frictionless ice. As shown in the top view above, a skater skates to the
board and hops onto one end, causing the board to slide and rotate. In this situation, which of the following
occurs?
(A) Linear momentum is converted to angular momentum.
(B) Kinetic energy is converted to angular momentum.
(C) Rotational kinetic energy is conserved.
(D) Translational kinetic energy is conserved.
(E) Linear momentum and angular momentum are both conserved.

Top View
7. A disk sliding on a horizontal surface that has negligible friction collides with a rod that is free to move and
rotate on the surface, as shown in the top view above. Which of the following quantities must be the same for
the disk-rod system before and after the collision?
I. Linear momentum
II. Angular momentum
III. Kinetic energy
(A) I only
(B) II only
(C) I and II only
(D)
(E) I, II, and III

8. A figure skater goes into a spin with arms fully extended. Which of the following describes the changes in the
rotational kinetic energy and angular momentum of the skater as the skater's arms are brought toward the body?
Rotational
Kinetic Energy Angular Momentum
(A) Remains the same Increases
(B) Remains the same Remains the same
(C) Increases Remains the same
(D) Decreases Increases
(E) Decreases Remains the same

138
AP Physics C Free Response Practice – Rotation

SECTION A – Torque and Statics

2008M2. The horizontal uniform rod shown above has length 0.60 m and mass 2.0 kg. The left end of the rod is
attached to a vertical support by a frictionless hinge that allows the rod to swing up or down. The right end of
the rod is supported by a cord that makes an angle of 30° with the rod. A spring scale of negligible mass
measures the tension in the cord. A 0.50 kg block is also attached to the right end of the rod.
a. On the diagram below, draw and label vectors to represent all the forces acting on the rod. Show each force
vector originating at its point of application.

b. Calculate the reading on the spring scale.

1
The rotational inertia of a rod about its center is 12
ML2 , where M is the mass of the rod and L is its length.
c. Calculate the rotational inertia of the rod-block system about the hinge.
d. If the cord that supports the rod is cut near the end of the rod, calculate the initial angular acceleration of the
rod-block system about the hinge.

SECTION B – Rotational Kinematics and Dynamics

1973M3. A ball of mass m is attached by two strings to a vertical rod. as shown above. The entire system rotates at
constant angular velocity about the axis of the rod.
a. Assuming is large enough to keep both strings taut, find the force each string exerts on the ball in terms of ,
m, g, R, and .
b. Find the minimum angular velocity, min for which the lower string barely remains taut.

139
1976M2. A cloth tape is wound around the outside of a uniform solid cylinder (mass M, radius R) and fastened to
the ceiling as shown in the diagram above. The cylinder is held with the tape vertical and then released from
rest. As the cylinder descends, it unwinds from the tape without slipping. The moment of inertia of a uniform
solid cylinder about its center is ½MR2.

a. On the circle above draw vectors showing all the forces acting on the cylinder after it is released. Label each
force clearly.
b. In terms of g, find the downward acceleration of the center of the cylinder as it unrolls from the tape.
c. While descending, does the center of the cylinder move toward the left, toward the right, or straight down?
Explain.

1978M1. An amusement park ride consists of a ring of radius A from which hang ropes of length l with seats for
the riders as shown in Figure I. When the ring is rotating at a constant angular velocity each rope forms a
constant angle with the vertical as shown in Figure II. Let the mass of each rider be m and neglect friction,
air resistance, and the mass of the ring, ropes, and seats.
a. In the space below, draw and label all the forces acting on rider X (represented by the point below) under the
constant rotating condition of Figure II. Clearly define any symbols you introduce.

b. Derive an expression for in terms of A, l, and the acceleration of gravity g.

c. Determine the minimum work that the motor that powers the ride would have to perform to bring the system
from rest to the constant rotating condition of Figure II. Express your answer in terms of m, g, l, , and the
speed v of each rider.

140
1983M2. A uniform solid cylinder of mass m 1 and radius R is mounted on frictionless bearings about a fixed axis
through O. The moment of inertia of the cylinder about the axis is I = ½m 1 R2. A block of mass m 2 , suspended
by a cord wrapped around the cylinder as shown above, is released at time t = 0.
a. On the diagram below draw and identify all of the forces acting on the cylinder and on the block.

b. In terms of m l , m 2 , R. and g, determine each of the following.

i. The acceleration of the block
ii. The tension in the cord

141
1985M3. A pulley of mass 3m and radius r is mounted on frictionless bearings and supported by a stand of mass 4m
at rest on a table as shown above. The moment of inertia of this pulley about its axis is 1.5mr2.
Passing over the pulley is a massless cord supporting a block of mass m on the left and a block of mass 2m on
the right. The cord does not slip on the pulley, so after the block-pulley system is released from rest, the pulley
begins to rotate.
a. On the diagrams below, draw and label all the forces acting on each block.

b. Use the symbols identified in part a. to write each of the following.

i. The equations of translational motion (Newton's second law) for each of the two blocks
ii. The analogous equation for the rotational motion of the pulley
c. Solve the equations in part b. for the acceleration of the two blocks.
d. Determine the tension in the segment of the cord attached to the block of mass m.
e. Determine the normal force exerted on the apparatus by the table while the blocks are in motion.

142
1988M3. The two uniform disks shown above have equal mass, and each can rotate on frictionless bearings about a
fixed axis through its center. The smaller disk has a radius R and moment of inertia I about its axis. The larger
disk has a radius 2R
a. Determine the moment of inertia of the larger disk about its axis in terms of I.

The two disks are then linked as shown below by a light chain that cannot slip. They are at rest when, at time t =
0, a student applies a torque to the smaller disk, and it rotates counterclockwise with constant angular
acceleration . Assume that the mass of the chain and the tension in the lower part of the chain, are negligible.
In terms of I, R, , and t, determine each of the following:

b. The angular acceleration of the larger disk

c. The tension in the upper part of the chain
d. The torque that the student applied to the smaller disk
e. The rotational kinetic energy of the smaller disk as a function of time

143
1989M2. Block A of mass 2M hangs from a cord that passes over a pulley and is connected to block B of mass 3M
that is free to move on a frictionless horizontal surface, as shown above. The pulley is a disk with frictionless
bearings, having a radius R and moment of inertia 3MR2. Block C of mass 4M is on top of block B. The
surface between blocks B and C is NOT frictionless. Shortly after the system is released from rest, block A
moves with a downward acceleration a, and the two blocks on the table move relative to each other.

In terms of M, g, and a, determine the

a. tension T v in the vertical section of the cord
b. tension T h in the horizontal section of the cord

If a = 2 meters per second squared, determine the

c. coefficient of kinetic friction between blocks B and C
d. acceleration of block C

144
1991M2. Two masses. m 1 and m 2 are connected by light cables to the perimeters of two cylinders of radii r 1 and r 2 ,
respectively. as shown in the diagram above. The cylinders are rigidly connected to each other but are free to
rotate without friction on a common axle. The moment of inertia of the pair of cylinders is I = 45 kg•m2
Also r 1 = 0.5 meter, r 2 = 1.5 meters, and m 1 = 20 kilograms.
a. Determine m 2 such that the system will remain in equilibrium.

The mass m 2 is removed and the system is released from rest.

b. Determine the angular acceleration of the cylinders.
c. Determine the tension in the cable supporting m 1
d. Determine the linear speed of m 1 at the time it has descended 1.0 meter.

1993M3. A long, uniform rod of mass M and length lis supported at the left end by a horizontal axis into the page
and perpendicular to the rod, as shown above. The right end is connected to the ceiling by a thin vertical thread
so that the rod is horizontal. The moment of inertia of the rod about the axis at the end of the rod is Ml2/3.
Express the answers to all parts of this question in terms of M, l, and g.
a. Determine the magnitude and direction of the force exerted on the rod by the axis.

The thread is then burned by a match. For the time immediately after the thread breaks, determine each of the
following:
b. The angular acceleration of the rod about the axis
c. The translational acceleration of the center of mass of the rod
d. The force exerted on the end of the rod by the axis

The rod rotates about the axis and swings down from the horizontal position.
e. Determine the angular velocity of the rod as a function of , the arbitrary angle through which the rod has
swung.

145
1999M3 As shown above, a uniform disk is mounted to an axle and is free to rotate without friction. A thin
uniform rod is rigidly attached to the disk so that it will rotate with the disk. A block is attached to the end of
the rod. Properties of the disk, rod, and block are as follows.
Disk: mass = 3m, radius = R, moment of inertia about center I D = 1.5mR2
Rod: mass = m, length = 2R, moment of inertia about one end I R = 4/3(mR2)
Block: mass = 2m

The system is held in equilibrium with the rod at an angle 0 to the vertical, as shown above, by a horizontal
string of negligible mass with one end attached to the disk and the other to a wall. Express your answers to the
following in terms of m, R, 0 , and g.
a. Determine the tension in the string.

The string is now cut, and the disk-rod-block system is free to rotate.

b. Determine the following for the instant immediately after the string is cut.
i. The magnitude of the angular acceleration of the disk
ii. The magnitude of the linear acceleration of the mass at the end of the rod

As the disk rotates, the rod passes the horizontal position shown above.
c. Determine the linear speed of the mass at the end of the rod for the instant the rod is in the horizontal position.

146
2000M3. A pulley of radius R 1 and rotational inertia I 1 is mounted on an axle with negligible friction. A light cord
passing over the pulley has two blocks of mass m attached to either end, as shown above. Assume that the cord
does not slip on the pulley. Determine the answers to parts a. and b. in terms of m, R 1 , I 1 , and fundamental
constants.
a. Determine the tension T in the cord.
b. One block is now removed from the right and hung on the left. When the system is released from rest, the three
blocks on the left accelerate downward with an acceleration g/3 . Determine the following.
i. The tension T 3 in the section of cord supporting the three blocks on the left
ii. The tension T l in the section of cord supporting the single block on the right
iii. The rotational inertia I 1 of the pulley

c. The blocks are now removed and the cord is tied into a loop, which is passed around the original pulley and a
second pulley of radius 2R 1 and rotational inertia 16I 1 . The axis of the original pulley is attached to a motor
that rotates it at angular speed 1 , which in turn causes the larger pulley to rotate. The loop does not slip on
the pulleys. Determine the following in terms of I 1 , R I , and 1 .
i. The angular speed 2 of the larger pulley
ii. The angular momentum L 2 of the larger pulley
iii. The total kinetic energy of the system

147
2001M3. A light string that is attached to a large block of mass 4m passes over a pulley with negligible rotational
inertia and is wrapped around a vertical pole of radius r, as shown in Experiment A above. The system is
released from rest, and as the block descends the string unwinds and the vertical pole with its attached apparatus
rotates. The apparatus consists of a horizontal rod of length 2L, with a small block of mass m attached at each
end. The rotational inertia of the pole and the rod are negligible.
a. Determine the rotational inertia of the rod-and-block apparatus attached to the top of the pole.
b. Determine the downward acceleration of the large block.
c. When the large block has descended a distance D, how does the instantaneous total kinetic energy of the three
blocks compare with the value 4mgD? Check the appropriate space below and justify your answer.

Greater than 4mgD Equal to 4mgD __ Less than 4mgD _____

The system is now reset. The string is rewound around the pole to bring the large block back to its original
location. The small blocks are detached from the rod and then suspended from each end of the rod, using strings
of length l. The system is again released from rest so that as the large block descends and the apparatus rotates,
the small blocks swing outward, as shown in Experiment B above. This time the downward acceleration of the
block decreases with time after the system is released.
d. When the large block has descended a distance D, how does the instantaneous total kinetic energy of the three
blocks compare to that in part c.? Check the appropriate space below and justify your answer.

Greater before _ Equal to before __ Less than before ___

148
2003M3. Some physics students build a catapult, as shown above. The supporting platform is fixed firmly to the
ground. The projectile, of mass 10 kg, is placed in cup A at one end of the rotating arm. A counterweight
bucket B that is to be loaded with various masses greater than 10 kg is located at the other end of the arm. The
arm is released from the horizontal position, shown in Figure 1, and begins rotating. There is a mechanism (not
shown) that stops the arm in the vertical position, allowing the projectile to be launched with a horizontal
velocity as shown in Figure 2.
a. The students load five different masses in the counterweight bucket, release the catapult, and measure the
resulting distance x traveled by the 10 kg projectile, recording the following data.
Mass (kg) 100 300 500 700 900
x (m) 18 37 45 48 51
i. The data are plotted on the axes below. Sketch a best-fit curve for these data points.

ii. Using your best-fit curve, determine the distance x traveled by the projectile if 250 kg is placed in the
counterweight bucket.
b. The students assume that the mass of the rotating arm, the cup, and the counterweight bucket can be neglected.
With this assumption, they develop a theoretical model for x as a function of the counterweight mass using the
relationship x = v x t, where v, is the horizontal velocity of the projectile as it leaves the cup and t is the time after
launch.
i. How many seconds after leaving the cup will the projectile strike the ground?
ii. Derive the equation that describes the gravitational potential energy of the system relative to the ground
when in the position shown in Figure 1, assuming the mass in the counterweight bucket is M.
iii. Derive the equation for the velocity of the projectile as it leaves the cup, as shown in Figure 2.
c. i. Complete the theoretical model by writing the relationship for x as a function of the counterweight mass
using the results from b. i and b. iii.
ii. Compare the experimental and theoretical values of x for a counterweight bucket mass of 300 kg. Offer a
reason for any difference.

149
2004M2. A solid disk of unknown mass and known radius R is used as a pulley in a lab experiment, as shown
above. A small block of mass m is attached to a string, the other end of which is attached to the pulley and
wrapped around it several times. The block of mass m is released from rest and takes a time t to fall the distance
D to the floor.
a. Calculate the linear acceleration a of the falling block in terms of the given quantities.
b. The time t is measured for various heights D and the data are recorded in the following table.

i. What quantities should be graphed in order to best determine the acceleration of the block? Explain your
reasoning.
ii. On the grid below, plot the quantities determined in b. i., label the axes, and draw the best-fit line to the
data.

iii. Use your graph to calculate the magnitude of the acceleration.

c. Calculate the rotational inertia of the pulley in terms of m, R, a, and fundamental constants.
d. The value of acceleration found in b.iii, along with numerical values for the given quantities and your answer to
c., can be used to determine the rotational inertia of the pulley. The pulley is removed from its support and its
rotational inertia is found to be greater than this value. Give one explanation for this discrepancy.

150
SECTION C – Rolling

1974M2. The moment of inertia of a uniform solid sphere (mass M, radius R) about a diameter is 2MR²/5. The
sphere is placed on an inclined plane (angle ) as shown above and released from rest.
a. Determine the minimum coefficient of friction between the sphere and plane with which the sphere will roll
down the incline without slipping
b. If were zero, would the speed of the sphere at the bottom be greater, smaller, or the same as in part a.?
Explain your answer.

1977M2. A uniform cylinder of mass M, and radius R is initially at rest on a rough horizontal surface. The
moment of inertia of a cylinder about its axis is ½MR2. A string, which is wrapped around the cylinder, is
pulled upwards with a force T whose magnitude is 0.6Mg and whose direction is maintained vertically upward
at all times. In consequence, the cylinder both accelerates horizontally and slips. The coefficient of kinetic
friction is 0.5.
a. On the diagram below, draw vectors that represent each of the forces acting on the cylinder identify and clearly
label each force.

b. Determine the linear acceleration a of the center of the cylinder.

c. Calculate the angular acceleration of the cylinder.
d. Your results should show that a and R are not equal. Explain.

1980M3. A billiard ball has mass M, radius R, and moment of inertia about the center of mass I c = 2 MR²/5
The ball is struck by a cue stick along a horizontal line through the ball's center of mass so that the ball initially
slides with a velocity v o as shown above. As the ball moves across the rough billiard table (coefficient of sliding
friction k ), its motion gradually changes from pure translation through rolling with slipping to rolling without
slipping.
a. Develop an expression for the linear velocity v of the center of the ball as a function of time while it is rolling
with slipping.
b. Develop an expression for the angular velocity of the ball as a function of time while it is rolling with
slipping.
c. Determine the time at which the ball begins to roll without slipping.
d. When the ball is struck it acquires an angular momentum about the fixed point P on the surface of the table.
During the subsequent motion the angular momentum about point P remains constant despite the frictional
force. Explain why this is so.

151
1986M2. An inclined plane makes an angle of with the horizontal, as shown above. A solid sphere of radius R
and mass M is initially at rest in the position shown, such that the lowest point of the sphere is a vertical height
h above the base of the plane. The sphere is released and rolls down the plane without slipping. The moment of
inertia of the sphere about an axis through its center is 2MR2/5. Express your answers in terms of M, R. h, g,
and .
a. Determine the following for the sphere when it is at the bottom of the plane:
i. Its translational kinetic energy
ii. Its rotational kinetic energy
b. Determine the following for the sphere when it is on the plane.
i. Its linear acceleration
ii. The magnitude of the frictional force acting on it

The solid sphere is replaced by a hollow sphere of identical radius R and mass M. The hollow sphere, which is
released from the same location as the solid sphere, rolls down the incline without slipping.
c. What is the total kinetic energy of the hollow sphere at the bottom of the plane?
d. State whether the rotational kinetic energy of the hollow sphere is greater than, less than, or equal to that of the
solid sphere at the bottom of the plane. Justify your answer.

1990M2. A block of mass m slides up the incline shown above with an initial speed v O in the position shown.
a. If the incline is frictionless, determine the maximum height H to which the block will rise, in terms of the given
quantities and appropriate constants.
b. If the incline is rough with coefficient of sliding friction , determine the maximum height to which the block
will rise in terms of H and the given quantities.

A thin hoop of mass m and radius R moves up the incline shown above with an initial speed v O in the position
shown.
c. If the incline is rough and the hoop rolls up the incline without slipping, determine the maximum height to
which the hoop will rise in terms of H and the given quantities.
d. If the incline is frictionless, determine the maximum height to which the hoop will rise in terms of H and the
given quantities.

152
1994M2. A large sphere rolls without slipping across a horizontal surface. The sphere has a constant translational
speed of 10 meters per second, a mass m of 25 kilograms, and a radius r of 0.2 meter. The moment of inertia of
the sphere about its center of mass is I = 2mr2/5. The sphere approaches a 25° incline of height 3 meters as
shown above and rolls up the incline without slipping.
a. Calculate the total kinetic energy of the sphere as it rolls along the horizontal surface.
b. i. Calculate the magnitude of the sphere's velocity just as it leaves the top of the incline.
ii. Specify the direction of the sphere's velocity just as it leaves the top of the incline.
c. Neglecting air resistance, calculate the horizontal distance from the point where the sphere leaves the incline to
the point where the sphere strikes the level surface.
d. Suppose, instead, that the sphere were to roll toward the incline as stated above, but the incline were frictionless.
State whether the speed of the sphere just as it leaves the top of the incline would be less than, equal to, or
greater than the speed calculated in b. Explain briefly.

1997M3. A solid cylinder with mass M, radius R, and rotational inertia ½MR2 rolls without slipping down the
inclined plane shown above. The cylinder starts from rest at a height H. The inclined plane makes an angle
with the horizontal. Express all solutions in terms of M, R, H, , and g.
a. Determine the translational speed of the cylinder when it reaches the bottom of the inclined plane.
b. On the figure below, draw and label the forces acting on the cylinder as it rolls down the inclined plane Your
arrow should begin at the point of application of each force.

c. Show that the acceleration of the center of mass of the cylinder while it is rolling down the inclined plane is
(2/3)g sin .
d. Determine the minimum coefficient of friction between the cylinder and the inclined plane that is required for
the cylinder to roll without slipping.
e. The coefficient of friction is now made less than the value determined in part d., so that the cylinder both
rotates and slips.
i. Indicate whether the translational speed of the cylinder at the bottom of the inclined plane is greater than, less
than, or equal to the translational speed calculated in part a. Justify your answer.
ii. Indicate whether the total kinetic energy of the cylinder at the bottom of the inclined plane is greater than,
less than, or equal to the total kinetic energy for the previous case of rolling without slipping. Justify your
answer.

153
2002M2. The cart shown above is made of a block of mass m and four solid rubber tires each of mass m/4 and
radius r. Each tire may be considered to be a disk. (A disk has rotational inertia ½ ML2, where M is the mass
and L is the radius of the disk.) The cart is released from rest and rolls without slipping from the top of an
inclined plane of height h. Express all algebraic answers in terms of the given quantities and fundamental
constants.
a. Determine the total rotational inertia of all four tires.
b. Determine the speed of the cart when it reaches the bottom of the incline.
c. After rolling down the incline and across the horizontal surface, the cart collides with a bumper of negligible
mass attached to an ideal spring, which has a spring constant k. Determine the distance x m the spring is
compressed before the cart and bumper come to rest.
d. Now assume that the bumper has a non-negligible mass. After the collision with the bumper, the spring is
compressed to a maximum distance of about 90% of the value of x m in part c.. Give a reasonable explanation
for this decrease.

2006M3. A thin hoop of mass M, radius R, and rotational inertia MR2 is released from rest from the top of the ramp
of length L above. The ramp makes an angle with respect to a horizontal tabletop to which the ramp is fixed.
The table is a height H above the floor. Assume that the hoop rolls without slipping down the ramp and across
the table. Express all algebraic answers in terms of given quantities and fundamental constants.

a. Derive an expression for the acceleration of the center of mass of the hoop as it rolls down the ramp.
b. Derive an expression for the speed of the center of mass of the hoop when it reaches the bottom of the ramp.
c. Derive an expression for the horizontal distance from the edge of the table to where the hoop lands on the floor.
d. Suppose that the hoop is now replaced by a disk having the same mass M and radius R. How will the distance
from the edge of the table to where the disk lands on the floor compare with the distance determined in part c.
for the hoop?

Less than____ The same as____ Greater than____

Briefly justify your response.

154
2010M2. A bowling ball of mass 6.0 kg is released from rest from the top of a slanted roof that is 4.0 m long and
angled at 30 , as shown above. The ball rolls along the roof without slipping. The rotational inertia of a sphere
of mass M and radius R about its center of mass is 2MR2/5.
a. On the figure below, draw and label the forces (not components) acting on the ball at their points of application
as it rolls along the roof.

b. Calculate the force due to friction acting on the ball as it rolls along the roof. If you need to draw anything other
than what you have shown in part a. to assist in your solution, use the space below. Do NOT add anything to the
figure in part a.
c. Calculate the linear speed of the center of mass of the ball when it reaches the bottom edge of the roof.
d. A wagon containing a box is at rest on the ground below the roof so that the ball falls a vertical distance of 3.0
m and lands and sticks in the center of the box. The total mass of the wagon and the box is 12 kg. Calculate the
horizontal speed of the wagon immediately after the ball lands in it.

155
SECTION D – Angular Momentum

View From Above

1975M2. A bicycle wheel of mass M (assumed to be concentrated at its rim) and radius R is mounted horizontally
so it may turn without friction on a vertical axle. A dart of mass m o is thrown with velocity v o as shown above
and sticks in the tire.
a. If the wheel is initially at rest, find its angular velocity after the dart strikes.
b. In terms of the given quantities, determine the ratio:
final kinetic energy of the system
initial kinetic energy of the system

1978M2. A system consists of a mass M 2 and a uniform rod of mass M 1 and length l. The rod is initially rotating
with an angular speed on a horizontal frictionless table about a vertical axis fixed at one end through point P.
The moment of inertia of the rod about P is Ml²/3. The rod strikes the stationary mass M 2 . As a result of this
collision, the rod is stopped and the mass M 2 moves away with speed v.
a. Using angular momentum conservation determine the speed v in terms of M 1 , M 2 , l, and .
b. Determine the linear momentum of this system just before the collision in terms of M 1 , l, and .
c. Determine the linear momentum of this system just after the collision in terms of M 1 l, and .
d. What is responsible for the change in the linear momentum of this system during the collision?
e. Why is the angular momentum of this system about point P conserved during the collision?

156
1981M3. A thin, uniform rod of mass M 1 and length L , is initially at rest on a frictionless horizontal surface. The
moment of inertia of the rod about its center of mass is M 1 L2/12. As shown in Figure I, the rod is struck at
point P by a mass m 2 whose initial velocity v is perpendicular to the rod. After the collision, mass m 2 has
velocity
–½v as shown in Figure II. Answer the following in terms of the symbols given.
a. Using the principle of conservation of linear momentum, determine the velocity v’ of the center of mass of this
rod after the collision.
b. Using the principle of conservation of angular momentum, determine the angular velocity of the rod about its
center of mass after the collision.
c. Determine the change in kinetic energy of the system resulting from the collision.

1982M3. A system consists of two small disks, of masses m and 2m, attached to a rod of negligible mass of length
3l as shown above. The rod is free to turn about a vertical axis through point P. The two disks rest on a rough
horizontal surface; the coefficient of friction between the disks and the surface is . At time t = 0, the rod has an
initial counterclockwise angular velocity o about P. The system is gradually brought to rest by friction.
Develop expressions for the following quantities in terms of m, l, g, and o
a. The initial angular momentum of the system about the axis through P
b. The frictional torque acting on the system about the axis through P
c. The time T at which the system will come to rest.

157
1987M3. A l.0-kilogram object is moving horizontally with a velocity of 10 meters per second, as shown above,
when it makes a glancing collision with the lower end of a bar that was hanging vertically at rest before the
collision. For the system consisting of the object and bar, linear momentum is not conserved in this collision,
but kinetic energy is conserved. The bar, which has a length l of 1.2 meters and a mass m of 3.0 kilograms, is
pivoted about the upper end. Immediately after the collision the object moves with speed v at an angle
relative to its original direction. The bar swings freely, and after the collision reaches a maximum angle of 90°
with respect to the vertical. The moment of inertia of the bar about the pivot is I bar = ml²/3 Ignore all friction.
a. Determine the angular velocity of the bar immediately after the collision.
b. Determine the speed v of the l-kilogram object immediately after the collision.
c. Determine the magnitude of the angular momentum of the object about the pivot just before the collision.
d. Determine the angle .

158
1992M2. Two identical spheres, each of mass M and negligible radius, are fastened to opposite ends of a rod of
negligible mass and length 2l. This system is initially at rest with the rod horizontal, as shown above, and is
free to rotate about a frictionless, horizontal axis through the center of the rod and perpendicular to the plane of
the page. A bug, of mass 3M, lands gently on the sphere on the left. Assume that the size of the bug is small
compared to the length of the rod. Express your answers to all parts of the question in terms of M, l, and
physical constants.
a. Determine the torque about the axis immediately after the bug lands on the sphere.
b. Determine the angular acceleration of the rod-spheres-bug system immediately after the bug lands.

The rod-spheres-bug system swings about the axis. At the instant that the rod is vertical, as shown above,
determine each of the following.
c. The angular speed of the bug
d. The angular momentum of the system
e. The magnitude and direction of the force that must be exerted on the bug by the sphere to keep the bug from
being thrown off the sphere

159
1996M3. Consider a thin uniform rod of mass M and length l, as shown above.

a. Show that the rotational inertia of the rod about an axis through its center and perpendicular to its length is
Ml2/12.

The rod is now glued to a thin hoop of mass M and radius R/2 to form a rigid assembly, as shown above. The
centers of the rod and the hoop coincide at point P. The assembly is mounted on a horizontal axle through point
P and perpendicular to the page.
b. What is the rotational inertia of the rod-hoop assembly about the axle?

Several turns of string are wrapped tightly around the circumference of the hoop. The system is at rest when a
cat, also of mass M, grabs the free end of the string and hangs vertically from it without swinging as it unwinds,
causing the rod-hoop assembly to rotate. Neglect friction and the mass of the string.
c. Determine the tension T in the string.
d. Determine the angular acceleration a of the rod-hoop assembly.
e. Determine the linear acceleration of the cat.
f. After descending a distance H = 5l/3, the cat lets go of the string. At that instant, what is the angular momentum
of the cat about point P?

160
1998M2. A space shuttle astronaut in a circular orbit around the Earth has an assembly consisting of two small
dense spheres, each of mass m, whose centers are connected by a rigid rod of length l and negligible mass. The
astronaut also has a device that will launch a small lump of clay of mass m at speed v 0 . Express your answers
in terms of m, v 0 l. and fundamental constants.

a. Initially, the assembly is "floating" freely at rest relative to the cabin, and the astronaut launches the clay lump
so that it perpendicularly strikes and sticks to the midpoint of the rod, as shown above.
i. Determine the total kinetic energy of the system (assembly and clay lump) after the collision.
ii. Determine the change in kinetic energy as a result of the collision.

b. The assembly is brought to rest, the clay lump removed, and the experiment is repeated as shown above, with
the clay lump striking perpendicular to the rod but this time sticking to one of the spheres of the assembly.
i. Determine the distance from the left end of the rod to the center of mass of the system (assembly and clay
lump) immediately after the collision. (Assume that the radii of the spheres and clay lump are much smaller
than the separation of the spheres.)

ii. On the figure above, indicate the direction of the motion of the center of mass immediately after the
collision.

iii. Determine the speed of the center of mass immediately after the collision.

iv. Determine the angular speed of the system (assembly and clay lump) immediately after the collision.

v. Determine the change in kinetic energy as a result of the collision.

161
2005M3. A system consists of a ball of mass M 2 and a uniform rod of mass M 1 and length d. The rod is attached to
a horizontal frictionless table by a pivot at point P
1
left. The rotational inertia of the rod about point P is M 1 d2 . The rod strikes the ball, which is initially at rest.
3
As a result of this collision, the rod is stopped and the ball moves in the direction shown above right. Express all
answers in terms of M 1 , M 2 and fundamental constants.
a. Derive an expression for the angular momentum of the rod about point P before the collision.
b. Derive an expression for the speed v of the ball after the collision.
c. Assuming that this collision is elastic, calculate the numerical value of the ratio M 1 / M 2

d. A new ball with the same mass M 1 as the rod is now placed a distance x from the pivot, as shown above. Again
assuming the collision is elastic, for what value of x will the rod stop moving after hitting the ball?

162
 ANSWERS - AP Physics Multiple Choice Practice – Rotation

SECTION A – Torque and Statics

Solution Answer

1. Definition of Torque D

2. To balance the forces (Fnet=0) the answer must be A or D, to prevent rotation, obviously A A
would be needed
3. FBD T Since the rope is at an angle it has x and y B
components of force.
H Therefore, H would have to exist to counteract T x .
W Based on net = 0 requirement, V also would have
V to exist to balance W if we were to chose a pivot
point at the right end of the bar

4. Applying rotational equilibrium to each diagram gives E

DIAGRAM 1: (mg)(L 1 ) = (M 1 g)(L 2 ) DIAGRAM 2: (M 2 g)(L 1 ) = mg(L 2 )

L 1 = M 1 (L 2 ) /m M 2 (L 1 ) = m (L 2 )
(sub this L 1 ) into the Diagram 2 eqn, and solve.

5. Find the torques of each using proper signs and add up. 2 C
+ (1) – (2) + (3) + (4)
+F(3R) – (2F)(3R) + F(2R) +F(3R) = 2FR

3
1
4

6. Simply apply rotational equilibrium B

(m 1 g) • r 1 = (m 2 g) • r 2
m1 a = m2 b

SECTION B – Rotational Kinematics and Dynamics

1. = / t A
2
2. K rot = ½ I B
3. = d /dt = 6t + 2; v = r D
2
4. I tot = I = I 0 + I M = I 0 + M(½L) A
5. = I where = (3M 0 )(l) – (M 0 )(2l) = M 0 l and I = (3M 0 )(l)2 + (M 0 )(2l)2 = 7M 0 l 2 A

6. X = Fl; O = F O L O sin , solve for the correct combination of F O and L O C

163
 7. Just as the tension in a rope is greatest at the bottom of as vertical circle, the force needed to C
maintain circular motion in any vertical circle is greatest at the bottom as the applied force must
balance the weight of the object and additionally provide the necessary centripetal force
8. F bottom = F adhesion – mg = F centripetal = m 2r E
2 2
9. For one complete revolution =2 ; = 0 +2 C
10. The moment of inertia is least about the object’s center of mass. The greater the distance of the E
axis from the center of mass, the greater the moment of inertia (I = I cm + MD2). The center of
mass is at point B.
11. = L/ t = (I f – 0)/T E
12. P avg = avg = (I f /T)(½ f) or P avg = K/T B
13. D
= T2R – T1R = I

14. = 0t + ½ t2 D
15. If the cylinder is “suspended in mid air” (i.e. the linear acceleration is zero) then F = 0 E
2
16. = TR = I = ½ MR which gives = 2T/MR and since F = 0 then T = Mg so = 2g/R B
the acceleration of the person’s hand is equal to the linear acceleration of the string around the
rim of the cylinder a = R = 2g
2
17. In order that the mass not slide down f = F N N =m R B
2
solving for gives R

SECTION C – Rolling

1. K tot = K rot + K trans = ½ I 2 + ½ Mv2 = ½ (2/5)MR2 2

+ ½ Mv2 = (1/5)Mv2 + ½ Mv2 = D
(7/10)Mv2 = MgH, solving gives H = 7v2/10g
2. Rolling without slipping: v = R. Linear momentum = Mv = M R A
3. Mgh = K tot = K rot + K trans , however without friction, there is no torque to cause the sphere to A
rotate so K rot = 0 and Mgh = ½ Mv2
4. Mgh = K tot = K rot + K trans = ½ I 2 + ½ Mv2; substituting v/r for gives E
Mgh = ½(I/r2 + M)v2 and solving for v gives v2 = 2Mgh/(I/r2 + M), multiplying by r2/r2 gives
desired answer
5. v = d/t and = v/r C
6. The first movement of the point of contact of a rolling object is vertically upward as there is no A
side to side (sliding) motion for the point in contact

164
SECTION D – Angular Momentum

1. L i = L f so I i I = I f f and since I f < I i (mass more concentrated near axis), then f > i C
The increase in is in the same proportion as the decrease in I, and the kinetic energy is
proportional to I 2 so the increase in results in an overall increase in the kinetic energy.
Alternately, the skater does work to pull their arms in and this work increases the KE of the
skater
2. L=I D
3. L i = L f so I i I = I f f and since I f < I i (mass more concentrated near axis), then f > i E
The increase in is in the same proportion as the decrease in I, and the kinetic energy is
proportional to I 2 so the increase in results in an overall increase in the kinetic energy.
Alternately, the skater does work to pull their arms in and this work increases the KE of the
skater
4. L = mvr where r is the perpendicular line joining the origin and the line along which the B
particle is moving
5. L = I and since is uniform the ratio L upper /L lower = I upper /I lower = 2mL2/2(2m)(2L)2 = 1/8 E
6. Since it is a perfectly inelastic (sticking) collision, KE is not conserved. As there are no E
external forces or torques, both linear and angular momentum are conserved
7. As there are no external forces or torques, both linear and angular momentum are conserved. C
As the type of collision is not specified, we cannot say kinetic energy must be the same.
8. L i = L f so I i I = I f f and since I f < I i (mass more concentrated near axis), then f > i C
The increase in is in the same proportion as the decrease in I, and the kinetic energy is
proportional to I 2 so the increase in results in an overall increase in the kinetic energy.
Alternately, the skater does work to pull their arms in and this work increases the KE of the
skater

165
166
AP Physics Free Response Practice – Rotation – ANSWERS

SECTION A – Torque and Statics

2008M2
a.

b. =0
About the hinge: TL sin 30º – mgL – Mg(L/2) = 0 gives T = 29 N
c. I total = I rod + I block where I rod,end = I cm + MD2 = ML2/12 + M(L/2)2 = ML2/3
I total = ML2/3 + mL2 = 0.42 kg-m2
d. =I
mgL + MgL/2 = I gives = 21 rad/s2

SECTION B – Rotational Kinematics and Dynamics

1973M3
a. Define a coordinate system with the x-axis directed to the vertical rod and the y-axis directed upwards and
perpendicular to the first. Let T 1 be the tension in the horizontal string. Let T 2 be the tension in the string tilted
upwards.
Applying Newton’s Second Law: F x = T 1 + T 2 sin = m 2R; F y = T 2 cos – mg = 0
Solving yields: T 2 = mg/cos and T 1 = m( 2R – g tan )
b. Let T 1 = 0 and solving for gives = (g tan /R)1/2

1976M2
a.

b. = I (about center of mass) (one could also choose about the point at which the tape comes off the cylinder)
TR = ½ MR2×(a/R)
T = ½ Ma

F = ma
Mg – T = Ma
Mg = 3Ma/2
a = 2g/3
c. As there are no horizontal forces, the cylinder moves straight down.

167
1978M1
a.

b. F = ma; T cos = mg and T sin = m 2r = m 2

(A + l sin )

=
+
c. W = E = K + U = ½ mv2 + mgl(1 – cos ) for each rider
W = 6(½ mv2 + mgl(1 – cos ))

1983M2
a.

b. i./ii. On the disk: = I = TR = ½ m 1 R2

For the block a = R so = a/R and F = m 2 g – T = m 2 a
Solving yields
2 2 1 2
= =
2 2+ 1 2 2+ 1

168
1985M3
a.

b. i. F = ma; T 1 – mg = ma and 2mg – T 2 = 2ma

ii. = I ; (T 2 – T 1 )r = I
c. = a/r
Combining equations from b.i. gives T 2 – T 1 = mg – 3ma
Substituting for (T 2 – T 1 ) into torque equation gives a = 2g/9
d. T 1 = m(g + a) = 11mg/9
e. F N = 7mg + T 1 + T 2 (the table counters all the downward forces on the apparatus)
T 2 = 2m(g – a) = 14mg/9
F N = 88mg/9

1988M3
a. I is proportional to mR2; masses are equal and R becomes 2R
I 2R = 4I
b. The disks are coupled by the chain along their rims, which means the linear motion of the rims have the same
displacement, velocity and acceleration.
v R = v 2R ; R R = 2R 2R ; R t = 2R 2R t gives 2R = /2
c. 2R = T(2R) = I 2R 2R = (4I)( /2) = 2I giving T = I /R
d. = student – TR = I
student = I + TR = I + (I /R)R = 2I
e. K = ½ I 2 = ½ I( t)2

1989M2
a. F = ma; 2Mg – T v = 2Ma so T v = 2M(g – a)
b. = T v R – T h R = I = 3MR2(a/R)
T h = (T v R – 3MRa)/R = 2M(g – a) – 3Ma = 2Mg – 5Ma
c. F f = F N = (4Mg)
T h – F f = 3Ma
2Mg – 5Ma – 4 Mg = 3Ma
4 Mg = 2Mg – 8Ma
= (2g – 8a)/4g
plugging in given values gives = 0.1
d. F f = 4 Mg = ma C = 4Ma C
a C = 1 m/s2

1991M2
a. = 0; m 2 gr 2 = m 1 gr 1 ; m 2 = m 1 r 1 /r 2 = 6.67 kg
b./c. = I ; Tr 1 = (45 kg-m2)
F = ma; (20 kg)g – T = (20 kg)a
Combining with a = r gives = 2 rad/s2 and T = 180 N
d. mgh = ½ mv2 + ½ I 2 = ½ mv2 + ½ I(v2/r2) giving v = 1.4 m/s

169
1993M3
a. = F a l – Mgl/2 = 0 giving F a = Mg/2
b. = Mgl/2 = I = (Ml2/3) ; = 3g/2l
c. a = r where r = l/2
a = (3g/2l)(l/2) = 3g/4
d. F = Ma; Mg – F a = Ma = M(3g/4)
F a = Mg/4
e. U = K rot
mgh = mg(l/2)sin = ½ I 2 = ½ (Ml2/3) 2

solving gives = (3gsin /l)1/2

1999M3
a. = 0 so cw = ccw and cw = TR (from the string) so we just need to find ccw as the sum of the torques from
the various parts of the system
ccw = rod + block = mgR sin 0 + 2mg(2R)sin 0 = 5mgR sin 0 = TR so T = 5mg sin 0
b. i. I = I disk + I rod + I block = 3mR2/2 + 4mR2/3 + 2m(2R)2 = 65mR2/6
= /I = (5mgR sin 0 )/(65mR2/6) = 6g sin 0 /13R
ii. a = r where r = 2R so a = 12g sin 0 /13
c. U (from each component) = K = ½ I 2
mgR cos 0 + 2mg(2R) cos 0 = ½ (65mR2/6) 2
= (12g cos 0 /13R)1/2 and v = r = (2R) = 4(3gR cos 0 /13)1/2

2000M3
a.

F = ma = 0 so T = 2mg

170
b. i.

F = ma
3mg – T 3 = 3m(g/3)
T 3 = 2mg

ii.

F = ma
T 1 – mg = m(g/3)
T 1 = 4mg/3

iii. = (T 3 – T 1 )R 1 = I and = a/R 1 = g/3R 1

(2mg – 4mg/3)R 1 = I 1 (g/3R 1 )
I 1 = 2mR 1 2
c. i. Tangential speeds are equal; 1 R 1 = 2 R 2 = 2 (2R 1 ) therefore 2 = 1 /2
ii. L = I = (16I 1 )( 1 /2) = 8I 1 1
iii. K = ½ I 1 1 2 + ½ I 2 2 2 = (5/2)I 1 1 2

2001M3
a. I = mr2 = mL2 + mL2 = 2mL2
b. F = ma; 4mg – T = 4ma
= I ; Tr = I ; T = I /r = 4mg – 4ma and = a/r, solving gives a = 2gr2/(L2 + 2r2)
c. Equal, total energy is conserved
d. Less, the small blocks rise and gain potential energy. The total energy available is still 4mgD, therefore the
kinetic energy must be less than in part c.

2003M3
a. i.

ii. x = 33 m

171
b. i. y = ½ gt2; t = (2y/g)1/2 = 1.75 s
ii. U initial = U bucket + U projectile = M(9.8 m/s2)(3 m) + (10 kg)(9.8 m/s2)(3 m) = 29.4M + 294
iii. U initial = U final + K where U final = Mg(1 m) + (10 kg)g(15 m) = 9.8M + 1470
K projectile = ½ 10v x 2 and K bucket = ½ Mv b 2 where v b = v x /6
putting it all together gives 29.4M + 294 = 9.8M + 1470 + 5v x 2 + (M/72)v x 2
19.6 1176
𝑣𝑥 =
5+ /72
c. i. x = v x t
19.6 1176
𝑥 = 1.75
5+ /72
d. x(300 kg) = 39.7 m (greater than the experimental value)
possible reasons include friction at the pivot, air resistance, neglected masses not negligible

2004M2
a. x = v 0 t + ½ at2
x = D and v 0 = 0 so D = ½ at2 and a = 2D/t2
b. i. graph D vs. t2 (as an example)
ii.

iii. a = 2(slope) = 2.04 m/s2

c. = TR = I and = a/R so I = TR2/a
F = mg – T = ma so T = m(g – a)
I = m(g – a)R2/a = mR2((g/a) – 1)
d. The string was wrapped around the pulley several times, causing the effective radius at which the torque acted
to be larger than the radius of the pulley used in the calculation.

The string slipped on the pulley, allowing the block to accelerate faster than it would have otherwise, resulting
in a smaller experimental moment of inertia.

Friction is not a correct answer, since the presence of friction would make the experimental value of the
moment of inertia too large

172
SECTION C – Rolling
NOTE: Rolling problems may be solved considering rotation about the center of mass or the point of contact. The
solutions below will only show one of the two methods, but for most, if not all cases, the other method is
applicable. When considering rotation about the point of contact, remember to use the parallel axis theorem for
the moment of inertia of the rolling object.

1974M2
a. Torque provided by friction; at minimum , F f = F N = Mg cos
= F f R = I = (2/5)MR2(a/R); F f = (2/5)Ma = Mg cos giving a = (5/2) g cos
F = Ma; Mg sin – Mg cos = Ma = (5/2) Mg cos giving = (2/7) tan
b. Energy at the bottom is the same in both cases, however with = 0, there is no torque and no energy in rotation,
which leaves more (all) energy in translation and velocity is higher

1977M2
a.

b. F y = 0; T + N = W; N = W – T = Mg – (3/5)Mg = (2/5)Mg
F x = ma; F f = ma; N = ma; ½ (2/5)Mg = Ma; a = g/5
c. = I ; (T – F f )R = ½ MR2
(3/5)Mg – (1/5)Mg = ½ MR
(2/5)g = ½ R
= 4g/5R
d. The cylinder is slipping on the surface and does not meet the condition for pure rolling

1980M3
a. F = ma; F f = F N ;
– Mg = Ma
a=– g
v = v 0 + at
v = v 0 – gt
b. = I where the torque is provided by friction F f = Mg
MgR = (2MR2/5)
= (5 g/2R)
= 0 + t = (5 g/2R)t
c. Slipping stops when the tangential velocity si equal to the velocity of the center of mass, or the condition for
pure rolling has been met: v(t) = (t)R
v 0 – gt = R(5g/2R)t, which gives T = (2/7)(v 0 / g)
d. Since the line of action of the frictional force passes through P, the net torque about point P is zero. Thus, the
time rate of change of the angular momentum is zero and the angular momentum is constant.

173
1986M2
a. U = K
Mgh = ½ Mv2 + ½ I 2 and = v/R
Mgh = ½ Mv2 + ½ (2/5)MR2(v/R)2 = ½ Mv2 + (1/5)Mv2 = 7Mv2/10
v2 = 10gh/7
i. K trans = ½ Mv2 = (5/7)Mgh
ii. K rot = ½ I 2 = (2/7)Mgh (or Mgh – K trans )
b. i. = F f R = I = I(a/R)
F f R = (2/5)MR2(a/R)
F f = (2/5)Ma
F = ma
Mg sin – F f = Ma
Mg sin – (2/5)Ma = Ma
g sin = (7/5) a
a = (5/7) g sin

ii. F f = (2/5)Ma = (2/5)M(5/7)g sin = (2/7)Mg sin

c. K tot = Mgh
d. Greater, the moment of inertia of the hollow sphere is greater and will be moving slower at the bottom of the
incline. Since the translational speed is less, the translational KE is taking a smaller share of the same total
energy as the solid sphere.

1990M2
a. K = U
½ mv 0 2 = mgH; H = v 0 2/2g
b. K + W f = U where W f = –F f d and F f = mg cos and d = h/sin
½ mv 0 2 – ( mg cos )(h/sin ) = mgh
½ mv 0 2 = mgh( cot + 1)
h = v 0 2/(2g( cot + 1)) = H/( cot + 1)
c. K trans + K rot = U where K rot = ½ I 2 = ½ (mR2)(v/R)2 = ½ mv 0 2
½ mv 0 2 + ½ mv 0 2 = mgh
h = v 0 2/g = 2H
d. Rotational energy will not change therefore ½ mv 0 2 = mgh and h = v 0 2/2g = H

1994M2
a. K tot = K trans + K rot = ½ Mv2 + ½ I 2 and = v/R
K tot = ½ Mv2 + ½ (2/5)MR2(v/R)2 = ½ Mv2 + (1/5)Mv2 = 7Mv2/10 = 1750 J
b. i. K total,bottom = K top + U top = 7Mv top 2/10 + Mgh; v top = 7.56 m/s
ii. It is directed parallel to the incline: 25º
c. y = y 0 + v oy t + ½ a y t2
0 m = 3 m + (7.56 m/s)(sin 25º)t + ½ (–9.8 m/s2)t2 which gives t = 1.16 s (positive root)
x = v x t = (7.56 m/s)(cos 25º)(1.16 s) = 7.93 m
d. The speed would be less than in b.
The gain in potential energy is entirely at the expense of the translational kinetic energy as there is no torque to
slow the rotation.

1997M3
a. U = K
MgH = ½ Mv2 + ½ I 2 and = v/R
MgH = ½ Mv2 + ½ (½)MR2(v/R)2 = ½ Mv2 + ¼ Mv2 = 3Mv2/4
v = (4gH/3)1/2

174
b.

c. For a change of pace, we can use kinematics:

v f 2 = v i 2 + 2ad
4gH/3 = 0 + 2a(H/sin )
a = (2/3)g sin
d. F = Ma
Mg sin – F f = Ma = M(2/3)g sin
Mg sin – Mg cos = (2/3)Mg sin
cos = (1/3) sin
= (1/3) tan
e. i. The translational speed is greater, less energy is transferred to the rotational motion so more goes into the
translational motion. Additionally, with a smaller frictional force, the translational acceleration is greater.
ii. Total kinetic energy is less. Energy is dissipated as heat due to friction.

2002M2
a. For each tire: I = ½ ML2 = ½ (m/4)r2
I total = 4×I = ½ mr2
b. U = K; total mass = 2m
2mgh = ½ (2m)v2 + ½ I 2 and = v/R
2mgh = mv2 + ½ (½)mr2(v/r)2 = ½ mv2 + (¼)mv2 = 5mv2/4
v = (8gh/5)1/2
c. U g = U s
2mgh = ½ kx m 2; x m = 2(mgh/k)1/2
d. In an inelastic collision, energy is lost. With less energy after the collision, the spring is compressed less.

2006M3
a. =I
F f R = I = MR2(a/R) ; F f = Ma
F = ma
Mg sin – F f = Ma
Mg sin – Ma = Ma
a = ½ g sin
b. v f 2 = 2aL = gL sin
v f = (gL sin )1/2
c. H = ½ gt2; t = (2H/g)1/2
d = v x t = (gL sin )1/2(2H/g)1/2 = (2LH sin )1/2
d. Greater. A disk will have smaller rotational inertia and will therefore have a greater rotational velocity. This
will lead to a greater translational velocity, and a greater distance x.

175
2010M2
a.

b. Torque provided by friction; F f = F N = Mg cos

= F f R = I = (2/5)MR2(a/R); F f = (2/5)Ma; Ma = (5/2)F f
F = Ma
Mg sin – F f = (5/2)F f
F f = (2/7)Mg sin = 8.4 N
c. Mgh = ½ Mv2 + ½ I 2 and = v/R
Mgh = ½ Mv2 + ½ (2/5)MR2(v/R)2 = ½ Mv2 + (1/5)Mv2 = 7Mv2/10
v2 = 10gh/7; v = 5.3 m/s
d. The horizontal speed of the wagon is due to the horizontal component of the ball in the collision:
M i v ix = M f v fx ; where M f = M ball + M wagon = 18 kg
(6 kg)(5.3 m/s)(cos 30º) = (18 kg)v f
v f = 1.5 m/s

SECTION D – Angular Momentum

1975M2
a. L i = L f
m 0 v 0 R sin = I
= m 0 v 0 R sin /I; I = (M + m 0 )R2
= m 0 v 0 sin /(M + m 0 )R
b. K i = ½ m 0 v 0 2
K f = ½ I 2 = ½ (M + m 0 )R2(m 0 v 0 sin /(M + m 0 )R)2
K f /K i = m 0 sin2 /(M + m 0 )

1978M2
a. L i = L f
I = mvr
(1/3)M 1 l2 = M 2 vl
v = M 1 l /3M 2
b. p system = p cm of rod = M 1 v cm = M 1 (l/2)
c. P f = M 2 v f = M 1 l/3M 2
d. There is a net external force on the system from the axis at point P.
e. Since the net external force acts at point P (the pivot), the net torque about point P is zero, hence angular
momentum is conserved.

1981M3
a. m 2 v = m 2 (–v/2) + M 1 v
v = 3m 2 v/2M 1
b. L i = L f
m 2 v(L/3) = m 2 (–v/2)(L/3) + (1/12)M 1 L2
= 6m 2 v/M 1 L

176