AP Physics C Mechanics Quiz: Conservation Of Angular Momentum
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Conservation Of Angular MomentumQuestion 1 of 20

A diver performs a somersault after jumping from a diving board. They initially leave the board with their body extended, then tuck into a compact shape, and finally extend their body again before entering the water. Neglecting air resistance, which of the following correctly describes the changes in their rotational inertia II and angular momentum LL about their center of mass during this process?

II decreases, then increases; LL remains approximately constant.
II remains constant; LL decreases, then increases.
Both II and LL decrease, then increase.
Both II and LL remain approximately constant.
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AP Physics C Mechanics Quiz

AP Physics C Mechanics Quiz: Conservation Of Angular Momentum

Practice Conservation Of Angular Momentum in AP Physics C Mechanics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Conservation Of Angular Momentum, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics C Mechanics.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

A diver performs a somersault after jumping from a diving board. They initially leave the board with their body extended, then tuck into a compact shape, and finally extend their body again before entering the water. Neglecting air resistance, which of the following correctly describes the changes in their rotational inertia II and angular momentum LL about their center of mass during this process?

  1. II decreases, then increases; LL remains approximately constant. (correct answer)
  2. II remains constant; LL decreases, then increases.
  3. Both II and LL decrease, then increase.
  4. Both II and LL remain approximately constant.
Explanation: The only significant external force on the diver during flight is gravity, which acts on the center of mass. Therefore, there is no net external torque about the diver's center of mass. This means their angular momentum LL is conserved and remains approximately constant. When the diver tucks, they bring their mass closer to the axis of rotation, decreasing their rotational inertia II. When they extend their body, they move mass away from the axis, increasing II.

Question 2

A horizontal disk with rotational inertia I0I_0 is rotating freely with angular velocity ω0\omega_0 about a vertical axis. A small piece of putty of mass mm is dropped vertically and sticks to the disk at a distance rr from the axis. What is the final angular velocity of the disk-putty system?

  1. I0ω0I0+mr2\frac{I_0 \omega_0}{I_0 + mr^2} (correct answer)
  2. I0ω0I0−mr2\frac{I_0 \omega_0}{I_0 - mr^2}
  3. (I0+mr2)ω0I0\frac{(I_0 + mr^2) \omega_0}{I_0}
  4. ω0\omega_0
Explanation: Since the putty is dropped vertically, it carries no initial horizontal momentum and thus no initial angular momentum about the disk's axis. The forces during the collision are internal to the disk-putty system, so the total angular momentum is conserved. The initial angular momentum is Li=I0ω0L_i = I_0 \omega_0. The final rotational inertia of the system is If=I0+Iputty=I0+mr2I_f = I_0 + I_{putty} = I_0 + mr^2. The final angular momentum is Lf=Ifωf=(I0+mr2)ωfL_f = I_f \omega_f = (I_0 + mr^2)\omega_f. Setting Li=LfL_i = L_f gives I0ω0=(I0+mr2)ωfI_0 \omega_0 = (I_0 + mr^2)\omega_f. Solving for ωf\omega_f yields the correct answer.

Question 3

Two disks on a common frictionless axle are brought into contact. Disk 1 has rotational inertia I1I_1 and is initially spinning. Disk 2 has rotational inertia I2I_2 and is initially at rest. They eventually rotate together at a common final angular velocity. Why is the total angular momentum of the two-disk system conserved during this process?

  1. Because the kinetic energy dissipated by friction is converted perfectly into potential energy.
  2. Because the supporting axle is frictionless, preventing any external forces on the system.
  3. Because the frictional torques that bring them to a common speed are internal to the two-disk system. (correct answer)
  4. Because the total mass and total rotational inertia of the combined system remain constant.
Explanation: The key to conservation of angular momentum is the absence of a net external torque. The frictional forces between the two disks create torques, but these torques are internal to the system consisting of both disks. Disk 1 exerts a torque on Disk 2, and Disk 2 exerts an equal and opposite torque on Disk 1. These internal torques cancel out when considering the system as a whole, so the system's total angular momentum is conserved. The frictionless axle ensures there is no external torque from the support.

Question 4

A circular platform of mass MM and radius RR is free to rotate friction-free about its center. A person of mass mm stands at the edge. The system is initially at rest. The person begins to walk along the edge with a speed vv relative to the platform. What is the magnitude of the angular velocity of the platform relative to the ground? The rotational inertia of the platform is Ip=12MR2I_p = \frac{1}{2}MR^2.

  1. mvR(m)\frac{mv}{R(m)}
  2. mvR(12M)\frac{mv}{R(\frac{1}{2}M)}
  3. mvR(12M+m)\frac{mv}{R(\frac{1}{2}M + m)} (correct answer)
  4. mvR(M+m)\frac{mv}{R(M + m)}
Explanation: The system starts from rest, so the total initial angular momentum is zero. As there are no external torques, the total final angular momentum must also be zero. Let ωp\omega_p be the platform's angular velocity relative to the ground. The person's velocity relative to the ground is vg=vrel−vplat=v−Rωpv_g = v_{rel} - v_{plat} = v - R\omega_p. The total final angular momentum is Lf=Ipωp+Ipersonωperson=0L_f = I_p\omega_p + I_{person}\omega_{person} = 0. The person's angular velocity is vg/Rv_g/R. So, (12MR2)ωp+(mR2)(v−RωpR)=0(\frac{1}{2}MR^2)\omega_p + (mR^2)(\frac{v - R\omega_p}{R}) = 0. Simplifying: 12MR2ωp+mR(v−Rωp)=0⇒12MR2ωp+mRv−mR2ωp=0⇒ωp(12MR2+mR2)=mRv\frac{1}{2}MR^2\omega_p + mR(v-R\omega_p) = 0 \Rightarrow \frac{1}{2}MR^2\omega_p + mRv - mR^2\omega_p = 0 \Rightarrow \omega_p(\frac{1}{2}MR^2 + mR^2) = mRv. Solving for ωp\omega_p gives the result.

Question 5

A rotating, uniform spherical star of mass MM and radius RR has an initial angular velocity ωi\omega_i. It undergoes a gravitational collapse, shrinking to a final radius of R/2R/2 while retaining all its mass. Assuming the star's density remains uniform, what is its final angular velocity ωf\omega_f? The rotational inertia of a uniform sphere is I=25MR2I = \frac{2}{5}MR^2.

  1. ωi/4\omega_i / 4
  2. ωi/2\omega_i / 2
  3. 2ωi2\omega_i
  4. 4ωi4\omega_i (correct answer)
Explanation: The collapse is due to internal gravitational forces, so there is no net external torque. Thus, the star's angular momentum is conserved. Let Li=LfL_i = L_f. The initial angular momentum is Li=Iiωi=(25MR2)ωiL_i = I_i \omega_i = (\frac{2}{5}MR^2)\omega_i. The final angular momentum is Lf=Ifωf=(25M(R/2)2)ωf=(25MR24)ωfL_f = I_f \omega_f = (\frac{2}{5}M(R/2)^2)\omega_f = (\frac{2}{5}M\frac{R^2}{4})\omega_f. Setting them equal: (25MR2)ωi=(14)(25MR2)ωf(\frac{2}{5}MR^2)\omega_i = (\frac{1}{4})(\frac{2}{5}MR^2)\omega_f. This simplifies to ωi=14ωf\omega_i = \frac{1}{4}\omega_f, or ωf=4ωi\omega_f = 4\omega_i.

Question 6

A child is sitting on the edge of a merry-go-round that is rotating freely. The child then slowly walks toward the center of the merry-go-round. The angular momentum of the child-merry-go-round system is conserved because:

  1. the forces between the child and the merry-go-round are internal to the system, resulting in zero net external torque. (correct answer)
  2. the rotational kinetic energy of the system remains constant throughout the process.
  3. the child's linear velocity changes, but the merry-go-round's angular velocity compensates perfectly.
  4. the angular momentum of the child is conserved, and the angular momentum of the merry-go-round is also independently conserved.
Explanation: The conservation of angular momentum for a system depends on the absence of a net external torque. When the child walks toward the center, the forces they exert on the merry-go-round (and the forces the merry-go-round exerts on them) are internal to the defined system. Assuming a frictionless axle, there is no significant external torque, so the total angular momentum of the system is conserved.

Question 7

A thin uniform rod of mass MM and length LL is pivoted at one end and hangs vertically at rest. A small ball of mass mm traveling horizontally with speed vv strikes the rod at its bottom end and embeds itself. What is the angular velocity of the rod-ball system immediately after the collision? The rotational inertia of the rod about the pivot is Irod=13ML2I_{rod} = \frac{1}{3}ML^2.

  1. mvL13ML2+mL2\frac{mvL}{\frac{1}{3}ML^2 + mL^2} (correct answer)
  2. (M+m)v13M+m\frac{(M+m)v}{\frac{1}{3}M + m}
  3. mv13ML\frac{mv}{\frac{1}{3}ML}
  4. mvL12ML2+mL2\frac{mvL}{\frac{1}{2}ML^2 + mL^2}
Explanation: Angular momentum is conserved about the pivot point during the collision, as the collision forces are internal and the gravitational torque is negligible over the short impact time. The initial angular momentum is solely from the ball: Li=∣r⃗×p⃗∣=L(mv)L_i = |\vec{r} \times \vec{p}| = L(mv). The final rotational inertia of the system is the sum of the rod's inertia and the ball's inertia (treated as a point mass): If=Irod+Iball=13ML2+mL2I_f = I_{rod} + I_{ball} = \frac{1}{3}ML^2 + mL^2. The final angular momentum is Lf=Ifω=(13ML2+mL2)ωL_f = I_f \omega = (\frac{1}{3}ML^2 + mL^2)\omega. Equating Li=LfL_i = L_f and solving for ω\omega gives the result.

Question 8

A person stands at the center of a frictionless turntable that is initially at rest. They hold two heavy weights. They extend their arms horizontally and swing them in a clockwise circle. What happens to the turntable?

  1. It remains at rest because the person does not push off of anything external.
  2. It rotates clockwise along with the arms and weights to conserve kinetic energy.
  3. It rotates counter-clockwise to keep the total angular momentum of the system zero. (correct answer)
  4. It rotates clockwise, but slower than the arms, to conserve linear momentum.
Explanation: The system consists of the person, the weights, and the turntable. It starts from rest, so its initial total angular momentum is zero. All forces and torques involved in moving the arms and weights are internal to this system. With no net external torque, the total angular momentum must remain zero. As the person gives the weights a clockwise angular momentum, the person's body and the turntable must acquire an equal and opposite (counter-clockwise) angular momentum to keep the total angular momentum of the system at zero.

Question 9

A rigid body's angular momentum about a fixed axis changes from 4.0 kg⋅m2/s4.0 \, \text{kg} \cdot \text{m}^2/\text{s} to 10.0 kg⋅m2/s10.0 \, \text{kg} \cdot \text{m}^2/\text{s} in 2.02.0 seconds. What is the magnitude of the average net external torque that acted on the body during this time?

  1. 2.0 N⋅m2.0 \, \text{N} \cdot \text{m}
  2. 3.0 N⋅m3.0 \, \text{N} \cdot \text{m} (correct answer)
  3. 5.0 N⋅m5.0 \, \text{N} \cdot \text{m}
  4. 7.0 N⋅m7.0 \, \text{N} \cdot \text{m}
Explanation: The rotational form of Newton's second law states that the average net external torque is equal to the rate of change of angular momentum: τavg=ΔLΔt\tau_{avg} = \frac{\Delta L}{\Delta t}. In this case, ΔL=Lf−Li=10.0−4.0=6.0 kg⋅m2/s\Delta L = L_f - L_i = 10.0 - 4.0 = 6.0 \, \text{kg} \cdot \text{m}^2/\text{s}, and Δt=2.0 s\Delta t = 2.0 \, \text{s}. Therefore, τavg=6.02.0=3.0 N⋅m\tau_{avg} = \frac{6.0}{2.0} = 3.0 \, \text{N} \cdot \text{m}. This question highlights that a non-zero torque causes a change in angular momentum, which is the corollary to the conservation principle.

Question 10

An ice skater is spinning on a frictionless surface with her arms extended. She then pulls her arms in close to her body. Which of the following correctly describes the change in her angular velocity and her rotational kinetic energy?

  1. Her angular velocity increases, and her rotational kinetic energy increases because she does positive work to pull her arms inward. (correct answer)
  2. Her angular velocity increases, but her rotational kinetic energy remains constant because there is no external torque acting on her.
  3. Her angular velocity remains constant because angular momentum is conserved, but her rotational kinetic energy increases.
  4. Her angular velocity decreases because her moment of inertia decreases, and her rotational kinetic energy also decreases.
Explanation: Because there is no net external torque on the skater, her angular momentum L=IωL = I\omega is conserved. When she pulls her arms in, her moment of inertia II decreases. To keep LL constant, her angular velocity ω\omega must increase. Her rotational kinetic energy is given by K=12Iω2=L22IK = \frac{1}{2}I\omega^2 = \frac{L^2}{2I}. Since LL is constant and II decreases, her kinetic energy KK must increase. The increase in kinetic energy comes from the work she does to pull her arms inward against the centrifugal effects.

Question 11

A satellite in space has initial rotational inertia Ii=120 kg\cdotpm2I_i=120\,\text{kg·m}^2 and spins at ωi=1.5 rad/s\omega_i=1.5\,\text{rad/s}. It deploys solar panels, increasing inertia to If=300 kg\cdotpm2I_f=300\,\text{kg·m}^2 with negligible external torque. Angular momentum is conserved during deployment. Considering the given conditions, calculate the new speed of rotation when the configuration changes.

  1. 0.60 rad/s (correct answer)
  2. 3.75 rad/s
  3. 1.50 rad/s
  4. -0.60 rad/s
Explanation: This question tests AP Physics C: Mechanics skills, specifically the conservation of angular momentum. Angular momentum is conserved in a closed system without external torques; it is calculated as the product of rotational inertia and angular velocity (L = Iω). In this scenario, a satellite deploys solar panels, increasing its moment of inertia from 120 to 300 kg·m², which causes the angular velocity to decrease proportionally. Choice A is correct because L_initial = I_i·ω_i = 120·1.5 = 180 kg·m²/s, and ω_final = L_initial/I_final = 180/300 = 0.60 rad/s. Choice B is incorrect as it represents the inverse calculation (300/120)·1.5, a common error when students confuse the relationship between I and ω. To help students: Use the ice skater analogy - extending arms increases I and decreases ω. Practice problems with changing configurations, emphasizing that L = Iω remains constant while I and ω change inversely.

Question 12

A comet moves in a highly elliptical orbit around the Sun. The gravitational force exerted by the Sun on the comet is always directed towards the Sun. Which of the following quantities remains constant for the comet throughout its orbit?

  1. Its linear velocity, because the gravitational force is the only force acting on it.
  2. Its kinetic energy, because the work done by the gravitational force is zero over one full orbit.
  3. Its angular momentum about the Sun, because the gravitational force exerts no torque about the Sun. (correct answer)
  4. Its potential energy, because the gravitational force is a conservative force.
Explanation: The gravitational force exerted by the Sun on the comet is a central force, meaning it is always directed along the line connecting the two bodies. The torque about the Sun is given by τ⃗=r⃗×F⃗\vec{\tau} = \vec{r} \times \vec{F}. Since r⃗\vec{r} and F⃗\vec{F} are parallel (or anti-parallel), their cross product is zero. With zero net external torque, the comet's angular momentum about the Sun is conserved. Its speed and kinetic energy change, being maximum at perihelion and minimum at aphelion. Its potential energy also changes with distance from the Sun.

Question 13

A student sits at rest on a frictionless rotating stool, holding a bicycle wheel that is spinning with its axis vertical, creating an upward angular momentum L⃗w\vec{L}_w. The student flips the wheel over so its axis is vertical but pointing downward. What is the final angular velocity of the student-stool system?

  1. Zero, because the wheel's momentum is canceled by the flipping action.
  2. A non-zero value, causing rotation in the same direction as the wheel's initial spin. (correct answer)
  3. A non-zero value, causing rotation in the opposite direction to the wheel's initial spin.
  4. The student oscillates but does not acquire a net angular velocity.
Explanation: The total angular momentum of the student-stool-wheel system is conserved. Initially, L⃗total=L⃗student+L⃗wheel=0+L⃗w=L⃗w\vec{L}_{total} = \vec{L}_{student} + \vec{L}_{wheel} = 0 + \vec{L}_w = \vec{L}_w. After the flip, the wheel's angular momentum is −L⃗w-\vec{L}_w. To conserve total angular momentum, the student and stool must acquire an angular momentum L⃗student′\vec{L}_{student}' such that L⃗total=L⃗student′−L⃗w\vec{L}_{total} = \vec{L}_{student}' - \vec{L}_w. For this to equal the initial momentum L⃗w\vec{L}_w, we must have L⃗student′=2L⃗w\vec{L}_{student}' = 2\vec{L}_w. Thus, the student and stool rotate in the same direction as the wheel's original spin.

Question 14

A comet in an elliptical orbit around the Sun has a speed vAv_A at its farthest point from the Sun (aphelion), a distance rAr_A. What is its speed, vPv_P, at its closest point (perihelion), a distance rPr_P?

  1. vP=vArArPv_P = v_A \sqrt{\frac{r_A}{r_P}}
  2. vP=vArPrAv_P = v_A \frac{r_P}{r_A}
  3. vP=vA(rPrA)2v_P = v_A \left(\frac{r_P}{r_A}\right)^2
  4. vP=vArArPv_P = v_A \frac{r_A}{r_P} (correct answer)
Explanation: The torque on the comet about the Sun is zero, so its angular momentum is conserved. At aphelion and perihelion, the velocity vector is perpendicular to the radius vector. The angular momentum magnitude is L=rmvL = rmv. Conservation of angular momentum implies LA=LPL_A = L_P, so mrAvA=mrPvPm r_A v_A = m r_P v_P. Solving for vPv_P gives vP=vA(rA/rP)v_P = v_A (r_A / r_P).

Question 15

A solid cylinder is rotating freely about its symmetry axis. The cylinder suddenly breaks into two equal halves that fly apart tangentially. Which statement correctly describes the total angular momentum of the two-half-cylinder system about the original axis of rotation, immediately after the separation?

  1. The total angular momentum is conserved because the forces causing the separation are internal to the system. (correct answer)
  2. The total angular momentum decreases because the rotational inertia of the system has changed.
  3. The total angular momentum increases because the pieces gain significant translational kinetic energy.
  4. The total angular momentum becomes zero because the two halves have equal and opposite linear momenta.
Explanation: The forces that cause the cylinder to break apart are internal forces within the original object. These internal forces cannot produce a net external torque on the system. Therefore, the total angular momentum of the system (now consisting of two pieces) about the original axis of rotation must be conserved and remains equal to its value just before the break.

Question 16

A spinning top with a large angular momentum L⃗\vec{L} pointing nearly vertically is subject to a small torque τ⃗\vec{\tau} due to gravity, which points horizontally. This torque causes the top to precess. The change in the angular momentum vector, ΔL⃗\Delta\vec{L}, over a short time interval Δt\Delta t is best described as:

  1. a vector parallel to L⃗\vec{L}, causing the top to spin faster or slower.
  2. a vector opposite to L⃗\vec{L}, causing the top to slow down and fall.
  3. a vector parallel to τ⃗\vec{\tau}, causing the direction of L⃗\vec{L} to change. (correct answer)
  4. a zero vector, because angular momentum must be conserved in this system.
Explanation: The relationship between torque and angular momentum is τ⃗=dL⃗/dt\vec{\tau} = d\vec{L}/dt, which can be approximated as ΔL⃗≈τ⃗Δt\Delta\vec{L} \approx \vec{\tau} \Delta t. This means the change in the angular momentum vector, ΔL⃗\Delta\vec{L}, is in the same direction as the torque vector τ⃗\vec{\tau}. Since the torque is horizontal, the change in angular momentum is horizontal. Adding this small horizontal change to the large vertical original momentum causes the momentum vector's direction to shift, which is the phenomenon of precession.

Question 17

A child pushes tangentially on the edge of a stationary playground merry-go-round, causing it to spin. Considering the merry-go-round as the system, its angular momentum is not conserved during this process. The most direct reason for this is that:

  1. the child is not part of the defined system and applies a net external torque. (correct answer)
  2. the final kinetic energy of the merry-go-round is greater than its initial kinetic energy.
  3. the merry-go-round's angular velocity increases, which violates conservation principles.
  4. the force applied by the child causes both rotational and translational motion.
Explanation: Angular momentum of a system is conserved only when the net external torque on the system is zero. In this case, the system is defined as just the merry-go-round. The force from the child is external to this system, and since it is applied tangentially at a distance from the axis, it creates a net external torque. This external torque changes the angular momentum of the merry-go-round. If the system were defined as the child plus the merry-go-round (on frictionless ice, for example), the total angular momentum would be conserved.

Question 18

A uniform disk with rotational inertia II is spinning freely with angular velocity ω\omega. A uniform ring with the same mass and radius as the disk, having a rotational inertia of 2I2I, is dropped coaxially onto the spinning disk. If there are no external torques, what is the final common angular velocity of the combined system?

  1. ω/3\omega/3 (correct answer)
  2. ω/2\omega/2
  3. 2ω/32\omega/3
  4. 3ω/23\omega/2
Explanation: The total angular momentum of the system is conserved because there are no external torques. The initial angular momentum is Li=Idiskω=IωL_i = I_{disk}\omega = I\omega. The final rotational inertia of the combined system is If=Idisk+Iring=I+2I=3II_f = I_{disk} + I_{ring} = I + 2I = 3I. The final angular momentum is Lf=Ifωf=(3I)ωfL_f = I_f \omega_f = (3I)\omega_f. Setting Li=LfL_i = L_f gives Iω=(3I)ωfI\omega = (3I)\omega_f. Solving for the final angular velocity yields ωf=ω/3\omega_f = \omega/3.

Question 19

A sticky lump of clay and a high-bouncing rubber ball have identical masses and are thrown with identical horizontal velocities toward the edge of a hinged door that is initially at rest. The clay sticks to the door. The ball bounces straight back with nearly its original speed. Which object causes the door to acquire a greater angular velocity?

  1. The clay, because it transfers its entire kinetic energy to the door during the collision.
  2. The rubber ball, because it has a larger change in angular momentum, imparting a larger angular impulse. (correct answer)
  3. Both cause the same angular velocity because they have the same initial mass and velocity.
  4. It cannot be determined without the mass of the door, as a larger mass door would spin slower.
Explanation: The change in the door's angular momentum (and thus its final angular velocity) is equal to the angular impulse imparted to it. This angular impulse is equal and opposite to the change in the projectile's angular momentum. The clay's angular momentum changes from LiL_i to 0, a change of magnitude ∣Li∣|L_i|. The rubber ball's angular momentum changes from LiL_i to approximately −Li-L_i, a change of magnitude ∣−Li−Li∣=2∣Li∣|-L_i - L_i| = 2|L_i|. Since the ball experiences a larger change in angular momentum, it imparts a larger angular impulse to the door, causing a greater final angular velocity.

Question 20

A uniform solid sphere of mass MM and radius RR is initially at rest on a frictionless horizontal surface. A constant tangential force FF is applied to the equator of the sphere for a time interval Δt\Delta t. What is the magnitude of the sphere's angular momentum about its center of mass after this interval?

  1. FRΔtF R \Delta t (correct answer)
  2. FΔtF \Delta t
  3. FRΔt/(25MR2)F R \Delta t / (\frac{2}{5}MR^2)
  4. F/RF / R
Explanation: The tangential force FF creates a torque about the center of mass of magnitude τ=RF\tau = R F. The angular impulse delivered to the sphere is the product of the constant torque and the time interval, which is τΔt=FRΔt\tau \Delta t = F R \Delta t. According to the angular impulse-momentum theorem, the angular impulse is equal to the change in angular momentum, ΔL\Delta L. Since the sphere starts from rest (Li=0L_i = 0), the final angular momentum is Lf=ΔL=FRΔtL_f = \Delta L = F R \Delta t.