AP Physics 1 Quiz: Conservation Of Angular Momentum

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 1.

How to use this quiz

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.

Question 1

1. $2L_0$, because $I$ doubled
2. $L_0/2$, because $\omega$ decreases
3. $0$, because the student did internal work
4. $L_0$ (correct answer)

Explanation: This question evaluates whether angular momentum is conserved when internal changes occur with negligible external torque. Angular momentum remains L₀ because external torque is negligible, unaffected by the student's movement increasing I from I to 2I. The internal repositioning changes I but not the total L of the isolated system. The new angular momentum is still L₀, as conservation holds. Choice A (2L₀) could result from mistakenly thinking L increases with I, ignoring that ω decreases. A transferable strategy is to recognize that internal forces cannot change total L; always confirm external torque is zero before applying conservation.

Question 2

1. It increases because the dumbbells have more leverage
2. It stays constant because external torque is negligible
3. It decreases to keep angular momentum constant (correct answer)
4. It becomes zero because inertia increases

Explanation: This problem demonstrates conservation of angular momentum in everyday situations. When external torque is negligible, angular momentum L = Iω remains constant for the person-stool system. Moving dumbbells farther from the rotation axis increases the system's rotational inertia I because more mass is distributed at larger distances from the axis. Since L must remain constant and I increases, the angular speed ω must decrease proportionally. Choice B incorrectly claims angular speed stays constant, confusing it with angular momentum. The strategy is to recognize that extending mass outward always increases rotational inertia, requiring decreased angular speed to conserve angular momentum.

Question 3

1. $\omega = 2\omega_0$ because the radius halves
2. $\omega = \tfrac{1}{2}\omega_0$ because the masses move inward
3. $\omega = 4\omega_0$ because angular momentum is conserved (correct answer)
4. $\omega = \omega_0$ because negligible torque means constant angular speed

Explanation: This problem tests conservation of angular momentum with changing radial positions. As masses slide inward, the moment of inertia becomes 1/4 of initial (since I proportional to r², and r halves), and no external torque conserves L. Initial L = I₀ω₀ = final (I₀/4)ω, so ω = 4ω₀, speeding up. Reduced I demands higher ω for constant L. Choice D errs by claiming constant ω from negligible torque, but torque absence conserves L, not ω. A transferable tip: For variable radius systems, recall I ∝ r² and use L conservation to predict ω adjustments.

Question 4

1. The student-turntable begins rotating in the original wheel-spin direction to conserve angular momentum.
2. The student-turntable begins rotating opposite the original wheel-spin direction to conserve angular momentum. (correct answer)
3. Nothing changes because flipping the wheel requires torque, so angular momentum is not conserved.
4. The student-turntable stops rotating because net external torque is zero.

Explanation: This problem demonstrates conservation of angular momentum in a system with changing angular momentum directions. Initially, the system (student + turntable + wheel) has angular momentum equal to the wheel's spin angular momentum pointing upward. When the student flips the wheel, its angular momentum reverses to point downward. Since external torque is negligible, total system angular momentum must remain constant (pointing upward). To compensate for the wheel's downward angular momentum, the student-turntable must acquire upward angular momentum by rotating opposite to the wheel's original spin direction. Choice C incorrectly claims that internal torques violate conservation—internal forces cannot change total angular momentum. The strategy is to track angular momentum as a vector quantity and ensure the total remains constant.

Question 5

1. It increases because angular momentum is conserved. (correct answer)
2. It decreases because the gymnast’s muscles apply an internal torque.
3. It stays constant because no external torque acts.
4. It becomes zero because torque is negligible.

Explanation: This problem examines conservation of angular momentum during aerial motion. With negligible air resistance and external torque about the center of mass, the gymnast's angular momentum remains constant throughout the somersault. When tucking reduces the moment of inertia I about the rotation axis, the angular speed ω must increase to maintain constant angular momentum L = Iω. This allows gymnasts to control rotation rate without external forces. Choice B incorrectly claims internal muscle torques affect total angular momentum—internal forces cannot change a system's total angular momentum. The key insight is that athletes can redistribute their mass to change I and thereby control ω while conserving L.

Question 6

1. $\omega_0/4$
2. $4\omega_0$ (correct answer)
3. $\omega_0$
4. $2\omega_0$

Explanation: This question assesses the conservation of angular momentum in a system with negligible external torque. Angular momentum is conserved because no external torque acts on the system about the vertical axis, so the initial angular momentum I₀ω₀ equals the final angular momentum (I₀/4)ω_f. When the student pulls the dumbbells inward, the moment of inertia decreases to one-fourth, causing the angular speed to increase to maintain the same angular momentum. Therefore, solving I₀ω₀ = (I₀/4)ω_f gives ω_f = 4ω₀. A common distractor like choice A (ω₀/4) might result from mistakenly thinking angular speed is directly proportional to moment of inertia instead of inversely. To solve similar problems, remember that when external torque is negligible, angular momentum L = Iω is conserved, and changes in I lead to inverse changes in ω.

Question 7

1. $\omega_0/3$
2. $\sqrt{3}\,\omega_0$
3. $3\omega_0$ (correct answer)
4. $\omega_0$

Explanation: This question tests conservation of angular momentum when the moment of inertia changes in a rotating system with negligible external torque. Angular momentum is conserved since external torque is negligible, so initial $I \omega_0$ equals final $\frac{I}{3} \omega_f$. The rider's position change reduces the moment of inertia to one-third, increasing the angular speed to maintain L. Thus, $\omega_f = 3 \omega_0$ as the system spins faster. Choice A ($\omega_0 / 3$) might arise from inverting the relationship and thinking speed decreases with smaller I. In general, verify no external torque, then use the inverse proportionality of $\omega$ and I to predict rotational changes.

Question 8

1. It doubles
2. It becomes one-fourth
3. It quadruples (correct answer)
4. It stays constant because torque is negligible

Explanation: This question explores conservation of angular momentum during a star's contraction with negligible external torque. Angular momentum is conserved as the star contracts uniformly, so initial I_i ω_i equals final I_f ω_f. For a uniform sphere, I ∝ R², so halving the radius quarters I, requiring angular speed to quadruple to keep L constant. Thus, ω_f = 4 ω_i, meaning it quadruples. Choice D (stays constant) might be chosen by confusing negligible torque with constant speed instead of constant L. Remember to calculate I changes based on geometry and use ω_f / ω_i = I_i / I_f for conserved angular momentum problems.

Question 9

1. Angular momentum about the spin axis (correct answer)
2. Angular speed
3. Rotational kinetic energy
4. Net torque about the axis must increase to keep rotation going

Explanation: This problem tests understanding of which quantities are conserved during rotational motion. With negligible external torque in deep space, angular momentum about the spin axis must remain constant as the masses move inward. While angular momentum L = Iω is conserved, the angular speed ω increases as moment of inertia I decreases. Rotational kinetic energy K = ½Iω² actually increases because work is done pulling the masses inward against centrifugal effects. Choice D incorrectly suggests torque must increase—no external torque is needed to maintain rotation in the absence of friction. The fundamental principle is that angular momentum is the conserved quantity when external torque is negligible, not angular speed or energy.

Question 10

1. $2\omega_0$
2. $\omega_0/2$ (correct answer)
3. $\omega_0$
4. Zero, because sticking implies angular momentum is not conserved

Explanation: This question assesses conservation of angular momentum in a sticking collision between wheels with negligible external torque. Angular momentum remains constant because external torque is negligible, with initial L = I ω₀ + 0 equaling final 2I ω_f for identical wheels. The second wheel adding to the system doubles the moment of inertia, halving the angular speed. Solving yields ω_f = ω₀ / 2 as they rotate together slower. Choice D (zero) is a distractor assuming sticking violates conservation, but momentum is conserved in inelastic rotational collisions. A key strategy is to treat the final system as combined and set L_i = L_f, solving for ω_f using the total I.

Question 11

1. $\omega_f=\omega_0\left(\tfrac{I_f}{I_0}\right)$
2. $\omega_f=\omega_0\left(\tfrac{I_0}{I_f}\right)$ (correct answer)
3. $\omega_f=\omega_0\left(\tfrac{I_f}{I_0}\right)^2$
4. $\omega_f=\omega_0+\tfrac{\tau\Delta t}{I_f}$

Explanation: This question tests the mathematical relationship in angular momentum conservation. When external torque is negligible, angular momentum is conserved: L_initial = L_final. This gives I₀ω₀ = I_fω_f, which rearranges to ω_f = ω₀(I₀/I_f). The final angular speed equals the initial speed multiplied by the ratio of initial to final rotational inertia. Choice A incorrectly inverts this ratio, while choice D incorrectly introduces torque when the problem states it's negligible. The strategy for deriving conservation equations is to set initial and final values of the conserved quantity equal and solve algebraically.

Question 12

1. $3\omega_0$
2. $\tfrac{1}{3}\omega_0$ (correct answer)
3. $\omega_0$
4. Zero, because the added ring provides an opposing torque

Explanation: This question involves conservation of angular momentum during a collision. With negligible external torque, the system's angular momentum L = Iω is conserved. Initially, L = I₀ω₀ (disk alone), and after the ring sticks, L = 3I₀ω_final (disk plus ring). Setting these equal: I₀ω₀ = 3I₀ω_final, yielding ω_final = ω₀/3. The angular speed decreases to one-third its initial value when rotational inertia triples. Choice D incorrectly suggests the ring provides torque, but internal forces cannot change total angular momentum. Remember that angular momentum conservation applies even during collisions when external torque is negligible.

Question 13

1. It increases by a factor of 9 because the panels add torque.
2. It decreases to $\omega_0/9$ because angular momentum is conserved. (correct answer)
3. It stays the same because external torque is negligible.
4. It decreases to $\omega_0/3$ because rotational kinetic energy is conserved.

Explanation: This problem demonstrates conservation of angular momentum when a system's configuration changes. With negligible external torque in deep space, angular momentum L = Iω remains constant as the satellite extends its solar panels. The rotational inertia increases from I₀ to 9I₀, so applying L₀ = L_f gives I₀ω₀ = 9I₀ω_f, resulting in ω_f = ω₀/9. Choice A incorrectly suggests the panels add torque, but extending panels radially produces no torque about the rotation axis. Remember: when rotational inertia increases by a factor, angular speed decreases by the same factor to conserve angular momentum.

Question 14

1. $0.80\,\omega_0$ (correct answer)
2. $1.25\,\omega_0$
3. $\omega_0$
4. Cannot be determined without the torque during impact

Explanation: This problem involves conservation of angular momentum during a collision. With negligible external torque, the system's angular momentum L = Iω is conserved. Initially, L = I₀ω₀, and after clay sticks, L = (1.25I₀)ω_new. Setting equal: I₀ω₀ = (1.25I₀)ω_new, which gives ω_new = ω₀/1.25 = 0.80ω₀. Choice D incorrectly suggests we need torque information, but angular momentum is conserved regardless of internal forces during the collision. For problems involving objects sticking together, always apply L_initial = L_final using the total final rotational inertia.

Question 15

1. $\omega_0$
2. $\tfrac{1}{2}\omega_0$
3. $2\omega_0$ (correct answer)
4. Zero, because the net torque is negligible

Explanation: This problem tests conservation of angular momentum. When external torque is negligible, the total angular momentum L = Iω of the system remains constant. Initially, L = I₀ω₀, and after pulling arms in, L = (½I₀)ω_new. Setting these equal: I₀ω₀ = (½I₀)ω_new, which gives ω_new = 2ω₀. Choice D incorrectly assumes that negligible torque means the angular speed becomes zero, but negligible external torque actually means angular momentum is conserved, not that rotation stops. When solving angular momentum conservation problems, use L_initial = L_final and solve for the unknown quantity.

Question 16

1. $4\omega_0$
2. $\omega_0$
3. $\tfrac{1}{4}\omega_0$ (correct answer)
4. Cannot be determined because extending arms adds external torque.

Explanation: This problem tests conservation of angular momentum when moment of inertia increases. With negligible external torque about the rotation axis, angular momentum L = Iω must remain constant as the student extends arms. Initially: L = I₀ω₀. After extending arms: L = (4I₀)ω_final. Setting these equal: I₀ω₀ = (4I₀)ω_final, yielding ω_final = ω₀/4. The quadrupling of moment of inertia causes angular speed to decrease by a factor of four. Choice D incorrectly claims extending arms adds external torque—arm extension involves only internal forces that cannot change total angular momentum. The strategy is recognizing that I and ω change inversely to maintain constant L.

Question 17

1. $\omega = \tfrac{2}{3}\omega_0$ because the moment of inertia decreased
2. $\omega = \omega_0$ because external torque is negligible
3. $\omega = \tfrac{3}{2}\omega_0$ because angular momentum is conserved (correct answer)
4. $\omega = \tfrac{9}{4}\omega_0$ because rotational kinetic energy is conserved

Explanation: This problem explores angular momentum conservation about a specific axis during reconfiguration. Pulling the wheel closer reduces the moment of inertia to 2/3 I₀, and with negligible external torque about the vertical axis, L is conserved. Initial L = I₀ω₀ = final (2/3 I₀)ω, so ω = (3/2)ω₀, increasing speed. The decreased I requires faster rotation to maintain L. Choice D is wrong as it assumes kinetic energy conservation, but energy isn't conserved here due to internal work. When analyzing axial momentum, ensure torque is negligible and apply Iω constant for varying I.

Question 18

1. $\omega = 0.8\omega_0$ because $\omega$ scales with $I$
2. $\omega = \tfrac{1}{0.8}\omega_0$ because angular momentum is conserved (correct answer)
3. $\omega = \omega_0$ because external torque is negligible
4. $\omega = \tfrac{1}{(0.8)^2}\omega_0$ because rotational kinetic energy is conserved

Explanation: This question evaluates angular momentum conservation during slow radial motion. Pulling the cart inward changes I to 0.8I₀, and negligible external torque conserves L = Iω. Initial L = I₀ω₀ = final 0.8I₀ ω, so ω = ω₀ / 0.8 = (1/0.8)ω₀, increasing speed. Decreased I accelerates rotation to maintain L. Choice C incorrectly assumes constant ω from no torque, overlooking inertia's role in L. In similar cases, calculate the I ratio and inversely apply it to ω via conservation of angular momentum.

Question 19

1. $\tfrac{3}{4}\omega_0$
2. $\tfrac{4}{3}\omega_0$ (correct answer)
3. $\sqrt{\tfrac{3}{4}}\,\omega_0$
4. $\omega_0$ because torque is negligible

Explanation: This question requires applying conservation of angular momentum quantitatively. With negligible external torque, L = Iω remains constant for the platform-student system. Initially, L = I₀ω₀, and after the student moves, L = (3/4)I₀ω_final. Setting these equal: I₀ω₀ = (3/4)I₀ω_final, which gives ω_final = (4/3)ω₀. When rotational inertia decreases to 3/4 of its initial value, angular speed increases by the reciprocal factor 4/3. Choice C incorrectly takes the square root, which has no physical basis here. To solve these problems, use the inverse relationship between I and ω when L is conserved.

Question 20

1. $\omega_0$
2. $3\omega_0$
3. $\omega_0/3$ (correct answer)
4. Zero, because the collision removes angular momentum

Explanation: This question tests the conservation of angular momentum during an inelastic collision with negligible external torque. Angular momentum remains constant because the axle is nearly frictionless and no external torque acts, so initial L = I₀ω₀ equals final L = 3I₀ω_f. The collar sticking to the disk increases the total moment of inertia to three times the initial value, reducing the angular speed accordingly. Thus, ω_f = ω₀/3 as the system slows down to conserve momentum. Choice D (zero) is a distractor that incorrectly assumes the collision dissipates all angular momentum, ignoring conservation. A useful strategy is to apply L_i = L_f and solve for the unknown variable, ensuring you account for changes in total I for combined systems.