Conservation of Angular Momentum - AP Physics 1

A student stands on a frictionless turntable holding two 2.0 kg dumbbells at arm's length, rotating at angular speed $ omega_0$. External torque about the vertical axis is negligible. The student pulls the dumbbells close to the body, decreasing the system's rotational inertia to one-fourth its initial value. Which statement best describes the final angular speed?
Answer: It becomes $4\omega_0$ because angular momentum is conserved. This problem tests conservation of angular momentum. When external torque is negligible, the system's angular momentum L = Iω remains constant. Since the student pulls the dumbbells inward, reducing rotational inertia to I₀/4, we can apply L₀ = $L_f$ , giving I₀ω₀ = (I₀/4)ω_f, which yields ω_f = 4ω₀. Choice C incorrectly assumes angular speed stays constant when only angular momentum is conserved. To solve angular momentum problems: identify if external torque is negligible, then use L_initial = L_final to find the new angular speed.
A compact star rotates in space with negligible external torque. It slowly expands so its rotational inertia increases by a factor of 4. By what factor does its angular speed change?
Answer: Decreases by a factor of 4 This question tests quantitative application of angular momentum conservation. With negligible external torque, angular momentum L = Iω is conserved for the star. If rotational inertia increases by a factor of 4 (I_final = 4I_initial), then angular speed must decrease by the same factor to keep L constant. From I_initial × ω_initial = 4I_initial × ω_final, we get ω_final = ω_initial/4. The angular speed decreases by a factor of 4, not 2 as choice C suggests. When solving conservation problems, the factors by which I and ω change are reciprocals of each other.
A gymnast performs a somersault in midair after leaving the floor. Air resistance and external torque about the gymnast's center of mass are negligible. The gymnast tucks, reducing moment of inertia about the rotation axis. What happens to the rotation rate?
Answer: It increases because angular momentum is conserved. 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.
A disk on a frictionless axle rotates at $6\ \text{rad/s}$ . External torque on the disk–clay system is negligible. A small lump of clay is gently dropped onto the outer rim and sticks, increasing the total moment of inertia from $I_0$ to $1.5I_0$ . The disk and clay rotate together afterward. What is the final angular speed of the disk–clay system?
Answer: $4\ \text{rad/s}$ because angular momentum is conserved This question assesses understanding of angular momentum conservation when mass is added to a rotating system. The clay is dropped with zero initial angular momentum, and with negligible external torque, the system's total angular momentum is conserved. Initial angular momentum is I₀ × 6 rad/s, and after sticking, the moment of inertia becomes 1.5I₀, so ω = (I₀ × 6) / (1.5I₀) = 4 rad/s. The rotational speed decreases because the added mass increases inertia without adding momentum. Choice B is a distractor as it incorrectly states angular speed remains unchanged, confusing no torque with constant speed instead of constant momentum. A key strategy is to calculate initial L and set it equal to final Iω for systems with changing inertia.
A spacecraft rotates in deep space with negligible external torque. It reconfigures by extending two booms, increasing its rotational inertia. Which statement about its rotational kinetic energy is necessarily true?
Answer: It must decrease because angular momentum is conserved This problem examines energy changes during angular momentum conservation. When external torque is negligible, angular momentum L = Iω is conserved, but rotational kinetic energy K = ½Iω² is not necessarily conserved. As the booms extend, rotational inertia I increases, so angular speed ω must decrease to keep L constant. Since K = ½Iω² = L²/(2I), and L is constant while I increases, the kinetic energy must decrease. Choice B incorrectly assumes kinetic energy is conserved like angular momentum. The key insight is that extending booms requires work against centrifugal effects, converting rotational kinetic energy to other forms while conserving angular momentum.
Two small masses connected by a light rod rotate about the rod's midpoint on a frictionless pivot. External torque about the pivot is negligible. The masses are pulled inward along the rod so their distance from the pivot is halved. How does the system's angular speed change?
Answer: It quadruples because the rotational inertia decreases by a factor of 4. This problem tests conservation of angular momentum with point masses. For two masses at distance r from the pivot, rotational inertia is I = 2mr². When the distance halves to r/2, the new rotational inertia becomes $I_f$ = 2m(r/2)² = I₀/4. With negligible external torque, angular momentum is conserved: I₀ω₀ = (I₀/4)ω_f, giving ω_f = 4ω₀. Choice A only considers the radius change but misses that rotational inertia depends on r². Remember: for point masses, rotational inertia scales with the square of the distance, so halving distance reduces I by a factor of 4.
A rotating disk initially has rotational inertia $I_0$ and angular speed $ omega_0$. External torque is negligible. A second identical disk is gently lowered onto it and they stick, so the total rotational inertia becomes $2I_0$ . What is the final angular speed of the two-disk system?
Answer: $\omega_0/2$ because angular momentum is conserved. This problem tests conservation of angular momentum in a rotational collision. Initially, only the first disk rotates with angular momentum L₀ = I₀ω₀. When the second identical disk is added, the total rotational inertia becomes 2I₀, but the total angular momentum remains L₀ since external torque is negligible. Applying conservation: I₀ω₀ = 2I₀ω_f gives ω_f = ω₀/2. Choice A incorrectly assumes the added disk contributes angular momentum, but a disk lowered gently (not rotating initially) has zero angular momentum. When combining rotating and non-rotating objects, only the initially rotating object contributes to the system's angular momentum.
A student sits on a low-friction rotating stool holding two identical masses at arm's length. External torque on the student–stool–masses system is negligible. The student pulls the masses inward so the system's rotational inertia decreases from $I_0$ to $\tfrac{3}{4}I_0$ . What happens to the angular speed?
Answer: It increases to $\tfrac{4}{3}\omega_0$ to conserve angular momentum This problem tests conservation of angular momentum with changing rotational inertia. When external torque is negligible, angular momentum L = Iω remains constant. Initially, L = I₀ω₀, and after pulling masses inward, L = (¾I₀)ω_new. Setting these equal: I₀ω₀ = (¾I₀)ω_new, solving gives ω_new = (4/3)ω₀. Choice A incorrectly assumes angular speed decreases with radius, but fails to account for the full conservation principle. The key insight: decreasing rotational inertia requires increasing angular speed to maintain constant angular momentum.
A turntable rotates at $\omega_0$ with a small mass attached near its rim. External torque is negligible. The mass is moved to half its original distance from the axis, decreasing total rotational inertia to $0.90I_0$ . What is the new angular speed?
Answer: $\tfrac{1}{0.90}\,\omega_0$ This problem tests conservation of angular momentum with repositioned mass. When external torque is negligible, angular momentum L = Iω is conserved throughout the motion. Initially, L = I₀ω₀, and after moving the mass, L = (0.90I₀)ω_new. Setting these equal: I₀ω₀ = (0.90I₀)ω_new, which gives ω_new = ω₀/0.90 = (1/0.90)ω₀. Choice A incorrectly multiplies by 0.90 instead of dividing, reversing the relationship between I and ω. When rotational inertia decreases by a factor, angular speed increases by the reciprocal of that factor.
A bicycle wheel spins at $_0$ on a frictionless axle; external torque is negligible. A small mass is attached at the rim, changing the wheel's moment of inertia from $I$ to $1.2I$ without slipping. What is $_f$ ?
Answer: $\omega_f=\omega_0/1.2$ This question tests conservation of angular momentum when adding mass to a spinning wheel with negligible external torque. Angular momentum is conserved due to the frictionless axle and negligible torque, so I ω₀ = 1.2 I ω_f. Attaching the mass at the rim increases the moment of inertia to 1.2 times initial, decreasing the angular speed inversely. Thus, ω_f = ω₀ / 1.2 as the wheel slows slightly. Choice B (ω₀) is a distractor for those forgetting that added mass changes I and thus affects ω. For similar problems, compute the I ratio and apply the inverse relationship to find the new ω while ensuring L remains constant.
A platform rotates with angular momentum $L_0$ about a vertical axis. External torque is negligible. A student moves from the center to the edge, increasing the system's moment of inertia from $I$ to $2I$ . What is the new angular momentum?
Answer: $L_0$ 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.
A student spins on a stool with initial angular speed $\omega_0$ . External torque is negligible. The student's rotational inertia changes from $I_0$ to $I_f$ . Which expression gives the final angular speed $\omega_f$ ?
Answer: $\omega_f=\omega_0\left(\tfrac{I_0}{I_f}\right)$ 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 = $I_f$ ω_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.
A student on a frictionless rotating platform holds a 3.0 kg mass close to the axis while rotating at $ omega_0$. External torque about the axis is negligible. The student extends the mass so the system's rotational inertia increases from $I_0$ to $1.25I_0$ . What is the platform's final angular speed?
Answer: $\omega_0/1.25$ because angular momentum is conserved. This problem demonstrates conservation of angular momentum when configuration changes. With negligible external torque about the rotation axis, the system's angular momentum L = Iω remains constant as the student extends the mass. The rotational inertia increases from I₀ to 1.25I₀, so applying conservation gives I₀ω₀ = 1.25I₀ω_f, yielding ω_f = ω₀/1.25 = 0.8ω₀. Choice A incorrectly assumes larger rotational inertia means larger angular speed, but they are inversely related when angular momentum is conserved. Strategy: when rotational inertia increases by a factor, angular speed decreases by the same factor.
A solid cylinder rotates on a frictionless axle at angular speed $ omega_0$. External torque is negligible. Clay is gently pressed onto the rim and sticks, increasing the total rotational inertia to $1.5I_0$ . What is the cylinder–clay system's new angular speed?
Answer: $\omega_0/1.5$ because angular momentum is conserved. This problem involves conservation of angular momentum when mass is added to a rotating system. With negligible external torque, the cylinder's initial angular momentum L₀ = I₀ω₀ equals the final angular momentum of the cylinder-clay system $L_f$ = 1.5I₀ω_f. Solving I₀ω₀ = 1.5I₀ω_f gives ω_f = ω₀/1.5. Choice A incorrectly assumes the clay adds angular momentum, but clay added gently (with no tangential velocity) has zero initial angular momentum. When mass is added to a rotating system without initial angular momentum, the system slows down to conserve total angular momentum.
A figure skater spins upright with angular speed $\omega_0$ while holding two identical masses at arm's length. External torque on the skater–masses system is negligible. The skater pulls the masses inward so the system's moment of inertia decreases from $I_0$ to $\tfrac{3}{4}I_0$ . How does the skater's angular speed change?
Answer: It increases to $\tfrac{4}{3}\omega_0$ because angular momentum is conserved This problem evaluates the conservation of angular momentum during a change in mass distribution. As the skater pulls the masses inward, the moment of inertia decreases to 3/4 of its initial value, and with no external torque, angular momentum L = Iω is conserved. Initial L = I₀ω₀ equals final (3/4 I₀)ω, so ω = (4/3)ω₀, meaning speed increases. The decrease in I requires an increase in ω to maintain constant L. Choice B incorrectly claims angular speed stays constant, mistaking no torque for no change in ω rather than conserved L. Remember, for isolated systems, use conservation laws like L = constant to predict rotational dynamics when I varies.
A ring rotates on a frictionless axle at angular speed $\omega_0$ . External torque about the axle is negligible. The ring's radius is slowly decreased so its moment of inertia becomes $\tfrac{1}{9}$ of the original. How does the angular speed change?
Answer: It becomes $9\omega_0$ because angular momentum is conserved. This problem involves conservation of angular momentum with a dramatic change in moment of inertia. With negligible external torque, the ring's angular momentum L = Iω remains constant as its radius decreases. Initially: L = I₀ω₀. After shrinking, the moment of inertia becomes I_final = I₀/9. Conservation requires I₀ω₀ = (I₀/9)ω_final, giving ω_final = 9ω₀. The moment of inertia decreasing by a factor of 9 causes angular speed to increase by a factor of 9. Choice A incorrectly inverts the relationship—when I decreases, ω must increase to conserve L. The key principle is that angular momentum conservation produces dramatic speed changes when moment of inertia changes significantly.
A student sits on a rotating stool holding a spinning bicycle wheel so its axle is vertical; the stool–student system rotates at $\omega_0$ . External torque about the vertical axis is negligible. The student pulls the wheel closer to the rotation axis, reducing the total moment of inertia about the vertical axis from $I_0$ to $\tfrac{2}{3}I_0$ . What is the final angular speed of the stool–student system?
Answer: $\omega = \tfrac{3}{2}\omega_0$ because angular momentum is conserved 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.
A solid cylinder rotates at angular speed $\omega_0$ on a frictionless axle. External torque on the cylinder–putty system is negligible. A piece of putty is pressed onto the cylinder's surface and sticks, increasing the moment of inertia to $2.5I_0$ . What is the cylinder's final angular speed?
Answer: $\omega = \tfrac{1}{2.5}\omega_0$ because angular momentum is conserved This question assesses angular momentum conservation with added adhering mass. The putty sticks with zero initial angular momentum, and negligible torque conserves L as I₀ω₀. Final I = 2.5I₀, so ω = ω₀ / 2.5, slowing the cylinder. Increased inertia reduces speed to preserve momentum. Choice C mistakenly holds ω constant due to no torque, confusing it with conserved L. To tackle these, always verify initial L and set equal to final Iω for inelastic rotational additions.
A uniform merry-go-round rotates at angular speed $\omega_0$ about a vertical axis. External torque on the merry-go-round–child system is negligible. A child initially at the center walks to the edge, increasing the system's moment of inertia from $I_0$ to $2I_0$ . What happens to the angular speed of the system?
Answer: It becomes $\tfrac{1}{2}\omega_0$ because angular momentum is conserved This question examines conservation of angular momentum as mass redistributes in a rotating system. When the child moves to the edge, the moment of inertia doubles to 2I₀, and negligible external torque ensures angular momentum is conserved. Initial L = I₀ω₀ = final 2I₀ω, so ω = (1/2)ω₀, decreasing the speed. The increase in I slows the rotation to keep L constant. Choice B is misleading, suggesting speed stays constant due to no torque, but no torque conserves L, not ω. Apply the principle that in torque-free systems, changes in I inversely affect ω to maintain L.
A rotating stool and student have initial moment of inertia $I_0$ and angular speed $\omega_0$ . External torque about the axis is negligible. The student extends arms, increasing moment of inertia to $4I_0$ . What is the final angular speed?
Answer: $\tfrac{1}{4}\omega_0$ 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.

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