Conservation of Angular Momentum — AP Physics 1 Study Guide

For: AP Physics 1 candidates sitting AP Physics 1.

Covers: The net torque condition for angular momentum conservation, conservation with changing moment of inertia, angular momentum in rotational collisions, core formulas (L = I\omega) and (I_i\omega_i = I_f\omega_f), and problem-solving strategies for AP-style questions.

You should already know: Torque as the rate of change of angular momentum, moment of inertia calculation for rigid bodies and point masses, basic rotational kinematics.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics 1 style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Conservation of Angular Momentum?

Angular momentum (L) is the rotational analog of linear momentum, and conservation of angular momentum states that the total angular momentum of a system remains constant if and only if the net external torque acting on the system is zero. This is one of three fundamental conservation laws tested in AP Physics 1, and makes up roughly 15-20% of the Unit 6 Rotational Motion weighting, which itself contributes 14-18% of the total AP Physics 1 exam score. This topic appears in both multiple-choice (MCQ) and free-response (FRQ) sections, often combined with energy or collision concepts to test multi-concept reasoning.

Notation conventions for AP Physics 1: we use (L = I\omega) for rigid rotating systems, where (I) is moment of inertia about the axis of rotation and (\omega) is angular velocity. For translating point masses, (L = mvr_\perp), where (r_\perp) is the perpendicular distance from the axis to the mass’s line of motion. Unlike energy conservation, angular momentum is conserved even when internal forces do work to change the system’s kinetic energy, making it especially useful for collision and variable-mass-distribution rotation problems.

2. Core Condition for Angular Momentum Conservation

The defining condition for angular momentum conservation is net external torque (\tau_{net,ext} = 0), not net external force equal to zero. This is the most commonly confused rule for this topic: many students incorrectly transfer the linear momentum condition (net external force = zero) to angular momentum. It is entirely possible for net external force to be non-zero but net external torque to be zero, meaning angular momentum is still conserved. For example, a spinning figure skater on ice: gravity and normal force cancel their force contributions, and both exert zero torque about the spin axis, so angular momentum is conserved even as the skater pulls their arms inward.

We derive the conservation law directly from rotational Newton’s second law: $${\tau }_{net}=\frac{\Delta L}{\Delta t}$$ If (\tau_{net,ext} = 0), then (\Delta L = 0), so (L_{initial} = L_{final}), or for rigid systems: $$I_i{\omega }_i=I_f{\omega }_f$$ The key to applying this rule correctly is properly defining the system, which lets you separate internal and external torques: internal torques from forces between objects inside the system always sum to zero by Newton’s third law, so they never change total angular momentum.

Worked Example

Problem: A 62 kg ice skater spins with an initial angular speed of 1.2 rad/s when her moment of inertia about the spin axis is (5.0 , kg \cdot m^2). She pulls her arms inward, reducing her moment of inertia to (2.5 , kg \cdot m^2). Friction torque from the ice is negligible. What is her new angular speed?

1. Define the system as the skater + Earth, with the rotation axis along the skater’s spin line.
2. Confirm the conservation condition: friction torque is negligible, and gravity/normal force exert no torque about the spin axis, so (\tau_{net,ext} = 0), so angular momentum is conserved.
3. Write the core conservation equation: (I_i \omega_i = I_f \omega_f).
4. Solve for (\omega_f): $${\omega }_f=\frac{I_i{\omega }_i}{I_f}=\frac{(5.0kg\cdot m^2)(1.2rad/s)}{2.5kg\cdot m^2}=2.4rad/s$$

Exam tip: Always check the torque condition first, not the force condition, before applying angular momentum conservation. A system can have non-zero net external force but still have zero net external torque.

3. Conservation with Changing Moment of Inertia

Changing moment of inertia occurs when mass moves radially (closer to or farther from) the axis of rotation, which changes the total (I) of the system without adding any external torque. This is the most frequently tested application of angular momentum conservation on the AP exam, appearing in both conceptual MCQs and quantitative FRQs.

Because (L = I\omega) is constant, a decrease in (I) causes a proportional increase in (\omega) (and vice versa). This is the physics behind the classic “figure skater effect”, spinning divers, and merry-go-round problems where a person walks radially across the rotating platform. A critical point to remember: kinetic energy is not conserved in these problems. The force that moves the mass radially does internal work on the system, so kinetic energy changes: pulling mass inward increases rotational KE, while letting mass move outward decreases it.

Worked Example

Problem: A 25 kg child stands at the edge of a stationary merry-go-round with moment of inertia (1000 , kg \cdot m^2) and radius 2.0 m. The merry-go-round is initially spinning at 0.8 rad/s when the child is at the edge. The child walks to the center of the merry-go-round. The axle has negligible friction. What is the final angular speed of the system?

1. System = child + merry-go-round, axis at the axle. No external torque, so angular momentum is conserved.
2. Calculate initial total moment of inertia: the child acts as a point mass, so (I_{child,i} = mr^2 = 25(2.0)^2 = 100 , kg \cdot m^2). Total (I_i = 1000 + 100 = 1100 , kg \cdot m^2).
3. Calculate final total moment of inertia: the child is at (r=0) at the center, so (I_{child,f} = 0). Total (I_f = 1000 + 0 = 1000 , kg \cdot m^2).
4. Solve for (\omega_f): $${\omega }_f=\frac{I_i{\omega }_i}{I_f}=\frac{(1100)(0.8)}{1000}=0.88rad/s$$

Exam tip: Always sum the moment of inertia for every object in the system, not just the moving mass. It is easy to forget to add the moment of inertia of the rigid rotating structure (like the merry-go-round itself).

4. Angular Momentum Conservation in Rotational Collisions

Rotational collisions (between a translating object and a pivoted rotating rigid body) are another common AP problem type. In these problems, angular momentum conservation is always the right approach, because the fixed pivot exerts no torque about the pivot axis, so net external torque is zero even though the pivot exerts a non-zero external force (meaning linear momentum is not conserved).

For a translating point mass (e.g., a clay ball, bullet, or person jumping onto a merry-go-round) that hits and sticks to a pivoted object, we calculate the initial angular momentum of the point mass as (L = mvr_\perp), where (r_\perp) is the perpendicular distance from the pivot to the mass’s line of motion. If the mass hits the edge of the pivoted object, (r_\perp) equals the radius/length of the object. After collision, we calculate the total moment of inertia of the combined system, then use (L_i = L_f) to solve for final angular speed.

Worked Example

Problem: A uniform solid rod of mass 1.5 kg and length 1.0 m is pivoted at one end and is initially stationary. A 0.05 kg bullet moving at 200 m/s hits the free end of the rod and sticks. What is the angular speed of the rod after the collision? (Moment of inertia of a rod about one end is (I = \frac{1}{3}ML^2)).

1. System = bullet + rod, axis at the pivot. The pivot force exerts no torque about the pivot, so angular momentum is conserved.
2. Calculate initial total angular momentum: the rod is stationary, so only the bullet contributes: (L_i = mvr = (0.05 kg)(200 m/s)(1.0 m) = 10 , kg \cdot m^2/s).
3. Calculate final total moment of inertia: rod (I_{rod} = \frac{1}{3}ML^2 = \frac{1}{3}(1.5 kg)(1.0 m)^2 = 0.5 , kg \cdot m^2). Bullet (I_{bullet} = mr^2 = (0.05 kg)(1.0 m)^2 = 0.05 , kg \cdot m^2). Total (I_f = 0.5 + 0.05 = 0.55 , kg \cdot m^2).
4. Solve for (\omega_f): $${\omega }_f=\frac{L_i}{I_f}=\frac{10}{0.55}\approx 18.2rad/s$$

Exam tip: Never use linear momentum conservation for collisions with a fixed pivoted object. The external force from the pivot means linear momentum is not conserved, but angular momentum about the pivot always is.

Common Pitfalls (and how to avoid them)

• Wrong move: Applying angular momentum conservation to a falling rotating stick, claiming angular momentum is constant as it rotates from horizontal to vertical. Why: Students forget gravity exerts a non-zero torque about the pivot, so net external torque is not zero. Correct move: Always explicitly check for zero net external torque about your chosen axis before writing (L_i = L_f).
• Wrong move: Forgetting to include the moment of inertia of a fixed rotating structure (e.g., the merry-go-round) and only adding the moving mass’s (I) to the total. Why: Students focus on the changing mass position and overlook the constant contribution of the rigid body. Correct move: Always write total (I) as the sum of (I) for every object in the system before plugging into the conservation equation.
• Wrong move: Assuming kinetic energy is conserved when moment of inertia changes or in inelastic rotational collisions. Why: Students assume all conservation laws apply at the same time, forgetting internal work or collision energy loss. Correct move: Only apply kinetic energy conservation if the problem explicitly states the collision is elastic or no internal work is done.
• Wrong move: Using linear momentum conservation for a collision with a fixed pivoted object. Why: Students transfer their knowledge of linear collisions to rotational problems incorrectly. Correct move: For any collision with a fixed pivot, use angular momentum conservation about the pivot axis.
• Wrong move: Using the wrong (r) when calculating angular momentum of a translating point mass. Why: Students confuse the object’s own radius with the perpendicular distance from the pivot to the mass’s path. Correct move: Always draw the pivot and the mass’s path, then measure the perpendicular distance to get the correct (r) for (L = mvr).

Practice Questions (AP Physics 1 Style)

Question 1 (Multiple Choice)

A spinning ice skater pulls their arms inward, decreasing their moment of inertia by a factor of 3, with angular momentum conserved. Which of the following correctly describes the change in rotational kinetic energy? A) Rotational kinetic energy remains constant, because angular momentum is conserved. B) Rotational kinetic energy decreases by a factor of 3, because moment of inertia decreases. C) Rotational kinetic energy triples, because angular speed triples. D) Rotational kinetic energy increases by a factor of 9, because angular speed increases by a factor of 9.

Worked Solution: From conservation of angular momentum: (I_i \omega_i = I_f \omega_f). If (I_f = I_i/3), then (\omega_f = 3\omega_i). Rotational kinetic energy is (KE = \frac{1}{2}I\omega^2). Substituting (L = I\omega), we get (KE = \frac{L^2}{2I}). Since (L) is constant, (KE_f = \frac{L^2}{2(I_i/3)} = 3 \times \frac{L^2}{2I_i} = 3 KE_i). So rotational kinetic energy triples. The correct answer is C.

Question 2 (Free Response)

A uniform solid disk merry-go-round has mass 120 kg and radius 2.0 m, pivoted without friction about its central axis. It is initially spinning at 0.4 rad/s. A 60 kg student runs tangentially at 3.0 m/s, in the opposite direction of the merry-go-round’s spin, and jumps onto the edge. (a) Explain why angular momentum is conserved for this system, but linear momentum is not. (b) Calculate the angular speed of the merry-go-round after the student jumps on. (c) Is the total kinetic energy after the jump greater than, less than, or equal to the total kinetic energy before the jump? Justify your answer.

Worked Solution: (a) Net external torque about the pivot is zero, because gravity and the pivot force exert no torque about the central axis, so angular momentum is conserved. Linear momentum is not conserved because the fixed pivot exerts an external net force on the system, so the condition for linear momentum conservation is not met.

(b) First, calculate the merry-go-round’s moment of inertia: (I_m = \frac{1}{2}MR^2 = 0.5(120)(2.0)^2 = 240 , kg \cdot m^2). Take the merry-go-round’s initial spin direction as positive: initial angular momentum of the merry-go-round is (L_m = I_m \omega_i = (240)(0.4) = 96 , kg \cdot m^2/s). The student moves opposite the spin, so their angular momentum is negative: (L_s = -m v R = -(60)(3.0)(2.0) = -360 , kg \cdot m^2/s). Total initial (L_i = 96 - 360 = -264 , kg \cdot m^2/s). Final total moment of inertia: (I_f = I_m + mR^2 = 240 + (60)(2^2) = 480 , kg \cdot m^2). By conservation: (\omega_f = L_i/I_f = -264/480 = -0.55 , rad/s). The magnitude of the angular speed is (0.55 , rad/s), spinning in the direction the student was running.

(c) The total kinetic energy after the jump is less than before. This is an inelastic collision where the student sticks to the merry-go-round, so internal friction does non-conservative work, converting some kinetic energy to thermal energy. Calculation confirms: initial total KE = 25.92 J + 270 J = 295.92 J, final KE = 0.5(480)(0.55)^2 = 72.6 J < 295.92 J.

Question 3 (Application / Real-World Style)

A competitive diver of mass 70 kg leaves a 10 m platform with an initial angular speed of 0.15 rev/s when fully stretched, with a moment of inertia of (14 , kg \cdot m^2) about their center of mass. To complete a double somersault, the diver tucks, reducing their moment of inertia to (2.8 , kg \cdot m^2). They have 1.6 seconds between leaving the board and entering the water, with no external torque acting during the jump. Can the diver complete 2 full somersaults in this time?

Worked Solution: Net external torque is zero, so angular momentum is conserved: (I_i \omega_i = I_f \omega_f). Solve for final angular speed: (\omega_f = \frac{I_i \omega_i}{I_f} = \frac{(14 , kg \cdot m^2)(0.15 , rev/s)}{2.8 , kg \cdot m^2} = 0.75 , rev/s). Total number of revolutions is (N = \omega_f \Delta t = (0.75 , rev/s)(1.6 s) = 1.2) revolutions. The diver can only complete 1 full somersault, so they cannot complete 2 full somersaults in the available time. In context, this means the diver needs to tuck even tighter to reduce their moment of inertia further to hit the 2 somersault goal.

Quick Reference Cheatsheet

What's Next

Mastering conservation of angular momentum is a critical prerequisite for the remaining topics in Unit 6 Rotational Motion: rotational kinetic energy in multi-concept problems, and rolling without slipping, which often combines angular momentum with linear motion concepts. This topic also feeds into the broader theme of conservation laws across AP Physics 1, which makes up nearly half of the total exam score. Without being able to correctly identify when angular momentum is conserved and apply the formula correctly, you will struggle with multi-concept FRQs that combine rotation, collisions, and energy, which are very common on the AP exam. Next, you will extend this concept to rolling motion and combined rotational-translational dynamics.

• Rotational Kinetic Energy
• Torque and Rotational Newton's Second Law
• Rolling Without Slipping
• Conservation of Energy