Conservation of Angular Momentum

AP Physics 1· AP Physics 1 CED — Rotational Motion· 14 min read

1. What Is Conservation of Angular Momentum?★★☆☆☆⏱ 3 min

Angular momentum ( $L$ ) is the rotational analog of linear momentum. 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 contributes 14-18% of the total AP exam score.

📘 Definition

Conservation of Angular Momentum

$L_{ini t ia l} = L_{f ina l}$

Total angular momentum of a system is constant when net external torque on the system equals zero

Example:

A spinning ice skater conserves angular momentum when pulling their arms inward

For rigid rotating systems, angular momentum is given by $L = I ω$ , where $I$ is moment of inertia about the rotation axis and $ω$ is angular velocity. For translating point masses, angular momentum is $L = m v r_{⊥}$ , where $r_{⊥}$ 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 ideal for collision and variable-mass rotation problems.

2. Core Condition for Angular Momentum Conservation★★☆☆☆⏱ 3 min

The defining condition for angular momentum conservation is net external torque $τ_{n e t, e x t} = 0$ , not net external force equal to zero. This is the most commonly confused rule: 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.

🔬 Derivation

Goal:

Derive conservation of angular momentum from rotational Newton's second law

Starting from:

Rotational Newton's second law: $τ_{n e t} = \frac{Δ L}{Δ t}$

1
If net external torque is zero, we substitute:
2
$0 = \frac{\Delta L}{\Delta t}$
3
This means the total change in angular momentum $Δ L = 0$ , so angular momentum is constant.

Result:

For rigid rotating systems, this simplifies to the core conservation equation: $I_{i} ω_{i} = I_{f} ω_{f}$

📐 Worked Example

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 $5.0 k g \cdot m^{2}$ . She pulls her arms inward, reducing her moment of inertia to $2.5 k g \cdot m^{2}$ . Friction torque from the ice is negligible. What is her new angular speed?

1
1. Define the system as the skater + Earth, with rotation axis along the skater's spin line.
2
1. Confirm the conservation condition: friction torque is negligible, gravity and normal force exert no torque about the spin axis, so $τ_{n e t, e x t} = 0$ , meaning angular momentum is conserved.
3
1. Write the core conservation equation:
4
$I_i \omega_i = I_f \omega_f$
5
1. Solve for final angular speed $ω_{f}$ :
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$\omega_f = \frac{I_i \omega_i}{I_f} = \frac{(5.0 \, kg \cdot m^2)(1.2 \, rad/s)}{2.5 \, kg \cdot m^2} = 2.4 \, rad/s$

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★★★☆☆⏱ 4 min

Changing moment of inertia occurs when mass moves radially (closer to or farther from) the axis of rotation, changing the total $I$ of the system without adding any external torque. This is the most frequently tested application on the AP exam, appearing in both multiple-choice and free-response questions. Because $L = I ω$ is constant, a decrease in $I$ causes a proportional increase in $ω$ (and vice versa). A critical point to remember: kinetic energy is not conserved in these problems, because the force moving the mass radially does internal work on the system. Pulling mass inward increases rotational kinetic energy, while letting mass move outward decreases it.

📐 Worked Example

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

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★★★★☆⏱ 4 min

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 that hits and sticks to a pivoted object, we calculate the initial angular momentum of the point mass as $L = m v r_{⊥}$ , where $r_{⊥}$ is the perpendicular distance from the pivot to the mass's line of motion.

📐 Worked Example

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 $I = \frac{1}{3} M L^{2}$ .)

1
1. System = bullet + rod, axis at the pivot. The pivot force exerts no torque about the pivot, so angular momentum is conserved.
2
1. Calculate initial total angular momentum: the rod is stationary, so only the bullet contributes:
3
$L_i = mvr = (0.05 kg)(200 m/s)(1.0 m) = 10 \, kg \cdot m^2/s$
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1. Calculate final total moment of inertia:
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$I_{rod} = \frac{1}{3}ML^2 = 0.5 \, kg \cdot m^2, \quad I_{bullet} = mr^2 = 0.05 \, kg \cdot m^2$
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Total final $I_{f} = 0.5 + 0.05 = 0.55 k g \cdot m^{2}$ .
7
1. Solve for $ω_{f}$ :
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$\omega_f = \frac{L_i}{I_f} = \frac{10}{0.55} \approx 18.2 \, rad/s$

✓ Quick check

Test your understanding of kinetic energy change when moment of inertia changes:

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.
Reveal answer
C —
From conservation of angular momentum: $I_{i} ω_{i} = I_{f} ω_{f}$ , so if $I_{f} = I_{i} /3$ , then $ω_{f} = 3 ω_{i}$ . Using $K E = \frac{L ^{2}}{2 I}$ , since $L$ is constant, $K E_{f} = 3 K E_{i}$ , so C is correct.

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.

5. Common Pitfalls

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 $L = m v r_{⊥}$ .

6. Quick Reference Cheatsheet

Category	Formula	Notes
Conservation Condition	$τ_{n e t, e x t} = 0 ⟹ L_{ini t ia l} = L_{f ina l}$	Does not require net external force = 0 or KE conservation
Angular Momentum (Rigid Body)	$L = I ω$	$I$ = total moment of inertia about rotation axis
Angular Momentum (Point Mass)	$L = m v r_{⊥}$	$r_{⊥}$ = perpendicular distance from axis to mass path
Changing $I$ Conservation	$I_{i} ω_{i} = I_{f} ω_{f}$	Applies for radial mass movement; KE not conserved
Moment of Inertia (Point Mass)	$I = m r^{2}$	Point mass at distance $r$ from axis
Rotational Kinetic Energy	$K E_{r o t} = \frac{L ^{2}}{2 I}$	Useful for relating KE to constant $L$
Moment of Inertia (Rod at End)	$I = \frac{1}{3} M L^{2}$	Uniform rod pivoted at one end
Moment of Inertia (Solid Disk)	$I = \frac{1}{2} M R^{2}$	Uniform solid disk rotating about central axis

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Ice skater changing moment of inertia
2022 · FRQ
Rotational collision with merry-go-round

What's Next

Mastering conservation of angular momentum is a critical prerequisite for the remaining topics in Unit 6 Rotational Motion, including rotational kinetic energy in multi-concept problems and rolling without slipping, which often combines angular momentum with linear motion concepts. This topic also contributes to 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 free-response questions 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.