AP Physics 1 · Unit 6: Energy & Momentum of Rotating Systems ·  Lesson 6.4

Deep Dive: Conservation of Angular Momentum

🔬 Deep Dive
This is your textbook for this topic. Take your time. Read it more than once.
6.4.A.1Concept

The Conservation Law

When the net external torque on a system is zero, the total angular momentum of that system remains constant:

L_total = constant  when  Στ_ext = 0

Equivalently, L_i = L_f. This is the rotational analog of conservation of linear momentum. The condition is the same structure: replace "no net external force" with "no net external torque." Internal torques between parts of the system always cancel by Newton's Third Law and never affect the total L.

Linear:    ΣF_ext = 0 → p_total = constant
Rotational: Στ_ext = 0 → L_total = constant
🔑Before applying L_i = L_f, always verify the zero-torque condition. State it explicitly in any AP FRQ response: "Since the net external torque on the system is zero, angular momentum is conserved."
6.4.A.2Math

Shape-Changing Systems — The Skater Problem

For a nonrigid system that changes shape while isolated (no external torque), L is constant but I can change. Since L = Iω and L is fixed:

I_1 * omega_1 = I_2 * omega_2

The key insight: I and ω are inversely proportional when L is conserved. Halve I (pull mass toward axis) and ω doubles. Double I (extend mass outward) and ω halves. The product Iω never changes.

⚠️Kinetic energy is NOT conserved in shape-change problems.When a skater pulls their arms in, their muscles do internal work on the system. L is conserved (no external torque) but K_rot increases because the skater added energy. KE conservation requires no non-conservative forces — muscle forces are non-conservative.

Set the initial rotational inertia and angular velocity of a spinning system, then change I to a new value. L stays constant — watch omega respond. Note whether KE is conserved (it usually is not).

I_initial (kg·m²)(arms out)3.0
omega_i (rad/s)(initial spin)2.0
I_final (kg·m²)(arms in/out)1.2
L = I_i * omega_i
6.00 kg·m²/s
conserved
omega_f = L / I_f
5.00 rad/s
new spin rate
I ratio I_i / I_f
2.50×
inertia change factor
omega ratio
2.50×
speed change factor
KE initial
6.00 J
KE final
15.00 J
increased
L_i = 3×2 = 6.00 kg·m²/s
L_f = 1.2×5.00 = 6.00 kg·m²/s ✓

Pull I_final below I_initial — omega increases. Notice KE increases too when I decreases: the system does internal work (muscle energy for the skater). L is conserved but KE is not. This is a critical AP distinction.

ExampleGuided Example — The Figure Skater

A skater spins with arms extended (I = 4.0 kg·m²) at 2.0 rad/s. She pulls her arms in, reducing I to 1.0 kg·m². Find her new angular velocity and compare kinetic energies before and after.

Step 1Verify conservation condition
The ice exerts negligible torque on the skater's rotation axis. No external torque → angular momentum is conserved.
6.4.A.3Math

Rotational Collisions

When a spinning object makes contact with a stationary one — a disk landing on another disk, a ball dropped onto a turntable — angular momentum is shared between them after the collision. This is the rotational analog of a perfectly inelastic linear collision.

The system's total L before equals total L after. If the objects end up rotating together at the same final omega:

I_1*omega_1 = (I_1 + I_2)*omega_f
💡This mirrors the perfectly inelastic collision formula from Unit 4: m_1*v_1 = (m_1 + m_2)*v_f. The structure is identical — only the symbols change. If you can solve a linear inelastic collision, you can solve a rotational collision.
ExampleWorked Example — Disk Dropped on Disk

A spinning disk (I_1 = 0.8 kg·m², omega_1 = 10 rad/s CCW) has a stationary disk (I_2 = 0.4 kg·m²) dropped onto it. Friction quickly brings them to a common angular velocity. Find omega_f.

6.4.A.4Concept

System Selection — The Strategic Choice

Conservation of angular momentum only holds for a system with no net external torque. The choice of system determines whether the law applies. This is why system selection is an AP scientific skill tested explicitly in FRQs.

✓ Conservation valid
Ball + turntable (whole system)

Internal torques (ball on table, table on ball) cancel. No external torque about the vertical axis. L is conserved. L_i = L_f applies.

✗ Conservation not valid
Turntable only (sub-system)

The ball exerts an external torque on the turntable sub-system. Net external torque ≠ 0. L of the turntable alone is NOT conserved.

🔑AP FRQ strategy: When asked to "justify why angular momentum is conserved," name the system explicitly, identify that the net external torque on that system is zero, and cite Newton's Third Law to explain why internal torques cancel. Three sentences, full credit.
6.4.A.5Concept

Newton's Third Law in Rotation

When two rotating objects interact, the angular impulse one exerts on the other is equal in magnitude and opposite in sign. This is Newton's Third Law applied to rotation.

Linear N3L:   F_AB = -F_BA → J_AB = -J_BA
Rotational N3L: tau_AB = -tau_BA → J_ang_AB = -J_ang_BA

This is why internal torques always cancel within a system. Every torque object A exerts on object B is precisely cancelled by the torque B exerts on A. The net contribution of all internal torques to the total angular momentum is exactly zero — leaving only external torques to change L.

ExampleWorked Example — Two Disks, Newton's Third Law

Disk A (I = 1.0 kg·m², omega = 6 rad/s) contacts disk B (I = 2.0 kg·m², omega = 0). After friction acts for 2 s, they reach a common omega. Find the torque each disk exerts on the other.

← Back to Lesson 6.4Next: Lesson 6.5 →Rolling — where translation and rotation finally meet.