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Conservation of Angular Momentum

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Zero external torque conserves angular momentum

Since τ_net = dL/dt, whenever the net external torque is zero the angular momentum stays constant. Internal forces — a skater pulling in her arms, a star collapsing — cannot change the total L. If the moment of inertia changes from I₁ to I₂, the angular velocity must adjust to keep the product constant: I₁ω₁ = I₂ω₂.

Conservation of angular momentum
L = Iω = constant I₁ω₁ = I₂ω₂
Holds when the net external torque is zero. Reducing I (pulling mass inward) speeds up the spin; increasing I slows it.

A spinning skater and the energy puzzle

When a skater pulls in her arms, her moment of inertia drops and her spin rate rises to keep L fixed — the classic demonstration. But her kinetic energy increases: since KE_rot = ½Iω² = ½Lω, a larger ω at fixed L means more energy. That extra energy is not free — it comes from the work she does pulling her arms inward against the outward-tending motion. Angular momentum is conserved; kinetic energy is not.

Worked example

A skater spins at 2 rad/s with a moment of inertia of 6 kg·m². She pulls in her arms, reducing her moment of inertia to 2 kg·m². Find her new angular velocity and compare the kinetic energies.

  1. 1.No external torque acts, so angular momentum is conserved: I₁ω₁ = I₂ω₂.
  2. 2.Solve for the new spin rate: ω₂ = I₁ω₁/I₂ = (6 × 2)/2 = 6 rad/s.
  3. 3.Kinetic energy before: ½I₁ω₁² = ½(6)(2²) = 12 J. After: ½I₂ω₂² = ½(2)(6²) = 36 J.
  4. 4.The kinetic energy tripled — the extra 24 J is the work the skater did pulling her arms in.
Answer: ω₂ = I₁ω₁/I₂ = 6 rad/s; KE rises from 12 J to 36 J (the skater does work)
Checkpoint

A skater spinning at 3 rad/s with a moment of inertia of 8 kg·m² pulls in her arms, reducing her moment of inertia to 2 kg·m². What is her new angular velocity?

Checkpoint

As the skater pulls in her arms and speeds up (with no external torque), her rotational kinetic energy:

Watch out

Do not confuse conserved angular momentum with conserved energy. In a "pull-in" problem L is constant but KE_rot = ½Lω grows, and in a "let-out" problem it shrinks. Always solve for ω from I₁ω₁ = I₂ω₂ first, then compute energy separately.

Answer the 2 checkpoints as you read.

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