AP Physics C: Mechanics · Unit 6 · Topic 3 of 5 · 6 min lesson

Conservation of Angular Momentum

A short lesson with diagrams, a worked example, a problem to try, and a quick check. Then quiz yourself on 5 practice questions.

Quiz this topic · 5 questions

The big idea

If the net external torque on a system is zero, its total angular momentum stays constant: L_i = L_f. For one object changing shape, I₁ω₁ = I₂ω₂. Kinetic energy usually isn’t conserved.

1. When L is conserved

The pivot force acts at the axis, so it makes no torque: angular momentum about the pivot is conserved even though linear momentum isn’t.

Στ_ext = 0then L is constant

I₁ω₁ = I₂ω₂changing shape

Particle hits rodmvr_⊥ = I_totalω

Kinetic energynot conserved when things stick or pull in

2. Choosing the axis

Pick the axis where unknown external forces act, like a pivot or axle. Their torque about that point is zero, so L about it is conserved.

A skater pulling in her arms speeds up, and her kinetic energy rises: her muscles do work. A person walking outward on a merry-go-round slows it down.

For collisions with a pivoted object, conserve angular momentum about the pivot, not linear momentum. The pivot pushes during the impact.

3. Example, then try it

Example

A disk (I = 1.2 kg·m²) spins at 10 rad/s. A second disk (I = 0.80 kg·m²), not spinning, is dropped onto it on the same axle, and they stick. Find the final ω and the kinetic energy lost.

L = (1.2)(10) = 12 kg·m²/s = (2.0)ω, so ω = 6.0 rad/s.

K: ½(1.2)(10)² = 60 J before; ½(2.0)(6.0)² = 36 J after. 24 J is lost.

Try it · work it on paper first

A 0.010 kg bullet at 400 m/s strikes and embeds in the free end of a uniform 2.0 kg, 1.5 m rod pivoted at its other end on a frictionless table (the figure below). Find the rod’s angular velocity and the fraction of kinetic energy lost.

L before (about the pivot): mvL = (0.010)(400)(1.5) = 6.0 kg·m²/s.

I after: rod ⅓ML² = ⅓(2.0)(1.5)² = 1.5 kg·m², plus bullet (0.010)(1.5)² ≈ 0.023 kg·m²: I ≈ 1.52 kg·m².

ω: 6.0 ÷ 1.52 ≈ 3.9 rad/s.

Energy: ½(0.010)(400)² = 800 J before; ½(1.52)(3.9)² ≈ 12 J after: about 98.5% lost.

Answer: about 3.9 rad/s; nearly all the kinetic energy is lost.

4. Common mistakes

Conserving linear momentum with a pivot.The pivot exerts an external force; use angular momentum about the pivot.

Assuming kinetic energy is conserved.Only angular momentum is.

Forgetting the particle’s own I after it sticks.Add mr² to the rod’s I.

Check yourself

In the bullet-and-rod collision, why isn’t the system’s linear momentum conserved?

The pivot pushes on the rod during the impact. That external force changes the linear momentum, but it acts at the axis, so it makes no torque about the pivot.

Ready to practice?

5 questions on Conservation of Angular Momentum, each with a hint and an explanation for every wrong answer.

Quiz this topic All of Unit 6