Conservation of Angular Momentum
A short lesson with diagrams, a worked example, a problem to try, and a quick check. Then quiz yourself on 6 practice questions.
If no net external torque acts on a system, its total angular momentum stays the same: I1ω1 = I2ω2. Pull mass in toward the axis and I drops, so ω rises.
1. Pulling in speeds up the spin
2. Energy is a different story
When a skater pulls her arms in, her rotational kinetic energy increases. Her muscles do work to pull her arms in against their tendency to keep moving outward.
So angular momentum is conserved, but rotational kinetic energy is not.
Linear and angular momentum are separate quantities with separate conservation laws. A system can conserve one without the other: a platform on a fixed axle conserves L, but the axle’s push means its linear momentum isn’t conserved.
3. Example, then try it
A skater spins at 2.0 rad/s with her arms out (I = 4.0 kg·m²). She pulls them in (I = 1.6 kg·m²). Find her new spin rate and compare kinetic energies.
I1ω1 = I2ω2 → (4.0)(2.0) = (1.6)ω → ω = 5.0 rad/s.
K before = ½(4.0)(2.0)² = 8.0 J. K after = ½(1.6)(5.0)² = 20 J. The extra 12 J comes from her arm muscles.
A platform (I = 2.0 kg·m²) spins freely at 3.0 rad/s. A 0.50 kg lump of clay is dropped straight down onto it and sticks 1.0 m from the axis. Find the new spin rate. Is kinetic energy conserved?
Before: L = (2.0)(3.0) = 6.0 kg·m²/s. (The falling clay has no angular momentum about the axis.)
After: I = 2.0 + (0.50)(1.0)² = 2.5 kg·m².
Conserve L: 6.0 = 2.5ω → ω = 2.4 rad/s.
Energy: K before = ½(2.0)(3.0)² = 9.0 J; after = ½(2.5)(2.4)² = 7.2 J. Not conserved: the clay sticks, like a sticky collision.
Answer: 2.4 rad/s; 1.8 J of kinetic energy is lost.
4. Common mistakes
A student sits on a freely spinning stool, holding dumbbells out to her sides. She pulls the dumbbells in to her chest. What happens to her angular speed?
No external torque acts, so Iω stays constant. Pulling the dumbbells in lowers I, so ω increases.
Ready to practice?
6 questions on Conservation of Angular Momentum, each with a hint and an explanation for every wrong answer.