Angular Momentum Conservation & the Energy Puzzle
- State the condition for angular momentum conservation and apply it to a changing rotational inertia
- Explain why kinetic energy increases when a spinning skater pulls her arms in
- Compute the angular momentum of a particle moving in a straight line
Angular momentum is conserved when net external torque is zero
The rotational analogue of momentum is L = Iω, and it is conserved when the net external torque on the system is zero. Note the condition carefully: not zero force, zero torque. A system can have large external forces acting and still conserve angular momentum, provided those forces are directed through the axis and so produce no torque — which is exactly the situation for gravity acting on a spinning top about its own axis, or for the Sun's pull on an orbiting planet.
The skater: same L, more K
A spinning skater pulls her arms in, reducing I by roughly half. Angular momentum is conserved because no external torque acts about her vertical axis, so ω roughly doubles. Now check the energy: K = ½Iω², and with I halved and ω doubled, K = ½(I/2)(2ω)² = Iω², which is twice the original ½Iω². Kinetic energy has doubled while angular momentum stayed fixed. That is not a violation — she did work with her arm muscles pulling her arms inward against the outward tendency of the rotating mass. The energy came from her, and it is the reason the maneuver is physically tiring.
Why K = L²/2I is the useful form
Substituting ω = L/I into K = ½Iω² gives K = L²/2I. When L is fixed, kinetic energy is inversely proportional to rotational inertia — so any reduction in I raises K. The relation is the rotational twin of K = p²/2m from Unit 4, and it disposes of the skater paradox in one line without any argument about arms.
A particle moving in a straight line has angular momentum
This is deeply counterintuitive and appears on the exam regularly. Angular momentum is defined relative to a chosen point, and for a particle it is L = mvr_⊥, where r_⊥ is the perpendicular distance from that point to the particle's line of motion. A car driving in a straight line past you has nonzero angular momentum about you — and constant, since r_⊥ and v are both unchanging. Only about a point on its line of motion is its angular momentum zero. This is what makes problems where a ball strikes and sticks to a pivoted rod tractable: the incoming ball brings angular momentum with it.
A merry-go-round (a disk, I = 200 kg·m²) rotates at 2.0 rad/s. A 50 kg child steps onto the rim at radius 2.0 m and holds on. Find the new angular speed.
- 1.No external torque about the vertical axis, so angular momentum is conserved.
- 2.Before: L = Iω = (200)(2.0) = 400 kg·m²/s.
- 3.The child adds rotational inertia I_child = mr² = (50)(2.0)² = 200 kg·m².
- 4.After: total I = 200 + 200 = 400 kg·m², so ω = L/I = 400 ÷ 400.
- 5.This is a rotational perfectly-inelastic collision, so kinetic energy is lost in the process.
Angular momentum conservation and rotational kinetic energy conservation are different claims with different conditions. When a person steps onto a spinning platform, or two disks are pressed together until they spin as one, L is conserved and K is not — those are rotational inelastic collisions.
A skater spinning with arms extended pulls them in, halving her rotational inertia. What happens to her angular momentum and kinetic energy?
A 2.0 kg ball moves in a straight line at 3.0 m/s, passing 0.50 m from a fixed point P at closest approach. What is its angular momentum about P?
A child runs and jumps onto the rim of a stationary merry-go-round, holding on. Which quantity is conserved during the landing?
Say "no net external torque acts about the axis, so angular momentum is conserved" in words before writing I₁ω₁ = I₂ω₂. Naming the condition is a rubric point independent of the algebra, and it forces you to check whether the condition actually holds.
Answer the 3 checkpoints as you read.
Sign in to save your progress