The Skater's Spin: Pulling In and Slowing Down
A figure skater (modeled as a point mass plus arms) spins with arms extended at 1.5 rev/s. Her moment of inertia with arms out is 4.0 kg·m². She pulls her arms in, reducing her moment of inertia to 1.6 kg·m². During the pull-in, a coach notes her angular speed rises to about 3.75 rev/s. A student argues that because kinetic energy went up, 'the skater must have done work - so angular momentum can't be conserved.' There is negligible external torque about the vertical axis.
1. Compute the angular speed after pull-in using conservation of angular momentum and compare to the coach's reading.
Read the worked answer
No external torque about vertical → L conserved: I₁ω₁ = I₂ω₂.
- I₁ = 4.0, ω₁ = 1.5 rev/s → L = 6.0 (kg·m²·rev/s).
- ω₂ = L/I₂ = 6.0/1.6 = 3.75 rev/s ✓ matches the coach.
Conservation of L is satisfied. The student's inference that 'energy up ⇒ L not conserved' conflates two separate conserved quantities.
2. Show quantitatively that rotational kinetic energy increases, and identify the source of that energy.
Read the worked answer
K = ½Iω². Convert to rad/s: ω₁ = 1.5(2π) ≈ 9.42 rad/s; ω₂ = 3.75(2π) ≈ 23.6 rad/s.
- K₁ = ½(4.0)(9.42²) ≈ 177 J.
- K₂ = ½(1.6)(23.6²) ≈ 445 J.
ΔK ≈ +268 J. This energy comes from internal work the skater does pulling her arms inward against the centrifugal tendency - chemical energy converted to rotational KE. Angular momentum is conserved because the pull-in force is radial (no torque about the spin axis), even though energy is not conserved.
3. A skeptic says 'if she can gain energy by pulling in, she could build unlimited speed by pumping.' What stops this?
Read the worked answer
Each pull-in can only harvest energy equal to the decrease in I at fixed L: ΔK = ½L²(1/I₂ − 1/I₁). Once her arms are fully in, I cannot decrease further, so no more energy can be extracted - the gain is bounded by the finite change in moment of inertia. To repeat, she must extend her arms again, which requires doing work against the outward centrifugal effect and returns the energy (minus losses). There is no free-energy loop: the net work per cycle is ≤ 0 with friction. The judgment is recognizing that 'energy appears' is consistent with L conservation when internal work is accounted for.