A.4 · IB Physics
Angular momentum: the spinning skater
A skater spins with no external torque. Pull the arms in and the moment of inertia drops, so the angular velocity must rise to keep L = Iω constant. The bars compare each quantity with its starting value.
Predict first
Start from the default settings (Arm extension 100 %). Drag Arm extension down to 50 %. What happens to the angular velocity ω?
With no external torque L = Iω is constant. Pulling the arms in cuts I from 5.92 to 2.80 kg m², so ω = L ÷ I rises by 5.92 ÷ 2.80 ≈ 2.11 times, from 2.00 to 4.23 rad s⁻¹.
Check it in the model: Drag the Arm extension slider from 100 % to 50 % and read the Angular velocity ω value (L stays at 11.8 kg m² s⁻¹).
Left: view from above (dashed circle = arms fully out). Right: bars show each quantity as a multiple of its value with arms fully out.
I = I₀ + 2mr² (I₀ = 0.80 kg m², m = 4.0 kg per hand)
r = 0.20 m + 0.60 m × extension
Ek = ½Iω² = L² ÷ 2I
Try this
- Start with arms out, then press "Pull them in!". Which of I, ω and L change, and which stays exactly the same?
- Set the extension to 50 % and check that I × ω gives the same L as before. By what factor has ω risen?
- Compare Ek with arms out and with arms in. The "Work done by skater" readout shows the difference: where does that energy come from?
Common error: saying that angular momentum increases when the arms are pulled in; L stays constant, while ω and the kinetic energy rise because the skater does work pulling the weights inwards.