A.4.10 (HL)—Angular momentum conservation

Conservation condition

Angular momentum remains constant when the resultant external torque about the chosen axis is zero:

$$L_i=L_f$$

Redistribute the mass

When a skater pulls their arms inward, $I$ decreases. With angular momentum conserved, $\omega$ increases.

Worked example from local Question Bank row 34033

A $0.200\,\mathrm{kg}$ particle moving at $12.0\,\mathrm{m\,s^{-1}}$ strikes $0.60\,\mathrm{m}$ from an axis. Its initial angular momentum is

$$L_i=mvr=(0.200)(12.0)(0.60)=1.44\,\mathrm{kg\,m^2\,s^{-1}}$$

If the combined system has $I_f=0.252\,\mathrm{kg\,m^2}$ and external torque is negligible,

$$I_f\omega_f=L_i\Rightarrow\omega_f=\frac{1.44}{0.252}=5.71\,\mathrm{rad\,s^{-1}}$$

Common trap

Angular momentum conservation does not require rotational kinetic energy to remain constant when the moment of inertia changes.