Question

An ice skater spins at $4 \pi \,rad / s$ with her arms extended. If her moment of inertia with arms folded is $80 \%$ of that with arms extended, find the fractional change in kinetic energy.

Answer 1

Conservation of angular momentum gives
$I_{1} \omega_{1}=I_{2} \omega_{2} $
$\left(I_{1}\right)(4 \pi)=\frac{80}{100} I_{1} \omega_{2} $
$\therefore \omega_{2}=5 \pi $
$\frac{\Delta K}{K_{1}}=\frac{K_{2}-K_{1}}{K_{1}}=\frac{K_{2}}{K_{1}}-1 $
$=\frac{(1 / 2) I_{2} \omega_{2}^{2}}{(1 / 2) I_{1} \omega_{1}^{2}}-1$
$=(0.8)\left(\frac{5 \pi}{4 \pi}\right)^{2}-1=\frac{1}{4}$