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

Answer: 0.25

Conservation of angular momentum gives
$I_1{\omega }_1=I_2{\omega }_2$
$\left(I_1\right)(4\pi )=\frac{80}{100}{I}_1{\omega }_2$
$\therefore {\omega }_2=5\pi$
$\frac{\Delta K}{K_1}=\frac{K_2-K_1}{K_1}=\frac{K_2}{K_1}-1$
$=\frac{(1/2)I_2{\omega }_2^2}{(1/2)I_1{\omega }_1^2}-1$
$=(0.8){\left(\frac{5\pi }{4\pi }\right)}^2-1=\frac{1}{4}$