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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 her arms extended, what is her angular velocity when she folded her arms?

Discussion:

Conservation of angular momentum gives $$I_1\omega_1=I_2 \omega_2$$
$$(I_1) (4\pi)=0.8 I_1 \omega_2$$
Hence, (\omega_2=5\pi)