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)
