Question

A person spins with their arms extended at an angular velocity of 4 radians/second. When they bring their arms in, their angular velocity becomes 9 radians/second. Their moment of inertia with arms extended was I. What is the skater's moment of inertia with her arms drawn in?

Answer 1

Since given that the moment of inertia of the person when her arms was extended is

and angular velocity $\omega_1 = 4$ rad/sec.

and when arms is drawn in, the angular velocity becomes, $\omega_2 = 9$ rad/sec.

Since there is no any external torque is applied.

hence the angular momentum will be conserved in both cases. i.e,

$I_1 \omega_1 = I_2 \omega_2$

$I \times 4 = I_2 \times 9$

$I_2 = \frac{4}{9} I$

hence the Skater's moment of inertia with her arms drawn in is $I_2 = \frac{4}{9} I$ .