Answer
The angular velocity of the turntable when the block is on the outer edge is 5.2 rad/s
Work Step by Step
We can find the initial moment of inertia of the turntable. Let $M_t$ be the turntable's mass.
$I_0 = \frac{1}{2}M_tR^2$
$I_0 = \frac{1}{2}(0.20~kg)(0.20~m)^2$
$I_0 = 4.0\times 10^{-3}~kg~m^2$
We can find the final moment of inertia of the turntable and the block when the block is on the outer edge. Let $M_t$ be the turntable's mass and let $M_b$ be the block's mass.
$I_f = \frac{1}{2}M_tR^2 + M_b~R^2$
$I_f = \frac{1}{2}(0.20~kg)(0.20~m)^2+(0.020~kg)(0.20~m)^2$
$I_f = 4.8\times 10^{-3}~kg~m^2$
The initial angular velocity is 60 rpm which is 1 rev/s. Then $\omega_0 = 2\pi~rad/s$.
We can use conservation of angular momentum to find the final angular velocity of the turntable when the block is on the outer edge. The final angular momentum of the turntable with the block on the outer edge will be equal to the initial angular momentum of the turntable with the block at the center.
$L_f = L_0$
$I_f~\omega_f = I_0~\omega_0$
$\omega_f = \frac{I_0~\omega_0}{I_f}$
$\omega_f = \frac{(4.0\times 10^{-3}~kg~m^2)(2\pi~rad/s)}{4.8\times 10^{-3}~kg~m^2}$
$\omega_f = 5.2~rad/s$
The angular velocity of the turntable when the block is on the outer edge is 5.2 rad/s.
