Answer
The turntable's angular velocity just after this event is 50 rpm
Work Step by Step
We can find the moment of inertia of the disk before the blocks land on it.
$I_0 = \frac{1}{2}MR^2$
$I_0 = \frac{1}{2}(2.0~kg)(0.10~m)^2$
$I_0 = 0.010~kg~m^2$
We can find the moment of inertia of the system after the blocks land on it. Let $M_b$ be the mass of the blocks.
$I_f = I_0 + M_b~R^2+M_b~R^2$
$I_f = 0.010~kg~m^2 + (0.50~kg)(0.10~m)^2+(0.50~kg)(0.10~m)^2$
$I_f = 0.020~kg~m^2$
We can use conservation of angular momentum to solve this question.
$L_f = L_0$
$I_f~\omega_f = I_0~\omega_0$
$\omega_f = \frac{I_0~\omega_0}{I_f}$
$\omega_f = \frac{(0.010~kg~m^2)(100~rpm)}{0.020~kg~m^2}$
$\omega_f = 50~rpm$
The turntable's angular velocity just after this event is 50 rpm.
