## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

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.