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

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 12 - Rotation of a Rigid Body - Exercises and Problems: 48

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.
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