College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 8 - Problems - Page 317: 80


The new angular speed of the disk is $16.9~Hz$

Work Step by Step

We can find the initial rotational inertia of the disk: $I_0 = \frac{1}{2}MR^2$ $I_0 = \frac{1}{2}(0.80~kg)(0.170~m)^2$ $I_0 = 0.01156~kg~m^2$ We can find the final rotational inertia of the disk and clay: $I_f = \frac{1}{2}MR^2+mr^2$ $I_f = \frac{1}{2}(0.80~kg)(0.170~m)^2+(0.120~kg)(0.080~m)^2$ $I_f = 0.01233~kg~m^2$ We can use conservation of angular momentum to find the new angular speed: $L_f = L_0$ $I_f~\omega_f = I_0~\omega_0$ $\omega_f = \frac{I_0~\omega_0}{I_f}$ $\omega_f = \frac{(0.01156~kg~m^2)(18.0~Hz)}{0.01233~kg~m^2}$ $\omega_f = 16.9~Hz$ The new angular speed of the disk is $16.9~Hz$.
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