Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 12 - Rotation of a Rigid Body - Exercises and Problems - Page 332: 43


The angular momentum is $0.025~kg~m^2/s$.

Work Step by Step

We can express the angular velocity in units of rad/s as: $\omega = (600~rpm)(\frac{2\pi~rad}{1~rev})(\frac{1~min}{60~s})$ $\omega = 20\pi~rad/s$ We then find the moment of inertia of the disk; $I = \frac{1}{2}MR^2$ $I = \frac{1}{2}(2.0~kg)(0.020~m)^2$ $I = 4.0\times 10^{-4}~kg~m^2$ Next, we find the angular momentum about the axle: $L = I~\omega$ $L = (4.0\times 10^{-4}~kg~m^2)(20\pi~rad/s)$ $L = 0.025~kg~m^2/s$ The angular momentum is $0.025~kg~m^2/s$.
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