College Physics (4th Edition)

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

Chapter 8 - Problems - Page 317: 73


The magnitude of the angular momentum is $~1.39\times 10^7~kg~m^2/s$

Work Step by Step

We can express the angular speed in units of rad/s: $350~rpm~\times \frac{2\pi~rad}{1~rev}\times \frac{1~min}{60~s} = 36.65~rad/s$ We can find the angular momentum: $L = I~\omega$ $L = MR^2~\omega$ $L = (5.6\times 10^4~kg)(2.6~m)^2~(36.65~rad/s)$ $L = 1.39\times 10^7~kg~m^2/s$ The magnitude of the angular momentum is $~1.39\times 10^7~kg~m^2/s$.
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