College Physics (4th Edition)

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

Chapter 8 - Problems - Page 316: 71


The angular momentum is $~0.0864~kg~m^2/s$

Work Step by Step

We can express the angular speed in units of rad/s: $0.550~rev/s\times \frac{2\pi~rad}{1~rev} = 3.456~rad/s$ We can find the angular momentum: $L = I~\omega$ $L = \frac{1}{2}MR^2~\omega$ $L = \frac{1}{2}(5.00~kg)(0.100~m)^2~(3.456~rad/s)$ $L = 0.0864~kg~m^2/s$ The angular momentum is $~0.0864~kg~m^2/s$.
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