Essential University Physics: Volume 1 (4th Edition)

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 11 - Exercises and Problems - Page 208: 43


$3.1\times10^{-16} \ Js$

Work Step by Step

We first find the mass of the disk: $M = \pi(150\times10^{-6})^2(2\times10^{-6})(2329)=3.293\times10^{-10}$ Thus, we find: $L=I\omega=\frac{1}{2}MR^2\omega=\frac{1}{2}(3.293\times10^{-10})(150\times10^{-6})^2(83.77)=3.1\times10^{-16} \ Js$
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