Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 10 - Exercises and Problems - Page 186: 65



Work Step by Step

We add negative area, so we subtract the two moment of inertias. Thus, we obtain: $I=\frac{1}{2}MR^2-\frac{3}{2}M(\frac{1}{4}R)^2(\frac{1}{4})^2=.494MR^2$ (Note, we multiply by the $(1/4)^2$ due to the smaller mass of the missing hole.)
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