University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 9 - Rotation of Rigid Bodies - Problems - Exercises - Page 297: 9.50


(a) $\frac{1}{12}Ma^2$ (b) $\frac{1}{12}Mb^2$

Work Step by Step

(a) We can use the parallel axis theorem to solve this question. The moment of inertia with respect to the axis that passes through the center of the plate is the moment of inertia $I_{cm}$ for the center of mass. $\frac{1}{3}Ma^2 = I_{cm} + M(\frac{a}{2})^2$ $I_{cm} = \frac{1}{3}Ma^2- \frac{1}{4}Ma^2$ $I_{cm} = \frac{1}{12}Ma^2$ The moment of inertia about this axis is $\frac{1}{12}Ma^2$ . (b) If the axis is perpendicular to the axis in part (a), we can replace the length of $a$ with the length of $b$ in the expression for the moment of inertia. $I = \frac{1}{12}Mb^2$
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