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: 9.51

Answer

$I = \frac{1}{3}M(a^2+b^2)$

Work Step by Step

For a rectangular sheet rotating around the center, the moment of inertia is $I_{com} = \frac{1}{12}M(a^2+b^2)$. Let $d$ be the distance from the center of the sheet to one corner. $d^2 = (\frac{a}{2})^2+ (\frac{b}{2})^2$ We can use the parallel axis theorem to find the moment of inertia when the axis is at one corner. $I = I_{com}+Md^2$ $I = \frac{1}{12}M(a^2+b^2)+ M[(\frac{a}{2})^2+ (\frac{b}{2})^2]$ $I = \frac{1}{12}M(a^2+b^2)+\frac{1}{4}M(a^2+b^2)$ $I = \frac{1}{3}M(a^2+b^2)$
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