Essential University Physics: Volume 1 (4th Edition)

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

Chapter 10 - Exercises and Problems - Page 193: 54

Answer

$I=(\frac{Ma^2}{12})$

Work Step by Step

We know that the moment of inertia is $\frac{M}{a}x^2$. The book asks us to use integration, so we take the integral to obtain: $I = \int_{-a/2}^{a/2}\frac{M}{a}x^2 dx \\= \frac{M}{a}((\frac{a/2}{3})^3-(\frac{-a/2}{3})^3\\ = \frac{M}{a}(\frac{2a^3}{24})\\ =(\frac{Ma^2}{12})$
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