Engineering Mechanics: Statics & Dynamics (14th Edition)

Published by Pearson
ISBN 10: 0133915425
ISBN 13: 978-0-13391-542-6

Chapter 10 - Moments of Inertia - Section 10.8 - Mass Moment of Inertia - Problems - Page 574: 103

Answer

$I=\frac{1}{2}ma^2$

Work Step by Step

We can find the required moment of inertia as follows: $I=\Sigma (I^{\prime}+md^2)$ $\implies I=3(\frac{ma^2}{12}+m(\frac{\sqrt{3}}{6}a)^2$ $\implies I=3(\frac{ma^2}{12}+m\frac{3}{36}a^2)$ $\implies I=\frac{1}{2}ma^2$
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