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 573: 98


$\bar{y}=1.78m$ $I_G=4.45Kgm^2$

Work Step by Step

We know that $\bar{y}=\frac{\Sigma m\bar{y}}{\Sigma m}$ We plug in the known values to obtain: $\bar{y}=\frac{3(1)+5(2+0.25)}{3+5}$ $\implies \bar{y}=1.78m$ Now the required moment of inertia can be determined as: $I_G=\Sigma (I^{\prime}+Ad^2)$ We plug in the known values to obtain: $I_G=\frac{3(2)^2}{12}+3(1.78-1)^2+\frac{5(1)^2}{12}+5(2.251)^2$ This simplifies to: $I_G=4.45Kgm^2$
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