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 575: 106


$I_G=118slug\cdot ft^2$

Work Step by Step

We can determine the required moment of inertia as follows: The mass of the big cylinder is given as $m=\gamma V$ $\implies m=90\pi(2.5)^2(1)$ $\implies m=1767lb$ The mass of the cylinder number 1 is given as $m=\gamma(-V)$ $\implies m=90[-\pi(2)^2\times(0.375)]$ $\implies m=-424lb$ The mass of the cylinder 2 is $m=\gamma(-V)$ $\implies m=90[-\pi(1)^2(0.25)]$ $\implies m=-70.69lb$ and the mass of the cylinder number 3 is given as $m=\gamma(-V)$ $\implies m=90[-\pi(2)^2(0.375)]$ $\implies m=-424lb$ Now $I_G=\Sigma I^{\prime}$ We plug in the known values to obtain: $I_G=\frac{1767(2.5)^2}{2}+\frac{-424(2)^2}{2}+\frac{-70.69(1)^2}{2}+\frac{-424(2)^2}{2}$ This simplifies to: $I_G=3791lb ft^2$ $\implies I_G=118slug\cdot ft^2$
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