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: 107


$I_O=282slug\cdot ft^2$

Work Step by Step

We can determine the required moment of inertia as follows: The mass of the big cylinder is given as $m=\gamma V$ $\implies m=(90)\pi(2.5)^2(1)=1767lb$ The mass of the cylinder number 1 is given as $m=\gamma (-V)$ $\implies m=90(-\pi)(2)^2(0.375)=-424lb$ The mass of the cylinder number 2 is $m=\gamma(-V)$ $\implies m=90(-\pi) (1)^2(0.25)=-70.69lb$ and the mass of the cylinder number 3 is given as $m=\gamma(-V)$ $\implies m=90(-\pi)(2)^2(0.375)=-424lb$ Now $I_O=\Sigma (I^{\prime}+md^2)$ We plug in the known values to obtain: $I_O=\frac{1767(2.5)^2}{2}+1767(2.5)^2+\frac{-424(2)^2}{2}-424(2.5)^2+\frac{-70.69(1)^2}{2}-70.69(2.5)^2+\frac{(-424.1)(2)^2}{2}-424(2.5)^2$ This simplifies to: $I_O=9092lb ft^2$ $\implies I_O=282slug\cdot ft^2$
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