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: 102



Work Step by Step

We can determine the required moment of inertia as follows: The mass of the hemisphere is given as $m=\rho V$ $\implies m=7.85\frac{2}{3}(300)^3$ $\implies m=443.91Kg$ The mass of the big cone is $m=\rho V$ $\implies m=7.85(\frac{1}{3})\pi (675)(300)^2$ $\implies m=500Kg$ and the mass of the small cone is $m=\rho V$ $\implies m=785(-\frac{1}{3}\pi (225)(100)^2)$ $\implies m=-18.5Kg$ Now $I=\Sigma I^{\prime}$ $\implies I=\frac{2}{5}(443.91)(0.3)^2+\frac{3}{10}(500)(0.3)^2+\frac{3}{10}(-18.5)(0.1)^2$ This simplifies to: $\implies I=29.4Kgm^2$
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