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 10.3 - Radius of Gyration of an Area - Problems - Page 538: 11



Work Step by Step

We can find the required moment of inertia as follows: Given that $y=\frac{1}{8}x^3$ $\implies x=2y^{\frac{1}{3}}$ We know that $dA=xdy$ $\implies dA=2y^{\frac{1}{3}}dy$ Now $I_x=\int_0^8 y^2(2y^{\frac{1}{3}})dy$ $\implies I_x=2\int_0^8 y^{\frac{7}{3}}dy $ $\implies I_x=2(\frac{3}{10}y^{\frac{10}{3}})|_0^8$ After applying the limits, we obtain: $I_x=614m^4$
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