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 537: 7



Work Step by Step

We can find the moment of inertia about the x-axis as follows: $dA=xdy$ As given that $y^2=1-0.5x$ $\implies x=2(1-y^2)$ $\implies dA=2(1-y^2)dy$ Now $I_x=\int y^2 dA$ $\implies I_x=\int_0^1 y^2 \cdot 2(1-y^2)dy$ $\implies I_x=2\int_0^1 (y^2-y^4)dy$ $\implies I_x=2(\frac{y^3}{3}|_0^1-\frac{y^5}{5}|_0^1)$ After applying the limits, we obtain: $I_x=2(\frac{1}{3}-\frac{1}{5})$ This simplifies to: $I_x=0.267m^4$
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