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 571: 86



Work Step by Step

We can find the required radius of gyration as follows: $k^2_x=\frac{I_{mx}}{m}$ We plug in the known values to obtain: $k^2_x=\frac{\int_{0}^{200}\rho \pi \frac{250}{2}x^2 dx}{\int _0 ^{200}\rho \pi 50 x dx}$ After taking the integral and applying the limits, we obtain: $k^2_x=\frac{\frac{50}{2}\frac{(200)^3}{3}}{\frac{(200)^2}{2}}$ This simplifies to: $k^2_x=\frac{1000}{3}$ Which further simplifies to: $k_x=57.7mm$
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