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 573: 97



Work Step by Step

We can determine the required moment of inertia as follows: $I_z=\frac{3\rho}{10}\Sigma \frac{1}{3}\pi r_i^2 h_i r_i^2$ $\implies I_z=\frac{\rho \pi}{10}\Sigma h_i r_i^4$ We plug in the known values to obtain: $I_z=\frac{200}{10}(1.6(0.4)4-0.6(0.4)^4-0.8(0.2)^4)$ This simplifies to: $I_z=1.53Kgm^2$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.