Essential University Physics: Volume 1 (4th Edition) Clone

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 10 - Exercises and Problems - Page 193: 54

Answer

$I=(\frac{Ma^2}{12})$

Work Step by Step

We know that the moment of inertia is $\frac{M}{a}x^2$. The book asks us to use integration, so we take the integral to obtain: $I = \int_{-a/2}^{a/2}\frac{M}{a}x^2 dx \\= \frac{M}{a}((\frac{a/2}{3})^3-(\frac{-a/2}{3})^3\\ = \frac{M}{a}(\frac{2a^3}{24})\\ =(\frac{Ma^2}{12})$
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.