Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 10 - Exercises and Problems - Page 187: 78


a) The proof is below. b) $I =M(h^2+r_{cm}^2)$

Work Step by Step

a) We know that the Law of Cosines is: $C^2 = A^2 + B^2 -2AB cos \gamma$ Where $\gamma$ is the angle opposite to side C. Thus, plugging in $r_{cm}$ for A, $h$ for B, and $r$ for C, we find: $r^2 =r_{cm}^2 + h^2 -2hr_{cm} cos \gamma$ Looking at the value of gamma, this simplifies to: $r^2 =r_{cm}^2 + h^2 -2hr_{cm} $ b) As the book recommends in the problem, we take the integral of $r^2$ and use substitution to solve: $I=\int r^2 dm$ $I=\int (r_{cm}^2 + h^2 -2hr_{cm}) dm$ Taking the integral, this becomes: $I =M(h^2+r_{cm}^2)$
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.