University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 9 - Rotation of Rigid Bodies - Problems - Exercises - Page 297: 9.47


(a) $I_2 = f^5~I_1$ (b) $KE_2 = 6.4\times 10^8~J$

Work Step by Step

(a) $I_2 = \sum m_2r_2^2$ $I_2 = \sum (f^3~m_1)~(f~r_1)^2$ $I_2 = f^5~\sum (m_1)~(r_1)^2$ $I_2 = f^5~I_1$ The original moment of inertia will be multiplied by $f^5$. (b) $KE_1 = \frac{1}{2}I_1\omega^2$ $KE_2 = \frac{1}{2}I_2\omega^2$ $KE_2 = \frac{1}{2}(f^5~I_1)\omega^2$ $KE_2 = f^5~\frac{1}{2}I_1\omega^2$ $KE_2 = f^5~KE_1$ $KE_2 = (48)^5~(2.5~J)$ $KE_2 = 6.4\times 10^8~J$
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.