Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 12 - Rotation of a Rigid Body - Exercises and Problems - Page 330: 11


(a) $I = 0.032~kg~m^2$ (b) The rotational kinetic energy is 15.8 J

Work Step by Step

(a) We can find $R$, the distance from each ball to the axis of rotation. $\frac{0.20~m}{R} = cos(30^{\circ})$ $R = \frac{0.20~m}{cos(30^{\circ})}$ $R = 0.23~m$ We can find the moment of inertia of the system. $I = (3)(M~R^2)$ $I = (3)(0.20~kg)(0.23~m)^2$ $I = 0.032~kg~m^2$ (b) We can find the angular velocity. $\omega = (5.0~rev/s)(2\pi~rad/rev)$ $\omega = 10\pi~rad/s$ We can find the rotational kinetic energy. $KE_{rot} = \frac{1}{2}I~\omega^2$ $KE_{rot} = \frac{1}{2}(0.032~kg~m^2)(10\pi~rad/s)^2$ $KE_{rot} = 15.8~J$ The rotational kinetic energy is 15.8 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.