Essential University Physics: Volume 1 (4th Edition)

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

Chapter 10 - Exercises and Problems - Page 193: 61



Work Step by Step

We use conversation of energy to find: $mgh = \frac{1}{2}mv^2 + \frac{1}{2}I \omega^2$ $mgdsin\theta = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{3}mR^2) \frac{v^2}{R^2}$ $gdsin\theta = \frac{1}{2}v^2 + \frac{1}{3}v^2$ $v=\sqrt{\frac{6}{5}gdsin\theta}$
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