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: 61

Answer

$\sqrt{\frac{6}{5}gdsin\theta}$

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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