Essential University Physics: Volume 1 (4th Edition) Clone

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

Chapter 10 - Section 10.5 - Rolling Motion - Example - Page 188: 10.12

Answer

$v = \sqrt{\frac{10}{7}gh}$

Work Step by Step

We use conservation of energy to find: $Mgh= \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2 $ $Mgh= \frac{1}{2}Mv^2 + \frac{1}{2}(\frac{2}{5}MR^2)\omega^2 $ $Mgh= \frac{1}{2}Mv^2 + \frac{1}{2}(\frac{2}{5}MR^2)(\frac{v}{R})^2 $ $Mgh = \frac{7}{10}Mv^2$ $v = \sqrt{\frac{10}{7}gh}$
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