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: 64



Work Step by Step

We use conservation of energy to obtain: $mgh = \frac{1}{2}mv^2+\frac{1}{2}\frac{2}{5}mr^2\omega^2 \\mgh = \frac{1}{2}mv^2+\frac{2}{10}mr^2\omega^2 \\mgh = \frac{1}{2}mv^2+\frac{2}{10}mr^2\frac{v^2}{r^2} \\ gh = \frac{1}{2}v^2+\frac{2}{10}v^2\\ gh = \frac{7}{10}v^2 \\ v = \sqrt{\frac{10gh}{7}}$ Plugging in this value of v and using conservation of energy again, we find: $\frac{1}{2}mv^2=mgh' \\ \frac{10}{14}gh=gh'\\ h'=\frac{5h}{7}$
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