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 - Exercises and Problems - Page 193: 62

Answer

1.163 meters

Work Step by Step

We use conservation of energy to obtain: $mgh = \frac{1}{2}mv^2+\frac{1}{2}\frac{2}{3}mr^2\omega^2 \\mgh = \frac{1}{2}mv^2+\frac{2}{6}mr^2\omega^2 \\mgh = \frac{1}{2}mv^2+\frac{2}{6}mr^2\frac{v^2}{r^2} \\ 2gh = \frac{1}{2}v^2+\frac{2}{6}v^2\\ h = \frac{\frac{5}{6}v^2}{g}= \frac{\frac{5}{6}(3.7)^2}{9.81}=\fbox{1.163 meters}$
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