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 - Page 186: 67


32.85 meters

Work Step by Step

We use conservation of energy to find: $mgh = \frac{1}{2}mv^2+I\omega^2 $ $mgh = \frac{1}{2}mv^2+I\frac{v^2}{r^2} $ $h = \frac{\frac{1}{2}mv^2+I\frac{v^2}{r^2}}{mg} $ $h = \frac{\frac{1}{2}(395)(23.6)^2+(2.1)\frac{23.6^2}{.26^2}}{(395)(9.81)}=\fbox{32.85 meters} $
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