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: 60

Answer

12.2 radians per second

Work Step by Step

We first find the moment of inertia: $I=mr^2=(380)(1.1)^2=459.8 \ kgm^2$ 'We now use conservation of energy to obtain: $mgh=\frac{1}{2}mv^2+\frac{1}{2}I\omega^2 \\ mgh=\frac{1}{2}mr^2\omega^2+\frac{1}{2}I\omega^2\\ \omega = \sqrt{\frac{2Mgh}{I+Mr^2}} \\\omega = \sqrt{\frac{2(509)(9.81)(16)}{459.8+(509)(1.1)^2}} =\fbox{12.2 radians per second}$
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