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 10.4 - Rotational Energy - Example - Page 187: 10.11

Answer

$46 \ Nm$

Work Step by Step

We find: $\tau = \frac{\frac{1}{2}I\omega_f^2}{\Delta\theta}= \frac{\frac{1}{2}(2.7)(73.3)^2}{157} = 46 \ Nm$
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