Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 10 - Rotation - Problems - Page 291: 68c

Answer

$\tau = 9.84~N \cdot m$

Work Step by Step

We can find the angular acceleration: $\alpha = \frac{\omega}{t}$ $\alpha = \frac{317~rad/s}{15.5~s}$ $\alpha = 20.45~rad/s^2$ We can find the required torque: $\tau = I~\alpha$ $\tau = \frac{2}{5}MR^2~\alpha$ $\tau = (\frac{2}{5})(1.65~kg)(0.854~m)^2~(20.45~rad/s^2)$ $\tau = 9.84~N \cdot m$
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