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: 59


2.088 radians per second

Work Step by Step

In order to solve this problem, we first need to find the torque: $\tau = Fr =75 \times .45 = 33.75 \ Nm$ We now find the moment of inertia: $ I = \frac{1}{2} mr^2 = \frac{1}{2} (120)(.45)^2 =12.15 \ kgm^2$ Thus, we find the final angular velocity: $ \omega_f =\sqrt{\frac{2\tau\theta}{I}}=\sqrt{\frac{2(33.75 \ Nm)(.785 \ rads)}{12.15 \ kgm^2}}=\fbox{2.088 radians per second}$
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