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: 58

Answer

171.41 rotations per minute

Work Step by Step

We know from Newton's second law that $\tau=I\alpha$, and we know that the moment of inertia of a wheel is given by $I=Mr^2$. Thus, we find: $\alpha=\frac{\tau}{Mr^2}$ $\alpha=\frac{-\mu_k F_n r}{Mr^2}$ $\alpha=\frac{-\mu_k F_n }{Mr}$ $\alpha=\frac{-(.46)(2.7)}{(1.9)(.33)}=-1.98 \ rads/s^2 = 18.9\ rpm/s$ Thus, it follows: $\omega_f = \omega_0 +\alpha t$ $\omega_f = 230-(18.9)(3.1)=\fbox{171.41 rotations per minute} $
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