## Essential University Physics: Volume 1 (4th Edition) Clone

Published by Pearson

# Chapter 10 - Exercises and Problems - Page 193: 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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