Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 11 - Exercises and Problems: 59

Answer

a) $\omega=\frac{2\omega_0}{7}$ b) $t=\frac{2R\omega_0}{\mu_k g}$

Work Step by Step

We first find torque: $\tau = R\mu_k mg$ Thus, we find alpha: $\alpha=\frac{\tau}{I}=\frac{R\mu_kmg}{\frac{2}{5}mR^2}=\frac{5mu_kg}{2R}$ Thus, because $v=\mu_kgt$, it follows: $v=\omega_0R+\alpha t$ $\mu_kgt=\omega_0R+\frac{5mu_kg}{2R} t$ $t=\frac{2R\omega_0}{\mu_k g}$ We now find the angular velocity. Using $v=\omega_0R+\alpha t$ and plugging in the values that we have solved for, we obtain: $\omega=\frac{2\omega_0}{7}$
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