## University Physics with Modern Physics (14th Edition)

(a) $\omega = 300~rpm$ (b) The flywheel would have stopped after 75.0 seconds. The flywheel would have made 313 revolutions.
(a) $\Delta \theta = (\frac{\omega + \omega_0}{2})~t$ $\omega = \frac{2~\Delta \theta}{t}-\omega_0$ $\omega = \frac{(2)(200~rev)}{30.0~s}-(500~rpm/60~s)$ $\omega = 5~rev/s$ $\omega = 300~rpm$ (b) $\alpha = \frac{\omega - \omega_0}{t}$ $\alpha = \frac{300~rpm/60~s - 500~rpm/60~s}{30.0~s}$ $\alpha = -\frac{1}{9}~rev/s^2$ We can find the time for the flywheel to stop. $t = \frac{\omega_0}{\alpha} = \frac{500~rpm/60~s}{\frac{1}{9}~rev/s^2}$ $t = 75.0~s$ The flywheel would have stopped after 75.0 seconds. We can find the number of revolutions $n$ in 75.0 seconds. $n = \omega_0~t + \frac{1}{2}\alpha~t^2$ $n = (500~rpm/60~s)(75.0~s) - \frac{1}{2}(\frac{1}{9}~rev/s^2)(75.0~s)^2$ $n = 313~revolutions$ The flywheel would have made 313 revolutions.