Answer
(a) $\omega = 300~rpm$
(b) The flywheel would have stopped after 75.0 seconds.
The flywheel would have made 313 revolutions.
Work Step by Step
(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.
