Answer
The rotor will turn 481 revolutions before coming to rest and it will take 6.3 seconds to come to rest.
Work Step by Step
First we can find the moment of inertia of the system:
$I = \frac{1}{2}MR^2$
$I = \frac{1}{2}(3.10~kg)(0.0710~m)^2$
$I = 0.00781~kg\cdot m^2$
Then, we find the magnitude of angular deceleration $\alpha$:
$I \alpha = \tau$
$\alpha = \frac{\tau}{I} = \frac{1.20~m\cdot N}{0.00781~kg\cdot m^2}$
$\alpha = 153.65~rad/s^2$
We can convert from rpm to rad/s.
$\omega = (9200~rpm)(2\pi \frac{rad}{rev})(\frac{1~m}{60~s}) = 963.4~rad/s$
Then, we use $\alpha = -153.65~rad/s^2$ and $\omega$ to find the time to stop:
$t = \frac{\Delta \omega}{\alpha} = \frac{0-963.4~rad/s}{-153.65~rad/s^2}$
$t = 6.3~s$
Next, we find the number of revolutions the cylinder makes while it stops:
$\theta = \omega t + \frac{1}{2}\alpha t^2$
$\theta = (963.4~rad/s)(6.3~s) - (0.5)(153.65~rad/s^2)(6.3~s)^2$
$\theta = 3020~rad$
The number of revolutions is $\frac{3020~rad}{(2\pi)~rad/rev}$ which is 481 revolutions.
