Answer
The rod makes $~~624~~$ revolutions.
Work Step by Step
We can find the angular acceleration:
$\omega = \omega_0+\alpha~t$
$0 = \omega_0+\alpha~t$
$\alpha = \frac{-\omega_0}{t}$
$\alpha = \frac{-(39.0~rev/s)(2\pi~rad)}{32.0~s}$
$\alpha = -7.66~rad/s^2$
We can find the number of revolutions:
$\theta = \omega_0~t+\frac{1}{2}\alpha~t^2$
$\theta = (39.0~rev/s)(2\pi~rad)(32.0~s)+(\frac{1}{2})(-7.66~rad/s^2)(32.0~s)^2$
$\theta = 3919.5~rad$
$\theta = (3919.5~rad)(\frac{1~rev}{2\pi~rad})$
$\theta = 624~rev$
The rod makes $~~624~~$ revolutions.
