Problem 10.7

10.7.

Let's break the motion into two stages, the acceleration period and the deceleration period.
While its speeding up,
$\theta_{1} = \frac{\omega_{f}+\omega_{i}}{2} t$
or $\theta_{1} = \frac{(10\pi)(8s)}{2} ~=~ 125.7 radians$.
While its slowing down we find
$\theta_{2} = \frac{\omega_{f}+\omega_{i}}{2} t$
or $\theta_{2} = \frac{(10\pi)(12s)}{2} ~=~ 188.5 radians$.
So, the total angle subtended is
$\theta_{tot} = \theta_{1}+\theta_{2} ~=~ 314.2 radians$
or $\theta_{tot} = 50.0 revolutions$.