Answer
(a) The total angle through which the wheel turns is 540 radians.
(b) The time when the wheel stopped is $t = 12.3 ~seconds$.
(c) $\alpha = -8.16~rad/s^2$
Work Step by Step
(a) We can find the angle through which the wheel turns during the first 2.00 seconds.
$\theta = \omega_0 ~t + \frac{1}{2}\alpha ~t^2$
$\theta = (24.0~rad/s)(2.00~s) + \frac{1}{2}(30.0~rad/s^2)(2.00~s)^2$
$\theta = 108~rad$
The total angle through which the wheel turns is 108 rad + 432 rad, which is 540 radians.
(b) We can find $\omega$ at t = 2.00 s:
$\omega = \omega_0 + \alpha ~t$
$\omega = (24.0~rad/s) + (30.0~rad/s^2)(2.00~s)$
$\omega = 84.0~rad/s$
We can find the time $t$ it took to stop after the 2.00-second mark. We can let $\omega_0 = 84.0~rad/s$ for this part of the question.
$\theta = (\frac{\omega+\omega_0}{2})~t$
$t = \frac{2~\theta}{\omega+\omega_0}$
$t = \frac{(2)(432~rad)}{0+84.0~rad/s}$
$t = 10.3~s$
The time when the wheel stopped is 10.3 s + 2.00 s, which is 12.3 seconds.
(c) $\omega = \omega_0 + \alpha ~t$
$\alpha = \frac{\omega-\omega_0}{t}$
$\alpha = \frac{0-84.0~rad/s}{10.3~s}$
$\alpha = -8.16~rad/s^2$
