Answer
(a) $\alpha = -50.0~rad/s^2$
(b) At t = 3.00 s, $\omega = 250~rad/s$
$\omega_0 = 400~rad/s$
(c) $\theta = 975~rad$
(d) At t = 7.86 seconds, the radial acceleration will be equal to g.
Work Step by Step
(a) $\alpha = \frac{a}{r} = \frac{-10.0~m/s^2}{0.200~m}$
$\alpha = -50.0~rad/s^2$
(b) At t = 3.00 s:
$\omega = \frac{v}{r} = \frac{50.0~m/s}{0.200~m}$
$\omega = 250~rad/s$
We can find the angular velocity $\omega_0$ at t = 0:
$\omega = \omega_0 + \alpha ~t$
$\omega_0 = \omega - \alpha ~t$
$\omega_0 = (250~rad/s)- (-50.0~rad/s^2)(3.00~s)$
$\omega_0 = 400~rad/s$
(c) $\theta = \omega_0 ~t + \frac{1}{2}\alpha ~t^2$
$\theta = (400~rad/s)(3.00~s) + \frac{1}{2}(-50.0~rad/s^2)(3.00~s)^2$
$\theta = 975~rad$
(d) $a_{rad} = \omega^2 ~r = g$
$\omega = \sqrt{\frac{g}{r}}$
$\omega = \omega_0 + \alpha ~t$
$t = \frac{\omega - \omega_0}{\alpha}$
$t = \frac{ \sqrt{\frac{g}{r}} - \omega_0}{\alpha}$
$t = \frac{ \sqrt{\frac{9.80~m/s^2}{0.200~m}} - 400~rad/s}{-50.0~rad/s^2}$
$t = 7.86~s$
At t = 7.86 seconds, the radial acceleration will be equal to g.
