#### Answer

(a) The magnitude of the angular acceleration is $1.96~rad/s^2$
(b) The train takes 1.6 seconds to stop.

#### Work Step by Step

(a) We can find the magnitude of deceleration of the train as:
$F_f = ma$
$mg~\mu = ma$
$a = g~\mu$
$a = (9.80~m/s^2)(0.10)$
$a = 0.98~m/s^2$
We can find the magnitude of the angular acceleration as:
$\alpha = \frac{a}{r}$
$\alpha = \frac{0.98~m/s^2}{0.50~m}$
$\alpha = 1.96~rad/s^2$
The magnitude of the angular acceleration is $1.96~rad/s^2$.
(b) We can find the speed $v$ of the train when it is released.
$v = (30~rpm)(\frac{2\pi~r}{1~rev})(\frac{1~min}{60~s})$
$v = (30~rpm)(\frac{(2\pi)(0.50~m)}{1~rev})(\frac{1~min}{60~s})$
$v = 1.57~m/s$
We can find the time it takes the train to stop.
$t = \frac{v-v_0}{a}$
$t = \frac{0-1.57~m/s}{-0.98~m/s^2}$
$t = 1.6~s$
The train takes 1.6 seconds to stop.