## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

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