Answer
(a) $\alpha = 4.2~rad/s^2$
(b) $a_R = 130~m/s^2$
$a_{tan} = 1.3~m/s^2$
Work Step by Step
(a) $\omega_0 = (120~rpm)(2\pi \frac{rad}{rev})(\frac{1~m}{60~s}) = 4\pi~rad/s$
$\omega_f = (280~rpm)(2\pi \frac{rad}{rev})(\frac{1~m}{60~s}) = \frac{28\pi}{3}~rad/s$
$\alpha = \frac{\Delta \omega}{t} = \frac{\frac{28\pi}{3}~rad/s ~- ~4\pi~rad/s}{4.0~s}$
$\alpha = \frac{4\pi}{3}~rad/s^2 = 4.2~rad/s^2$
(b) 2.0 seconds after the wheel started accelerating:
$\omega = \omega_0 + \alpha ~t$
$\omega = (4\pi~rad/s) + (\frac{4\pi}{3}~rad/s^2)(2.0~s)$
$\omega =\frac{20\pi}{3} ~rad/s$
We can find the radial acceleration $a_R$.
$a_R = \omega^2 ~r = (\frac{20\pi}{3} ~rad/s)^2(0.61~m/2)$
$a_R = 130~m/s^2$
We can find the tangential acceleration $a_{tan}$.
$a_{tan} = \alpha ~r = (\frac{4\pi}{3}~rad/s^2)(0.61~m/2)$
$a_{tan} = 1.3~m/s^2$