Answer
$\omega_2=21.0~rad/s$
Work Step by Step
The angular velocity required can be determined as follows:
According to the principle of impulse and momentum
$I_{\circ}\omega_1+\Sigma\int_{t_1}^{t_2}M_Adt=I_{\circ}\omega_2+mv_{A_2}r$ [eq(1)]
We know that
$I_{\circ}=mk_{\circ}^2=30(0.125)^2=0.4688Kg\cdot m^2$
and $\omega_1=0$
$\omega_2=\frac{v_{\circ 2}}{r}=\frac{v_{O_2}}{0.15}$
This can be rearranged as:
$v_{O_2}=0.15\omega_2$
Similarly $\Sigma\int_{t_1}^{t_2}M_Adt=\int_{0}^{4} 0.15\times 20tdt=24N\cdot s$
We plug in the known values in eq(1) to obtain:
$0+24=0.4688\omega_2+30(0.15)(0.15\omega_2)$
This simplifies to:
$\omega_2=21.0~rad/s$
