Answer
$\omega=12.72 ~rad/s$
Work Step by Step
We can determine the required angular velocity as follows:
We know that
$I_A\omega_1+\Sigma \int_{t_1}^{t_2} M_A dt=I_A\omega_2~~~$[eq(1)]
Now $I_A=m(K_{\circ}^2+r^2)$
$\implies I_A=(\frac{30}{32.2})[(0.45)^2+(0.9)^2]$
$\implies I_A=0.9433 slug\cdot ft^2$
and $\omega_1=0$
Similarly $\Sigma \int_{t_1=0}^{t_2=4} M_A dt=0.6 Pt$
$\Sigma \int_{t_1=0}^{t_2=4} M_A dt=(0.6)(5)(4)=12lb\cdot s$
We plug in the known values in eq(1) to obtain:
$0+12=0.9433\omega$
This simplifies to:
$\omega=12.72 rad/s$
