Answer
$\omega_2=53.7~ rad/s$
Work Step by Step
We can determine the required angular velocity as follows:
We know that
$I_{\circ}\omega_1+\Sigma \int_{t_1}^{t_2} M_{\circ} dt=I_{\circ}\omega_2~~~$[eq(1)]
As $I_{\circ}=\frac{1}{2}mr^2$
$\implies I_{\circ}=\frac{1}{2}(\frac{8}{32.2})(0.6)^2=0.0447slug\cdot ft^2$
and $\omega_1=0$
Now $\int_{t_1}^{t_2} M_{\circ}=T_{B}rt-T_A rt=(5)(4)(0.6)-(4)(4)(0.6)=2.4lb\cdot ft/s$
We plug in the known values in eq(1) to obtain:
$0+2.4=0.0447\omega_2$
This simplifies to:
$\omega_2=53.7 rad/s$
