Answer
$\omega_2=103~ rad/s$
Work Step by Step
The required angular velocity can be determined 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 $I_{\circ}=\frac{1}{2}mr^2$
$\implies I_{\circ}=\frac{1}{2}(\frac{10}{32.2})(0.5)^2=0.0388slug\cdot ft^2$
and $\omega_1=0$
Now $\int_{t_1}^{t_2}M_{\circ}=Prt=2(0.5)(4)=4lb\cdot ft/s$
We plug in the known values in eq(1) to obtain:
$0+4=(0.0388)\omega_2$
This simplifies to:
$\omega_2=103~ rad/s$
