Answer
$v_{G_2}=26.8~ft/s$
Work Step by Step
The required velocity can be determined as follows:
According to the principle of impulse and momentum
$I_A\omega_1+\Sigma \int_{t_1}^{t_2}M_A dt=I_A\omega_2$ [eq(1)]
We know that
$I_A=mK_G^2+mr^2=(\frac{100}{32.2})(0.75)^2+(\frac{100}{32.2})(0.5)^2=2.5233 slug\cdot ft$
and $\omega_2=\frac{v}{r}=\frac{v}{0.5}=2v$
$\omega_1=0$
$\implies v_{G_1}=0$
Similarly $\Sigma \int_{t_1=0}^{t_2=3}M_A dt=Tt(2r)-\mu_k Ntr$
$\implies \Sigma \int_{t_1=0}^{t_2=3}M_A dt= 50(3)(2\times 0.5)-0.1(100)(3)(0.5)=135lb\cdot s$
We plug in the known values in eq(1) to obtain:
$0+135=2.5233(2v_{G_2})$
This simplifies to:
$v_{G_2}=26.8~ft/s$
