Answer
$\omega_2=0.01778rad/s$
Work Step by Step
We can determine the required angular velocity as follows:
We apply the principle of impulse and momentum
$I_G\omega_1+\Sigma\int_{t_1}^{t_2}M_G dt=I_G\omega_2~~$[eq(1)]
As $I_G=mk_G^2$
$I_G=200(10^3)(15)^2=45\times 10^6Kg\cdot m^2$
and $\omega_1=0$
Similarly, $\Sigma\int_{t_1}^{t_2}M_G=T_A tr_{G/A}-T_B tr_{G/B}$
$\Sigma\int_{t_1}^{t_2}M_G=40\times 10^3\times5\times8-20\times 10^3 \times 5\times 8=8\times 10^5N\cdot m\cdot s$
We plug in the known values in eq(1) to obtain:
$0+8\times 10^5=45\times 10^6\omega_2$
This simplifies to:
$\omega_2=0.01778rad/s$
