Answer
$\omega_2=36.3 rad/s$
Work Step by Step
We can determine the required angular velocity as follows:
First, we apply the principle of impulse and momentum in the x-direction
$mv_{x_1}+\Sigma\int_{t_1}^{t_2}F_xdt=mv_{x_2}$
We plug in the known values to obtain:
$0+Nt-(\frac{5}{13})F_{OB}t=0$
$6N=6(\frac{5}{13})F_{OB}$
This simplifies to:
$N=\frac{5}{13}F_{OB}~~~$[eq(1)]
Now, we apply the impulse and momentum principle in the y-direction
$mv_{y_1}+\Sigma\int_{t_1}^{t_2}F_y dt=mv_{y_2}$
$\implies 0+\mu_k Nt+(\frac{12}{13})F_{OB}t+Pt-mgt=0$
We plug in the known values to obtain:
$0.2N+6(\frac{12}{13})F_{OB}+40-40\times 9.81=0$
This simplifies to:
$F_{OB}=352.4N$
We plug in this value in eq(1) to obtain:
$N=134.5385N$
We know that
$I_{\circ}\omega_1+\Sigma\int_{t_1}^{t_2}M_{\circ}dt=I_{\circ}\omega_2~~~$[eq(2)]
As $I_{\circ}=mk_{\circ}=40(0.08)^2=0.256Kg\cdot m^2$
Now, from eq(2)
$0+Ptr-\mu_kNrt=0.256\omega_2$
We plug in the known values to obtain:
$40(0.12)(6)-0.2(135.54)(0.12)(6)=0.256\omega_2$
This simplifies to:
$\omega_2=36.3 rad/s$
