Answer
$\omega=18.39 rad/s \circlearrowright$
Work Step by Step
We can determine the required angular velocity as follows:
First, we apply the principle of impulse and momentum in normal direction
$mv_{Gn1}+\Sigma\int_{t_1}^{t_2} F_n dt=mv_{Gn2}$
$\implies 0+Nt-mgcos30t=0$
$\implies 4N-100(9.81)cos 30(4)=0$
This simplifies to:
$N=849.571N$
Now, we apply the impulse momentum principle in tangential direction
$mv_{Gt1}+\Sigma\int_{t_1}^{t_2} F_t dt=mv_{Gt2}$
$\implies 0+Tt-mgsin30t+\mu_k Nt=mv_{G2}$
$\implies 4T-100(9.81)sin30(4)+0.1(849.571)(4)=100v_G$
This simplifies to:
$T-25v_{G}=405.5429$ [eq(1)]
Now $I_G\omega_1+\Sigma\int_{t_1}^{t_2}M_Gdt=I_G\omega_2$ [eq(2)]
As $I_G=mk_G^2=100(0.25)^2=6.25Kg\cdot m^2$
and $\omega_1=0$
$\Sigma\int_{t_1=0}^{t_2=4}M_Gdt=\mu_kNtr_A-Ttr_G=0.1(849.57)\times 4\times 0.4-0.2(T)\times 4=135.9312-0.8TN\cdot s$
We plug in the known values in eq(2)to obtain:
$0+135.9312-0.8T=6.25\omega$ [eq(3)]
and we know that $v_G=0.2\omega$ [eq(4)]
We plug in this value from eq(3) into eq(1) to obtain:
$T-5\omega=405.5429$ [eq(5)]
After solving eq(3) and eq(5), we obtain:
$T=313.59N$
and $\omega=18.39 rad/s\circlearrowright$
