Answer
$\omega=3.90 rad/s$
Work Step by Step
We can determine the required angular velocity as follows:
First, we apply the impulse momentum principle in y-direction
$mv_{Gy1}+\Sigma \int_{t_1}^{t_2} F_y dt=mv_{Gy2}$
$\implies mv_{Gy1}+Isin60=mv_{Gy2}$
$\implies 0+8sin60=4v_{Gy2}$
This simplifies to:
$v_{Gy2}=\sqrt{3}m/s$
Now we apply the impulse momentum principle in x-direction
$mv_{Gx1}+\Sigma \int_{t_1}^{t_2} F_x dt=mv_{Gx2}$
$\implies mv_{Gx1}+Icos60=mv_{Gx2}$
This simplifies to:
$v_{Gx2}=1m/s$
Thus, $v_G=\sqrt{v_{Gx}^2+v_{Gy}^2}=\sqrt{(1)^2+\sqrt{3}^2}=2m/s$
Now, we can calculate the angular velocity
$I_A\omega_1+\Sigma\int_{t_1}^{t_2} M_Gdt=I_A\omega_2$ [eq(1)]
As $I_A=\frac{1}{12}mL^2=\frac{1}{12}(4)(2)^2=\frac{4}{3}Kg\cdot m^2$
and $\omega_1=0$
Similarly $\Sigma\int_{t_1}^{t_2} M_Gdt=Isin60\times 0.75=3\sqrt {3} N\cdot s$
We plug in the known values in eq(1) to obtain:
$0+3\sqrt{3}=\frac{4}{3}\omega$
This simplifies to:
$\omega=3.90 rad/s$
