Answer
$\omega=9 rad/s$
Work Step by Step
We can determine the required angular velocity as follows:
We apply the principle of impulse and momentum
$0+\Sigma\int_{t_1}^{t_2}Mdt=I_{AB}\omega+mv_{AB}r_{AB}+I_{BC}\omega+mv_{BC} r_{BC}$ [eq(1)]
We know that
$I_{AB}=I_{BC}=\frac{1}{12}ml^2=(\frac{1}{12})(9)(1)^2=0.75Kg\cdot m^2$
and $r_{AB}=\frac{1}{2}=0.5m$
Similarly $r_{BC}=\sqrt{(1)^2+(0.5)^2}=1.118m$
$v_{AB}=r_{AB}\omega=0.5\omega$
$v_{BC}=r_{BC}\omega=1.118\omega$
Now, we plug in the known values in eq(1) to obtain:
$\int_{0}^{t}15t^2dt=0.75\omega+9(0.5\omega)(0.5)+0.75\omega+9(1.118\omega)(1.118)$
This simplifies to:
$\omega=9 rad/s$
