Answer
$\omega_1=1 ~rad/s$
Work Step by Step
The required angular velocity can be determined as follows:
We apply the conservation of angular momentum
$I_{P_{\circ}}\omega_{\circ}=I_{P_{1}}\omega_1~~~$[eq(1)]
We know that
$I_{P_1}=\frac{ml^2}{12}+mr^2$
$\implies I_{P_{1}}=\frac{ml^2}{12}+m(\frac{l}{2})^2$
$\implies I_{P_{1}}=\frac{ml^2}{12}+\frac{ml^2}{4}=\frac{ml^2}{3}$
We plug in the known values in eq(1) to obtain:
$\frac{ml^2}{12}\omega_{\circ}=\frac{ml^2}{3}\omega_1$
This simplifies to:
$\omega_1=\frac{\omega_{\circ}}{4}$
$\implies \omega_1=\frac{4}{4}= 1 ~rad/s$