Let the rod be deviated through an angle $\varphi$ 'from its initial position at an arbitrary instant of time, measured relative to the initial position in the positive direction. From the equation of the increment of the mechanical energy of the system.
or,
$$
\begin{gathered}
\Delta T=A_{e x t} \\
\frac{1}{2} I \omega^{2}=\int N_{z} d \varphi
\end{gathered}
$$
or,
$$
\frac{1}{2} \frac{M l^{2}}{3} \omega^{2}=\int_{n}^{\varphi} F l \cos \varphi d \varphi=F l \sin \varphi
$$
Thus,
$$
\omega=\sqrt{\frac{6 F \sin \varphi}{M l}}
$$