Physics - Rotation Question with Solution | TestHub

When the rod makes an angle $\alpha$ $$\begin{array}{rl}& \text{ Displacement of centre of mass }=\frac{l}{2}\cos \alpha \\ & mg\frac{l}{2}\cos \alpha =\frac{l}{2}I{\omega }^2\\ & mg\frac{l}{2}\cos \alpha =\frac{m{l}^2}{6}{\omega }^2(\because \text{ M.I. of thin }\end{array}$$ uniform rod about an axis passing through its centre of mass and perpendicular to the $$\begin{array}{rl}& \mathrm{rod}I=\frac{m{l}^2}{12})\\ & \Rightarrow \omega =\sqrt{\frac{3g\cos \alpha }{l}}\end{array}$$ Speed of end $=\omega \times l=\sqrt{3g\cos \alpha l}$ i.e., Speed of end, $\omega \propto \sqrt{\cos \alpha }$