Physics - W.E.P. Question with Solution | TestHub

When the rod rotates through an angle $\theta$, the centre of gravity falls through a distance h. From $△BC{G}^{\prime }$ $\begin{array}{rl}& \cos \theta =\frac{(l/2)-h}{{\mu }_2}\\ & or,h=\frac{1}{2}(1-\cos \theta )\end{array}$ Decrease in PE $=mg\frac{1}{2}(1-\cos \theta )...(i)$ The decrease in P.E. is equal to the kinetic energy of rotation $\left(\frac{1}{2}I{\omega }^2\right)$ $(\mathrm{KE}{)}_{\text{rotational }}=\frac{1}{2}\left(\frac{m{l}^2}{3}\right){\omega }^2...(ii)$ From Eqs. (i) and (ii), we get $\begin{array}{rl}\frac{1}{2}\left(\frac{m{l}^2}{3}\right){\omega }^2& =mg\frac{1}{2}(1-\cos \theta )\\ \omega & =\sqrt{\frac{6g}{l}}\sin \frac{\theta }{2}\end{array}$