Question

A thin uniform rod of mass m and length l is free to rotate about an horizontal axis is shown in figure. The minimum initial angular velocity imparted to rod so that it becomes horizontal is

• A. $\sqrt{\frac{6g}{l}}\sin \theta /2$
• B. $\sqrt{\frac{3g}{l}}\sin \theta /2$
• C. $\sqrt{\frac{2g}{l}}\cos \theta /2$
• D. $\sqrt{\frac{g}{l}}\cos \theta /2$

Answer 1

The correct option is A $\sqrt{\frac{6g}{l}}\sin \theta /2$

$\begin{array}{c}Sincethereactionforceattheheightisnotduringanywork,\\ totalmechanicalenerguyisconserved.\\ \Delta PE=mgl/2(1-\cos \theta )\\ \Delta K=\frac{1}{2}{I}_0.w^2\\ =\frac{1}{2}\frac{m{l}^2}{3}.w^2\\ \frac{1}{2}\frac{m{l}^2}{3}{w}^2=mgl.{\sin }^2\theta /2\\ w=\sqrt{\frac{6g}{l}}\sin \theta /2\end{array}$

Hence,

option $\left(A\right)$ is correct answer.