question_answer

A uniform rod of length L and mass M is held vertical, with its bottom end pivoted to the floor. The rod falls under gravity, freely turning about the pivot. If acceleration due to gravity is g, what is the instantaneous angular speed of the rod when it makes an angle \[{{60}^{o}}\] with the vertical?

A)  \[{{\left( \frac{g}{L} \right)}^{1/2}}\]

B)  \[{{\left( \frac{3g}{4L} \right)}^{1/2}}\]

C)  \[{{\left( \frac{3\sqrt{3}g}{2L} \right)}^{1/2}}\]

D)  \[{{\left( \frac{3g}{2L} \right)}^{1/2}}\]

Correct Answer: D

Solution :

The fall of centre of gravity h \[\frac{\left( \frac{1}{2}-h \right)}{\frac{1}{2}}=\cos \,\,{{60}^{o}}\] \[h=\frac{1}{2}(1-\cos \,{{60}^{o}})\] Decrease in potential energy \[Mgh=Mg\frac{1}{2}(1-\cos {{60}^{o}})\] Kinetic energy of rotation \[=\frac{1}{2}I{{\omega }^{2}}\] \[=\frac{1}{2}\times \frac{M{{I}^{2}}}{3}{{\omega }^{2}}\] \[Mg=\frac{1}{2}(1-\cos \,{{60}^{o}})=\frac{M{{I}^{2}}}{6}{{\omega }^{2}}\] \[\Rightarrow \] \[\omega =\frac{\sqrt{3g}}{2L}\]