question_answer

A uniform rod of length L and mass M is held vertical, with its bottom end pivote 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 \[60{}^\circ \] 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 his given by \[\frac{\left( \frac{L}{2}-h \right)}{\left( \frac{L}{2} \right)}=\cos \,60{}^\circ \] or \[h=\frac{L}{2}(1-cos\,60{}^\circ )\] \[\therefore \] Decrease in potential energy \[=\,\,Mgh=Mg\frac{L}{2}\,\left( 1-cos\,60{}^\circ  \right)\] Kinetic energy of rotation = \[=\,\,\frac{1}{2}1{{\omega }^{2}}=\frac{1}{2}\times \frac{M{{L}^{2}}}{3}\omega \] \[l=\frac{M{{L}^{2}}}{3}\] (because rod is rotating about an axis passing through its one end)] According to law of conservation of energy \[Mg\frac{L}{2}(1-cos\,60{}^\circ )=\frac{M{{L}^{2}}}{6}{{\omega }^{2}}\Rightarrow \omega =\sqrt{\frac{3g}{2L}}\]