question_answer

A uniform rod of mass m and length \[{{l}_{0}}\]is pivoted at one end and is hanging in the vertical direction. The period of small angular oscillations of the rod (if displaced slightly from its position) is

A) \[3\pi \sqrt{\frac{2{{l}_{0}}}{3g}}\]      

B) \[4\pi \sqrt{\frac{{{l}_{0}}}{g}}\]

C) \[2\pi \sqrt{\frac{{{l}_{0}}}{3g}}\]        

D) \[2\pi \sqrt{\frac{2{{l}_{0}}}{3g}}\]

Correct Answer: D

Solution :

[d] : Here the rod is oscillating about an end point O. Hence,  moment of inertia of rod about the point of oscillation is\[I=\frac{1}{3}ml_{0}^{2}\]   Moreover,  length \[l\] of the pendulum = distance from the oscillation axis to centre of mass of rod\[={{l}_{0}}/2\] \[\therefore \]Time period of oscillation \[T=2\pi \sqrt{\frac{I}{mgl}}=2\pi \sqrt{\frac{\frac{1}{3}ml_{0}^{2}}{mg\left( \frac{{{l}_{0}}}{2} \right)}},2\pi \sqrt{\frac{2{{l}_{0}}}{3g}}\]