Question

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 is

• A. $T=3 \pi \sqrt{\frac{2 l_{0}}{3 g}}$
• B. $T=4 \pi \sqrt{\frac{l_{0}}{3 g}}$
• C. $T=4 \pi \sqrt{\frac{2 l_{0}}{3 g}}$
• D. $T=2 \pi \sqrt{\frac{2 l_{0}}{3 g}}$

Answer 1

Answer: (D)

Here, the rod is oscillating about an end point $O$. Hence, moment of inertia of rod about the point of oscillation is $I=\frac{1}{3} m l_{0}^{2}$
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,
$T=2 \pi \sqrt{\frac{I}{m g l}}=2 \pi \sqrt{\frac{\frac{1}{3} m l_{0}^{2}}{m g\left(\frac{l_{0}}{2}\right)}}$
$\Rightarrow T=2 \pi \sqrt{\frac{2 l_{0}}{3 g}}$