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. $T=3\pi \sqrt{\frac{2l_0}{3g}}$
• B. $T=4\pi \sqrt{\frac{l_0}{3g}}$
• C. $T=4\pi \sqrt{\frac{2l_0}{3g}}$
• D. $T=2\pi \sqrt{\frac{2l_0}{3g}}$

Answer 1

The correct option is D $T=2\pi \sqrt{\frac{2l_0}{3g}}$

here the rod is oscilliting about an end point $O$. Hence, moment of inertia of rod about the point of oscillating is
$1=\frac{1}{3}m{l}_0^2$
moreover, length length of perpendicular $=$ distance from the oscillation axis to centre of mass of rod $\frac{l_0}{2}$
$\therefore$ Time period of oscillation
$T=2\pi \sqrt{\frac{1}{mgl}}=2\pi \frac{\sqrt{\frac{1}{3}m{l}_0^2}}{mg\left(\frac{l_0}{2}\right)}$
$\Rightarrow T=2\pi \sqrt{\frac{2l_0}{3g}}$