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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
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Verified Answer
The correct answer is:
$T=2 \pi \sqrt{\frac{2 l_0}{3 g}}$
Here, the rod is oscillating about an end point $O$. Hence, moment of inertia of rod about the point of oscillating is
$\mathrm{I}=\frac{1}{3} \mathrm{~m} l_0^2$
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,
$\begin{aligned} & \mathrm{T}=2 \pi \sqrt{\frac{\mathrm{I}}{\mathrm{mg} l}}=2 \pi \sqrt{\frac{\frac{\mathrm{l}}{3} \mathrm{~m} l_0^2}{m g\left(\frac{l_0}{2}\right)}} \\ & \Rightarrow \quad \mathrm{T}=2 \pi \sqrt{\frac{2 l_0}{3 \mathrm{~g}}} \\ & \end{aligned}$
$\mathrm{I}=\frac{1}{3} \mathrm{~m} l_0^2$
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,
$\begin{aligned} & \mathrm{T}=2 \pi \sqrt{\frac{\mathrm{I}}{\mathrm{mg} l}}=2 \pi \sqrt{\frac{\frac{\mathrm{l}}{3} \mathrm{~m} l_0^2}{m g\left(\frac{l_0}{2}\right)}} \\ & \Rightarrow \quad \mathrm{T}=2 \pi \sqrt{\frac{2 l_0}{3 \mathrm{~g}}} \\ & \end{aligned}$
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