Question

A thin rod of length $l$ in the shape of a semicircle is pivoted at one of its ends such that it is free to oscillate in its own plane. Find the frequency $f$ of small oscillations of the semicircular rod.

Answer 1

Let mass of rod be $m$, radius of semicircle $r=\frac{l}{\pi }$

Moment of inertia of semicircular ring $I_{CM}=\frac{m{\pi }^2}{4}$ (about axis passing through its centre and perpendicular to the plane of it)

$=\frac{m(\frac{l}{\pi }{)}^2}{4}$

$=\frac{m{l}^2}{4\pi }$

Period of oscillation $T=2\pi \sqrt{\frac{I}{mg\left(d\right)}}$

Where, $d$=distance between point of suspension and C.M

$I$=Moment of inertia about point of suspension

$=I_{CM}+m{p}^2$

$=\frac{m{l}^2}{4{\pi }^2}+\frac{m{l}^2}{{\pi }^2}$

$=\frac{5m{l}^2}{4{\pi }^2}$

$\Rightarrow 2\pi \sqrt{\frac{\frac{5m{l}^2}{4{\pi }^2}}{mgr}}$

$\Rightarrow 2\pi \sqrt{\frac{\frac{5m{l}^2}{4{\pi }^2}}{mg\left(\frac{l}{\pi }\right)}}$

$\Rightarrow 2\pi \sqrt{\frac{5l}{4g}}$

So, frequency $f=\frac{1}{T}=\frac{1}{2\pi }\sqrt{\frac{4g}{5l}}$