NSEP 2015 Problem 5 | Rotational Mechanics & Small Oscillation

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 place. The frequency f of small oscillation of the semicircular rod is

(a) $\frac{1}{2\pi} \sqrt{\frac{g\pi}{2l}}$ (b) $\frac{1}{2\pi} \sqrt{\frac{g\sqrt{\pi^2+4}}{2l}}$ (c) $\frac{1}{2\pi} \sqrt{\frac{g(\pi+4)}{l}}$ (d) $\frac{1}{2\pi} \sqrt{\frac{g(\pi^2+1)}{2\pi l}}$

Concept of Physics H.C. Verma

University Physics by H. D. Young and R.A. Freedman

Fundamental of Physics D. Halliday, J. Walker and R. Resnick

National Standard Examination in Physics(NSEP) 2015-2016

Option-(b) $\frac{1}{2\pi} \sqrt{\frac{g\sqrt{\pi^2+4}}{2l}}$

Suppose, we have $\Gamma = -k\theta $, where $\theta $ is the angular displacement and $\Gamma $ is the torque.

Then, we can write the angular acceleration $\alpha $ as,

$$ \alpha = \frac{\Gamma}{I}= -\frac{k}{I}\theta = w^2 \theta$$

where I is the moment of inertia, i.e., mass's counterpart in rotational mechanics.

Finally we have, $w = 2\pi f= \sqrt{\frac{\alpha }{\theta}} = \sqrt{\frac{\Gamma }{I\theta}} $, where f is the frequency of the oscillation.

Now for a force $\vec{F}$, $\Gamma = F r \sin(\theta)$. Now using small angle approximation $\sin(\theta) = \theta $. Hence,

$$ f = \frac{1}{2\pi }\sqrt{\frac{Fr}{I}} $$

Now, we can consider the semi-circle of mass m as a point of mass m concentrated into it's centre of mass. This will give us as shown in the figure.

Now, for a semi-circle the centre of mass is at a height of $\frac{2R}{\pi}$ distance above the radius, i.e., $AO = \frac{2R}{\pi}$, where $AP=R=\frac{l}{\pi}$.

Now, $F r = mg l_{c} = mg \sqrt{R^2 + (\frac{2R}{\pi})^2}$

and $I = 2mR^2$

Putting these values gives us,

$f=\frac{1}{2 \pi} \sqrt{\frac{F r}{I}}$

=$\frac{1}{2 \pi} \sqrt{\frac{g \sqrt{\pi^{2}+4}}{2 \pi R}}$

=$\frac{1}{2 \pi} \sqrt{\frac{g \sqrt{\pi^{2}+4}}{2 l}}$

Physics Olympiad Program at Cheenta

Subscribe to Cheenta at Youtube