Question

The string of a simple pendulum is replaced by a thin uniform rod of mass $M$ and length $L$. The mass of bob is $m$. If it is allowed to oscillate with small amplitude. the period of oscillation is:

• A. $2\pi \sqrt{\frac{2(M+3m)L}{3(M+2m)g}}$
• B. $2\pi \sqrt{\frac{3(M+2m)L}{2(M+3m)g}}$
• C. $2\pi \sqrt{\frac{(M+2m)L}{(M+3m)g}}$
• D. $2\pi \sqrt{\frac{2(M+m)L}{3(M+2m)g}}$

Answer 1

The correct option is A $2\pi \sqrt{\frac{2(M+3m)L}{3(M+2m)g}}$

$\theta =\frac{x}{L/2}$

$x=\theta \frac{L}{2}$
$x^{\prime }=\theta .L$

$I\alpha +Mg\times \frac{\theta L}{2}+mg\theta .L=0$

$\frac{M{L}^2}{3}+M{L}^2\theta +\theta L[\frac{Mg}{2}+mg]=0$

$\omega =\sqrt{\frac{\frac{Mg+2mg}{2}}{\frac{ML+3ML}{3}}}$

$\omega =\sqrt{\frac{3}{2}\frac{(Mg+2mg)}{ML+3mL)}}$

$T=2\pi \sqrt{\frac{2}{3}\frac{(M+3m)}{M+2m)}\frac{L}{g}}$