Question

As shown in the figure, a smooth chain $AB$ of mass $m$ is resting on a surface in the form of a quarter of a circle of radius $R$ . After releasing it from rest, find the velocity of the chain after it comes over the horizontal part of the surface.

• A. $\sqrt{2 g R}$
• B. $\sqrt{g R}$
• C. $\sqrt{2 g R \left(1 - \frac{2}{\pi }\right)}$
• D. $\sqrt{2 g R \left(2 - \pi \right)}$

Answer 1

Answer: (C)

Consider an elementary mass $dm$ which makes an angle $d\theta $ at the centre of ring as in figure.


Mass of element is, $dm=\frac{m}{\pi / 2}.d\theta $
Gravitational potential energy of the element is,
$d U_i=(d m) g h=\left(\frac{2 m}{\pi} d \theta\right) g R(1-\cos \theta)$
So initial potential energy of the chain is
$U_i=\int d U_i$
$U_i=\frac{2 m g R}{\pi} \int\limits_0^{\frac{\pi}{2}}(1-\cos \theta) d \theta$
$U_i=\frac{2 m g R}{\pi}[\theta-\sin \theta]_0^{\frac{\pi}{2}}$
So, $U_i=\frac{2 m g R}{\pi}\left(\frac{\pi}{2}-1\right)$
and, $U_f=0$
Let speed of chain on horizontal part is $v$.
By conservation of mechanical energy
$M . E_i=M . E_f$
$\frac{2 m g R}{\pi}\left(\frac{\pi}{2}-1\right)=\frac{1}{2} m v^2$
$v=\sqrt{2 g R\left(1-\frac{2}{\pi}\right)}$