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

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,
$d{U}_i=(dm)gh=\left(\frac{2m}{\pi }d\theta \right)gR(1-\cos \theta )$
So initial potential energy of the chain is
$U_i=\int d{U}_i$
$U_i=\frac{2mgR}{\pi }\underset{0}{\overset{\frac{\pi }{2}}{\int }}(1-\cos \theta )d\theta$
$U_i=\frac{2mgR}{\pi }[\theta -\sin \theta {]}_0^{\frac{\pi }{2}}$
So, $U_i=\frac{2mgR}{\pi }\left(\frac{\pi }{2}-1\right)$
and, $U_f=0$
Let speed of chain on horizontal part is $v$.
By conservation of mechanical energy
$M.E_i=M.E_f$
$\frac{2mgR}{\pi }\left(\frac{\pi }{2}-1\right)=\frac{1}{2}m{v}^2$
$v=\sqrt{2gR\left(1-\frac{2}{\pi }\right)}$