Question

A chain $AB$ of mass $m$ rests against a smooth surface in the form of a quarter of a circle of radius $R$. If it is released from rest, the velocity of the chain after it comes over the horizontal part of the surface is

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

Answer 1

Answer: (C)

$\sqrt{2gR(1-\frac{2}{\pi })}$

$dm=\left(\frac{m}{\pi /2}\right)d\theta =\left(\frac{2m}{\pi }\right)d\theta$

$h=R(1-cos\theta )$

potential energy of small mass $dm$

$d{U}_i=\left(dm\right)gh=\frac{2mgR}{\pi }(1-cos\theta )d\theta$

Integrating to find total potential energy of chain

$\therefore U_i={\int }_0^{\pi /2}d{U}_i=\frac{2mgR}{\pi }(\frac{\pi }{2}-1)$

$=mgR(1-\frac{2}{\pi })$

Now, $U_i+K_i=U_f+K_f$

$\therefore mgR(1-\frac{2}{\pi })=0+\frac{1}{2}m{v}^2$

or $v=\sqrt{2gR(1-\frac{2}{\pi })}$