Question

A smooth chain $AB$ of mass $m$ rests against a 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

The correct option is C $\sqrt{2gR(1-\frac{2}{\pi })}$

$h=R(1-cos\theta )$
$dm=\left(\frac{m}{\pi /2}\right).d\theta$
$=\frac{2md\theta }{\pi }$
$d{U}_i=\left(dm\right)gh$
$=\frac{2mgR(1-cos\theta )d\theta }{\pi }$
$\therefore U_i={\int }_0^{{90}^{\circ }}d{U}_i$
$=mgR(1-\frac{2}{\pi })$
$U_f=0$
Decrease in PE = increase in KE.
$U_i-U_f=\frac{1}{2}m{v}^2$
$mgR(1-\frac{2}{\pi })=\frac{1}{2}m{v}^2$
$v=\sqrt{2gR(1-\frac{2}{\pi })}$