Question

A smooth chain $\text{AB}$ of mass $m$ rests against a surface in the form of a quarter-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. $\sqrt{3gR}$
• C. $\sqrt{2gR(1-\frac{2}{\pi })}$
• D. $\sqrt{2gR(2-\pi )}$

Answer 1

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

Let us consider a small elemental mass $dm$ subtended at an angle $d\theta$ from centre at a height $h$ above from the horizontal surface.

Let the angle subtended by the element $dm$ at the centre be $\theta$.



From the image we get,

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

The mass of the element $dm$ is,

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

$\therefore dm=\frac{2md\theta }{\pi }$

Initial potential energy of the element $dm$ is,

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

Integrating with proper limits, we get,

$\therefore U_i={\int }_0^{{90}^{\circ }}d{U}_i$

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

Finally, it becomes horizontal and final potential energy will be zero.

$U_f=0$

If $v$ is the velocity of the chain at the bottom, so from the conservation of energy,

Decrease in PE = increase in KE

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

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

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option $\left(c\right)$ is correct.

Why this question ? If work done by net external force acting on the system is zero, then mechanical energy is conserved. $K_i+U_i=K_f+U_f$