Question

A uniform chain of length $\pi r$ lies inside a smooth semicircular tube AB of radius $r$. Assuming a slightly disturbance to start the chain in motion, the velocity with which it will emerge from the end B of tube will be:

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

Answer 1

The correct option is B $\sqrt{2gr(\frac{2}{\pi }+\frac{\pi }{2})}$

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

$\therefore dU=\left(dm\right)gh=\left(\frac{m}{\pi }d\theta \right)gr\sin \theta$

$U_i={\int }_0^{\pi }dU=\frac{2mgr}{\pi }$

Now, $K_i+U_i=K_f+U_f$

$\therefore 0+\frac{2mgr}{\pi }=\frac{1}{2}m{v}^2-mg\left(\frac{\pi r}{2}\right)$

$\therefore v=\sqrt{2gr(\frac{2}{\pi }+\frac{\pi }{2})}$