Question

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

Answer: (B)

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

$h_{COM}=2r/\pi$

$d_{COM-final}=\pi r/2$

$\Delta totalpot.energy=mg(h+d)$

$0.5m{v}^2=mg(h+d)$

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