Question

The flexible bicycle type chain of length $\frac{\pi r}{2}$ and mass per unit length p is released from rest with $\theta =0^{\circ }$ in the smooth circular and falls through the hole in the supporting surface. Determine the velocity v of the chain as the last link leaves the slot.

• A. $\sqrt{gr(\frac{\pi }{3}+\frac{4}{\pi })}$
• B. $\sqrt{gr(\frac{\pi }{2}+\frac{4}{\pi })}$
• C. $\sqrt{gr(\frac{\pi }{2}+\frac{3}{\pi })}$
• D. $\sqrt{gr(\frac{\pi }{3}+\frac{3}{\pi })}$

Answer 1

The correct option is B $\sqrt{gr(\frac{\pi }{2}+\frac{4}{\pi })}$

Taking the horizontal plane as the reference, gravitational potential energy of the chain is

${=}_0^{\pi /2}\int \left(prd\theta \right)g\left(rcos\theta \right)$ $=pg{r}^2$

When the chain leaves the slot, the center of mass of the chain is at a distance of $\frac{\pi r}{4}$ from the reference.

Thus the gravitational potential energy of the chain is $-\left(p\frac{\pi r}{2}\right)g\left(\frac{\pi r}{4}\right)$

$=-\frac{pg{\pi }^2{r}^2}{8}$

Thus loss in potential energy of chain=$pg{r}^2(1+\frac{{\pi }^2}{8})$

Gain in kinetic energy of the chain=$\frac{1}{2}m{v}^2$

$=\frac{1}{2}\left(p\frac{\pi r}{2}\right)v^2=pg{r}^2(1+\frac{{\pi }^2}{8})$

$⟹v=\sqrt{gr(\frac{\pi }{2}+\frac{4}{\pi })}$