Question

The flexible bicycle-type chain of length $\frac{\pi r}{2}$ m and mass per unit length $\rho$ is released from rest with angle $\theta =0^{\circ }$ in the smooth circular channel of radius r = $10\pi$ m and falls through the hole in the supporting surface as shown in figure . If v is the velocity of chain in m/s as the last link leaves the slot, then v is (take ${\pi }^2=10,g=10m/s^2$).

Answer 1

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=$\frac{1}{2}m{v}^2$

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

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

Putting the values we get, $v$ = 30 $m/s$