Physics - W.E.P. Question with Solution | TestHub

Since friction absent, we can apply the law of conservation of energy. Centre of gravity of a semicircular arc is at a distance$\left(\frac{2\mathrm{\pi }}{r}\right)$from the centre.

$\text{Initial potential energy}=\left(\mathit{\text{λ}}\text{π}\mathit{\text{r}}\right)g\left(\frac{2\mathit{\text{r}}}{\text{π}}\right)$

$\text{Final potential energy}=\left(\mathit{\text{λ}}\text{π}\mathit{\text{r}}\right)g\left(\frac{-\text{π}\mathit{\text{r}}}{2}\right)$

When the chain is completely slipped off the tube, all the links of the chain have the same velocity$v$.



$\text{Kinetic energy of chain}=\frac{1}{2}\left(\mathit{\text{λ}}\text{π}\mathit{\text{r}}\right)v^2$

$\text{From COE,}\mathit{\text{λ}}\text{π}\mathit{\text{rg}}\left(\frac{2\text{π}}{\text{r}}\right)=\left(\mathit{\text{λ}}\text{π}\mathit{\text{r}}\right)\mathit{\text{g}}\left(\frac{-\text{π}\mathit{\text{r}}}{2}\right)+\frac{1}{2}\left(\mathit{\text{λ}}\text{π}\mathit{\text{r}}\right)v^2$

$\text{From which we find,}v=\sqrt{2gr\left(\frac{2}{\text{π}}+\frac{\text{π}}{2}\right)}$