Question

A uniform rope of mass $m$ and length $l$ is held on cylindrical edge of a platform. The upper end of the rope coincides with the boundary between the top of cylindrical edge and horizontal top of the platform. Radius of the cylindrical edge is $r$. The rope is pulled gradually on the horizontal top of the platform. Find work done by the agency pulling the rope.

Answer 1

mg \left[ \frac{l}{2} + r \left(1 - \frac{\pi}{2}\right) + \frac{r^2}{l} \left(1 - \frac{\pi}{2} + \frac{\pi^2}{8}\right) \right]

Answer 2

The work done by the agency is equal to the change in the potential energy of the rope, as it is pulled gradually (no change in kinetic energy) and there is no friction. The potential energy is calculated with respect to the horizontal platform as the reference ($U=0$).

1. Initial Potential Energy ($U_i$): The rope is divided into two parts: a segment on the curved cylindrical edge and a segment hanging vertically.

   • The potential energy of the curved part (length $\frac{\pi r}{2}$) is calculated using the y-coordinate of its center of mass, which is $-\frac{2r}{\pi}$. This gives $U_1 = -\frac{mgr^2}{l}$.

   • The potential energy of the vertically hanging part (length $l - \frac{\pi r}{2}$) is calculated using the y-coordinate of its center of mass, which is $-r - \frac{1}{2}(l - \frac{\pi r}{2})$. This gives $U_2 = -\frac{mg}{l}\left(l - \frac{\pi r}{2}\right) \left(r + \frac{l}{2} - \frac{\pi r}{4}\right)$.

   • Total initial potential energy $U_i = U_1 + U_2$.

2. Final Potential Energy ($U_f$): When the entire rope is on the horizontal platform, its potential energy is $U_f = 0$.

3. Work Done ($W$): $W = U_f - U_i = -U_i$. Substituting the expressions for $U_1$ and $U_2$ and simplifying yields the final result.