Question

A chain of length l and mass m lies on the surface of a smooth sphere of radius R > l with one end tied to the top of the sphere.  Find the gravitational potential energy of the chain with reference level at the centre of the sphere.

Answer 1

Let us consider a small element, which makes angle 'dθ' at the centre.

\[\therefore dm = \rho \left( \frac{m}{L} \right) Rd\theta\]

Gravitational potential energy of 'dm' with respect to centre of the sphere

\[= \left( \text{ dm }\right) \text{ g R }\cos \theta\]
\[ = \left( \frac{\text{ mg }}{\text{ L}} \right) R^2 \cos \theta d\theta\]

\[\therefore \text{ Total gravitational potential energy, }E_P = \int\limits_0^{L/R} \text{ mg }\frac{R^2}{L} \cos \theta d\theta\]
\[ E_P = \frac{m R^2 g}{L}\left[ \sin \theta \right] \left[ \text{ As }, \theta = \frac{L}{R} \right]\]
\[ E_P = \frac{m R^2 g}{L}\sin \left( \frac{L}{R} \right)\]