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.

• A. $\frac{m{R}^{2}g}{l}\sin \left(\frac{l}{R}\right)$
• B. $\frac{m{R}^{2}g}{2l}\sin \left(\frac{l}{R}\right)$
• C. $\frac{2m{R}^{2}g}{l}\sin \left(\frac{l}{R}\right)$
• D. $\frac{m{R}^{2}g}{l}\sin \left(\frac{l}{2R}\right)$

Answer 1

The correct option is A $\frac{m{R}^{2}g}{l}\sin \left(\frac{l}{R}\right)$

Let $\theta$ be the angle the other end and the vertical from center subtends at the center.

Let $\lambda$ be the mass per unit length of the chain=$\frac{m}{l}$.

Gravitational potential energy of the chain=${\int }_0^{\theta }\lambda Rd\theta g\left(Rcos\theta \right)$

$=\frac{m{R}^{2}g}{l}sin\theta$

Since $R>l$, $\theta =\frac{l}{R}$

$⟹GPE=\frac{m{R}^{2}g}{l}sin\frac{l}{R}$