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. (a) Find the gravitational potential energy of the chain with reference level at the centre of the sphere. (b) Suppose the chain is released and slides down the sphere. Find the kinetic energy of the chain, when it has slid through an angle θ. (c) Find the tangential acceleration dνdt of the chain when the chain starts sliding down.

Answer 1

Explanation:

(a) The gravitational potential energy of the chain at the centre of sphere:

With reference to a small element of the chain, we can calculate the gravitational potential energy of the chain at an angle ‘dθ’ at the centre of the sphere,

$\mathrm{E}_{\mathrm{p}}=\int_{0}^{L / R} m g\left(R^{2} / L\right) \cos \theta \mathrm{d} \theta$,

As,$\theta=\frac{L}{R}$  Total gravitational potential energy, $\mathrm{E}_{\mathrm{p}}=\frac{m R^{2} g}{L} \sin \frac{L}{R}$.

(b) When the chain is released from the sphere, the kinetic energy of the chain through an angle θ:

Difference in kinetic energy of the chain = Difference in potential energy of the chain

=$=\frac{m R^{2} g}{L}\left[\sin \frac{L}{R}+\sin \theta-\sin \left\{\theta+\frac{L}{R}\right\}\right]$.

(c) The tangential acceleration of the chain when it starts sliding down:

θ=0°  and Kinetic energy is, $\theta=0^{\circ} \text { and Kinetic energy is, } \frac{1}{2} \mathrm{mv}^{2}=\frac{m R^{2} g}{L}\left[\sin \left(\frac{L}{R}\right)\right]$,

On taking derivative on both sides with respect to ‘t’, the solution is  

$\frac{d v}{d t}=\frac{R_{g}}{L}\left[1-\cos \frac{L}{R}\right]$.