Question

A chain of length ${}^{\prime }{l}^{\prime }$ mass $m$ lies on the surface of a frictionless sphere of radius ${}^{\prime }{R}^{\prime }(>\ell )$ with one end tied to the top of the sphere
Find the gravitational potential $e$ of the chain with reference level at the centre of the sphere
Suppose the chain is released to slide down the sphere. Find the kinetic $e$ of the chain when it has slid through an angle $\theta$.

Answer 1

Given,

Mass of chain is $m$

Length is $l$

Radius of sphere is $R$

$\Rightarrow h=Rsin\theta$

$\Rightarrow dl=Rd\theta$

The mass of dl length of the chain $dm=mdl$

Potential energy.

$dU=\left(dm\right)gh$

$\Rightarrow dU=\frac{mdl}{l}gh=\frac{mghdl}{l}=\frac{mg{h}^2\sin \theta d\theta }{l}$

$\Rightarrow dU=\frac{mg{R}^2\sin \theta d\theta }{l}$

On Integrating

$U={\int }_{\theta }^{\frac{x}{2}}\frac{mg{R}^2\sin \theta d\theta }{l}$

$U=-\frac{mg{R}^2}{l}[\cos \theta {]}_{\theta }^{\frac{\pi }{2}}=-\frac{mg{R}^2}{l}[\cos \left(\frac{\pi }{2}\right)-\cos \theta ]$

$U=-\frac{mg{R}^2}{l}\cos \theta =\frac{mg{R}^2}{l}\sin \left(\frac{l}{R}\right)$

Hence, potential energy is $\frac{mg{R}^2}{l}\sin \left(\frac{l}{R}\right)$