Question

A uniform chain of length $l$ and mass $m$ lies on the surface of a smooth hemisphere of radius $R(R>l)$ with one end tied to the top of the hemisphere as shown in the figure. Gravitational potential energy of the chain with respect to the base of the hemisphere is

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

Answer 1

Answer: (B)

Given,
Mass of chain is $m$
Length is 1
Radius of sphere is $R$
$<br/>\begin{array}{l}<br/>\Rightarrow h=R\sin \theta \\ <br/>\Rightarrow dl=Rd\theta <br/>\end{array}<br/>$
The mass of dl length of the chain $dm=mdl$
Potential energy.
$<br/>\begin{array}{l}<br/>dU=(dm)gh\\ <br/>\Rightarrow dU=\frac{mdl}{1}gh=\frac{mghdl}{1}=\frac{mg{h}^2\sin \theta d\theta }{1}\\ <br/>\Rightarrow dU=\frac{mg{R}^2\sin \theta d\theta }{1}<br/>\end{array}<br/>$
On Integrating
$<br/>\begin{array}{l}<br/>U={\int }_{\theta }^{\frac{x}{2}\frac{mg{R}^2\sin \theta d\theta }{1}}\\ <br/>U=-\frac{mg{R}^2}{1}[\cos \theta {]}_{\theta }^{\frac{\pi }{2}}=-\frac{mg{R}^2}{1}\left[\cos \left(\frac{\pi }{2}\right)-\cos \theta \right]\\ <br/>U=-\frac{mg{R}^2}{1}\cos \theta =\frac{mg{R}^2}{1}\sin \left(\frac{1}{R}\right)\\ <br/>\text{ Hence, potential energy is }\frac{mg{R}^2}{1}\sin \left(\frac{1}{R}\right)<br/>\end{array}<br/>$