A chain of length l and mass m lies of the surface of a smooth hemisphere of radius `Rgtl` with one end tied to the top of the hemisphere. Taking base of the hemisphere as reference line, find the gravitational potential energy of the chain.

Options

A. (a) `(mR^2g)/(l)(l/R-sinl/R)`
B. (b) `(mR^2g)/(2l)(l/R-si nl/R)`
C. (c) `(mR^2g)/(2l)(si n l/R-l/R)`
D. (d) `(mR^2g)/(l)(si n l/R-l/R)`

Answer

D

Explanation

The mass of elemental length of chain subtending an angle `dtheta`
at the centre `=m/l(Rdtheta)`.
`:.` Its `PE=m/l(Rd theta)g[-(R-Rcostheta)]`
The negative sign implies that the elemental length is below the datum level. Therefore,
`dU=(-mgR^2)/(l)(1-costheta)d theta`
Therefore, total PE of the chain
`U=underset0overset(l//R)int-(mgR^2)/(l)(1-cos theta)d theta`
`implies -(mgR^2)/(l)[theta-sin theta]_0^(l//R)=-(mgR^2)/(l)[1/R-si n l/R]`