1.

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.

Answer» The mass is distributed in chain uniformly along its length.
Choose a small element of chain of width `dtheta` at an angle `theta` from the vertical.
The mass of the element, `dm=(m/lRd theta)`
The gravitational potential energy of the element `dU=(dm)gy`
Thus, the gravitational potential energy of whole chain
`U=int(dm)gy`
`=underset0overset((l//R))int(m/lRd theta)g(Rcostheta)`
`=(mR^2g)/(l)underset0overset((l/R))intcosthetad theta`
`=(mgR^2)/(l)|sintheta|_0^(l//R)=(mgR^2)/(l)sin(l/R)`


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