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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. |
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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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