As shown in the figure, a chain of mass $m$ is placed on a smooth quarter circular portion of radius $R$ . End $A$ is tied with a wedge while the remaining chain is free, then the minimum work required to be done by the external agent to make the chain horizontal keeping point $A$ fixed, is

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Center of mass of the chain is at 2 R π from horizontal As shown in figure ....... Height gain by center of mass is 2 R π Wg = - m g 2 R π W ext. = - Wg = m g 2 R π Alternate solution U i = ∫ - dm gy = g ∫ R cos θ m π 2 d θ = - 2 Rg m π ∫ 0 π 2 cos θ d θ = - 2 Rg m π [ sin θ ] 0 π / 2 = - 2 m gR π U f = 0 ⇒ Δ U = w ext = 2 m gR π

Physics - Center of Mass Question with Solution | TestHub

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