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Question
A chain of length l and mass m lies on the surface of a smooth sphere of radius R > l with one end tied to the top of the sphere. Suppose the chain is released and slides down the sphere. Find the kinetic energy of the chain, when it has slid through an angle θ.
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Solution
Let us consider a small element, which makes angle 'dθ' at the centre.
\[\therefore dm = \rho \left( \frac{m}{L} \right) Rd\theta\]
When the chain is released from rest and slides down through an angle θ,
Change in K.E. of the chain = Change in potential energy of the chain
\[= \frac{m R^2 g}{L}\sin \left( \frac{L}{R} \right) - \int\frac{g R^2}{L} \cos \theta d\theta\]
\[ = \frac{m R^2 g}{L}\left[ \sin \left( \frac{L}{R} \right) + \sin \theta - \sin \left\{ \theta + \left( \frac{L}{R} \right) \right\} \right]\]
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