A particle of mass $$5$$ kg is free to slide on a smooth ring of radius r $$=$$ $$20$$ cm fixed in a vertical plane. The particle is attached to one end of a spring whose other end is fixed to the top point O of the ring. Initially, the particle is at rest at a point A of the ring such that ∠$$OCA=60$$∘, C being the center of the ring. The natural length of the spring is also equal to r $$=$$ $$20$$ cm. After the particle is released and slides down the ring, the contact force between the particle and the ring becomes zero when it reaches the lowest position B. Determine the force constant (in$$ ×$$102Nm$$−$$1) of the spring.

Question

A particle of mass $$5$$ kg is free to slide on a smooth ring of radius r $$=$$ $$20$$ cm fixed in a vertical plane. The particle is attached to one end of a spring whose other end is fixed to the top point O of the ring. Initially, the particle is at rest at a point A of the ring such that ∠$$OCA=60$$∘, C being the center of the ring. The natural length of the spring is also equal to r $$=$$ $$20$$ cm. After the particle is released and slides down the ring, the contact force between the particle and the ring becomes zero when it reaches the lowest position B. Determine the force constant (in$$ ×$$102Nm$$−$$1) of the spring.

Infinity Learn · Accepted Answer

According to conservation of energy between A and B, loss in gravitational potential energy $$=$$ gain in elastic potential energy $$+$$ gain in $$KEmgr2+r=12k(2r$$−$$r)2+12mv2$$                 …(i)At$$ point B, centripetal force equation is kr−$$mg=mv2r$$                         …$$(ii)Solving$$ Eqs. $$(i) and (ii), we get $$k=500Nm$$−$$1$$

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