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A particle of mass m slides down an inclined plane and reaches the bottom with linear velocity v. If the same mass is in the form of ring and rolls without slipping down the same inclined plane, its velocity will be: |
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Answer» Solution :Here in the first case `(1)/(2)mv^(2)=mghorv=sqrt(2gh)` In second case `v.[(2gh)/(1+(K^(2))/(R^(2)))]^((1)/(2))` As for the ring, `k=r,v.=sqrt(gh)` `therefore (v.)/(v)=(1)/(sqrt(2))ORV.=(v)/(sqrt(2))` |
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