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Question

A particle of mass m attached to a string of length l is describing circular motion on a smooth plane inclined at an angle α with the horizontal. For the particle to complete the loop, its velocity at the lowest point should exceed 


  1. 5gl

  2. 5gl(cosα+1)

  3. 5gltanα

  4. 5glsinα


Solution

The correct option is D

5glsinα


A is the lowest point and B the highest point.
At B, in critical case tension is zero.
Let velocity of particle at B at this instant be vB.
Then from Free Body diagram

mg sin α=mv2Bl
or v2B=gl sin α
Now as h=2l sin α 
v2A=v2B+2gh=(gl sin α)+2g(2l sin α 

therefore vA=5gl sin α

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