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
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