A is the bottom most point of a particle describing a vertical circle of radius R. At point B, the acceleration of the particle is g√11. Find the velocity of the particle at top point C :
A
[gR(√10−2)]12
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B
[gR]12
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C
[gR(√11−2)]12
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D
None of these
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Solution
The correct option is A[gR(√10−2)]12 Net acceleration at point B is given by a=(a2C+a2T)12=g√11
But aT=g a2C=11g2−g2=10g2 ac=g√10=V2BR V2B=gR√10
Using kinematics equation, V2C=V2B−2gR=gR(√10−2) VC=√gR(√10−2)