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A particle moves along a circle of radiu...

A particle moves along a circle of radius r with constant tangential acceleration. If the velocity of the particle is v at the end of second revolution, after the revolution has started, then the tangential acceleration is

A

`(v^2)/(8pir)`

B

`(v^2)/(6pir)`

C

`(v^2)/(4pir)`

D

`(v^2)/(2pir)`

Text Solution

Verified by Experts

The correct Answer is:
A
According to question, applying equation of motion ,
`v^2 =u^2 +2as`
[Symbols have their usual meanings ]
Given ,u=0
`S=2xx2pir=4pir`
`so , " "v^2 =2axx4pirrArr a =(v^2)/(8pir)`
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Knowledge Check

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