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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)/(8pi r)`

B

`(v^(2))/(6pi r)`

C

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

D

` (v^(2))/(2pi r)`

Text Solution

Verified by Experts

The correct Answer is:
A
In the equilibrium `v^(2) = u^(2) + 2as, `
we have a = tangential acceleration .
and for the circular motion `,s= r theta , u=0`
` therefore v^(2) =2 ar theta `
But ` theta = 4pi` in two revolutions,
` therefore v^(2) = 2ar ( 4pi ) = 8 pi ar `
` therefore a=(v^(2))/(8pi r)`
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