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.A particle moves along a circle of radius (20)/(pi) m with constant tangential acceleration.If the velocity of the particle is "80m/s" at the end of the second revolution after motion has begun,the tangential acceleration is :-

A

`640 pi m//s^(2)`

B

`160 pi m//s^(2)`

C

`40 pi m//s^(2)`

D

`40 m//s^(2)`

Text Solution

Verified by Experts

The correct Answer is:
D
`omega= (v)/(r)= (80)/(20//pi)= 4pi, omega_(0)= 0, theta= 2pi xx 2= 4pi` (Second revolution)
`omega^(2)= omega_(0)^(2) + 2 alpha theta`
`rArr alpha= (omega^(2)- omega_(0)^(2))/(2 theta)= (16pi^(2))/(2 xx 4pi)= 2pi`
Tangential acceleration `(a_(t))= r alpha= (20)/(pi) xx 2pi= 40 m//s^(2)`
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