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

Options

A. `640 pims^(-2)`
B. `160 pims^(-2)`
C. `400 pims^(-2)`
D. `40 ms^(-2)`

Answer

D

Explanation

`omega^2=omega_0^2+2alphatheta:. alpha=(omega^2-omega_0^2)/(2theta)(1)`
`omega=v/R=80/(20//pi)=4pi,omega_0=0=2pin=4pi" " :. [n=2]`
Substituting in eqn. (1)
`alpha= ((4pi)^(2)-0)/(2xx4pi)=(16pi^2)/(8pi)=2pi, a_r=Ralpha=20/pi.2pi=40 ms^(-2)`