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

Options

A. `40 m//s^(2)`
B. `20 m//s^(2)`
C. `10 m//s^(2)`
D. `5 m//s^(2)`

Answer

A

Explanation

`r=20/(pi)m, a_(1)` =cosntant
`n=2^(nd)`=revolution
v=80 m/s
`omega_(0)=0, omega_(f)=v/r=80/(20//pi)=4pi ` rad/sec
`theta=2pixx2=4pi`
from `3^(rd)`
`omega^(2)=omega_(0)^(2)+2 alpha theta`
`rArr (4pi^(2))=0^(2)+2xxalphaxx(4pi)`
`alpha=2pi rad//s^(2)`
`a_(1)=alphar=2pixx20/(pi)`
`=40 m//s^(2)`