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The frequency of a particle performing c...

The frequency of a particle performing circular motion changes from 60 rpm to 180 rpm in 20 s . Then the angular acceleration is

A

`- 2pi rad //s^(2)`

B

`-pi rad //s^(2)`

C

`-3pi rad//s^(2)`

D

`-(pi)/(2 ) rad//s^(2)`

Text Solution

Verified by Experts

The correct Answer is:
A
`alpha =(omega _(2)- omega_(1))/(t) =(2pi (n_(2)-n_(1)))/(t)=(2pi (0-10))/(10)`
`=-2 pi rad //s^2`
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