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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

`6.284 rad//s^(2)`

B

`3.142 rad /s^(2)`

C

`0.6284 rad//s^(2)`

D

`0.142 rad //s^(2)`

Text Solution

Verified by Experts

The correct Answer is:
C
`alpha =(2pi (n_(2) -n_(1)))/(20) =(2pi (3-1))/(20)`
`=(2pi)/(10)=0.6284 rad//s^(2)`
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