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A wheel of moment of inertia 5xx10^(-3)k...

A wheel of moment of inertia `5xx10^(-3)kgm^(2)` is making 20 rps. It is stopped in 20 s. The angular retardation is :

A

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

B

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

C

`4pi` rad `s^(-2)`

D

`8pi` rad `s^(-2)`

Text Solution

Verified by Experts

The correct Answer is:
B
`omega_(0)=2piv=40pis^(-1).omega=0`
`alpha=(omega-omega_(0))/(t)=(0-40pi)/(20)=-2pi" rad/"s^(2)`
`therefore` angular retardation = `2pi" rad/"s^(2)`
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