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A flywheel rotating at 420 rpm slows dow...

A flywheel rotating at `420 rpm` slows down at a constant rate of `2 rad s^-2` The time required to stop the flywheel is.

A

22 s

B

11 s

C

44 s

D

12 s

Text Solution

Verified by Experts

The correct Answer is:
A
Here, `v_(0) = 420 rpm = 7 rps`
`omega_(0)= 2 xx 22/7 xx 7 = 44 rads^(-1)`
`omega=0, alpha=-2 rad s^(-2)`
`therefore t=(omega-omega_(0))/(alpha) = -44/-2=22 s`
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