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A flywheel rotating at 420 rpm slows dow...

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

A

22 s

B

11 s

C

44 s

D

12 s

Text Solution

Verified by Experts

The correct Answer is:
A
(a) Here, `v_0 = 420 rpm = 7 rps`
:. `omega_0 = 2 pi v_0 = 2 xx (22)/(7) xx 7 = 44 rad s^-1`
`omega = 0, prop = -2 rad s^-2`
`t = (omega - omega_0)/(prop) = (-44)/(-2) = 22 s`.
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Knowledge Check

  • A flywheel is revolving at 150 revolutions per minutes .If deccelerates at a constant rate of 2 pi rad //s^(2) , then time requred to stop it is

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    C
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