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A fly wheel rotating at 420 rpm slows down at a constant rate of 2 $rad s^{- 2}$ The time required to stop the flywheel is

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22 s

12 s

44 s

11 s

answer is A.

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Detailed Solution

Here, $v_{o} = 420 rpm = 7 rps$

$ω_{o} = 2 {πv}_{o} = 2 \times \frac{22}{7} \times 7 = 44 rad s^{- 1}$

$ω = 0, α = - 2 rad s^{- 2}$

t = $\frac{ω - ω_{o}}{α} = \frac{- 44}{- 2} = 22 s$

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