A flywheel of mass $25 k g$ has a radius of $0.2 m$ . It is making $240 r p m$ . What is the torque necessary to bring it to rest in $20 s$ ? If the torque is due to a force applied tangentially on the rim of the flywheel, what is the magnitude of the force?

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$- (π / 5) N m, πN$

$(π / 5) N m, πN$

$- (π / 10) N m, 2 πN$

$(π / 10) N m, 2 πN$

answer is C.

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Detailed Solution

If $α$ be the necessary retardation, then

$0 = ω_{0} - α t a = \frac{ω_{0}}{t} = \frac{(2 π \times \frac{240}{60})}{} = \frac{2 π}{5} r a d / s^{2}$

The torque $(τ) = I α$

$= \frac{m R^{2}}{2} α = \frac{25 \times 0 . 2^{2}}{2} \times \frac{2 π}{5} \frac{π}{5} N - m$

If $F$ is the required force, then $F = \frac{τ}{R} = \frac{π / 5}{0.2} = πN$

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