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A flywheel, in the form of a solid disc, of mass 25 kg has a radius of 0.2 m. It is making 240 rpm. If the torque is due to a force applied tangentially on the rim of the fly wheel, what is the magnitude of the force to bring it to rest in 20 sec ?

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8.48 N

3.14 N

1.57 N

6.28 N

answer is B.

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Detailed Solution

$τ = F r = I α = \frac{{mr}^{2}}{2} \frac{ω}{t} F = \frac{mrω}{2 t} = \frac{25 \times 0.2 \times 2 π \times 240}{2 \times 20 \times 60} = π = 3.14 N$

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