A ring-type flywheel of mass 100 kg and diameter 2m is rotating at the rate of $(300 / π)$ revolutions per minute. Then:

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The kinetic energy of rotation of the flywheel is 5KJ

The moment of inertia of the flywheel is $100 kg - m^{2}$

The flywheel, if subjected to a rotating torque of 200N -m, will come to rest in 5sec

All of the above

answer is A, B, C, D.

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Detailed Solution

Moment of inertia $= {MR}^{2} = 100 kg \times 1 \times 1 m^{2} = 100 kg - m^{2}$

Energy of rotation $= \frac{1}{2} \times 100 \times {[2 π \frac{300}{π \times 60}]}^{2} = 5 KJ$

$ω_{0} = 2 π \times \frac{300}{π \times 60} = 10 rad / \sec : ω = 0,$

When a 200N- m torque is applied

$α = \frac{200}{100} rad / \sec^{2} Now, ω = ω_{0} - αt, t = 5 \sec$

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