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Q.

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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a

The kinetic energy of rotation of the flywheel is 5KJ

b

The moment of inertia of the flywheel is 100kg-m2

c

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

d

All of the above

answer is A, B, C, D.

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Detailed Solution

Moment of inertia = MR2
=100kg x 1 x 1m2 =10kg - m2
Energy of rotation =12×100×2π×300π×602=5KJ
=ω0=2π×300π×60=10rad/sec;ω=0 α=200100rad/sec2 Now, ω=ω0αt,t=5sec

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