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

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

The kinetic energy of rotation of the flywheel is $6 kJ$

The flywheel, if subjected to a rotating torque of 200Nm, will come to rest in $5 \sec$

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Detailed Solution

Moment of inertia $= {MR}^{2}$

$= 100 \times 1 \times 1 = 100 kg - m^{2}$

$ω = \frac{300}{π} \times \frac{2 π}{60} = 10 rad / s$

$KE = \frac{1}{2} {Iω}^{2} = \frac{1}{2} \times 100 \times {(10)}^{2} = 5 kJ$

$α = \frac{τ}{I} = \frac{200}{100} = 2 rad / \sec^{2}$

$Now, ω = ω_{0} - αt$

$\Rightarrow t = 5 \sec \begin{array}{r} \end{array}$

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