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A flywheel is attached to an engine to keep its rotational speed between 100 rad/s and 110 rad/s. If the energy fluctuation in the flywheel between these two speeds is 1.05 kJ then the moment of inertia of the flywheel is _______ kg.m2 (round off to 2 decimal places).

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Concept:

ω1 = maximum angular speed, ω2 = minimum angular speed,

ΔKE = Change in Kinetic energy, If= Moment of Inertia of Flywheel

Therefore, \({\rm{\Delta KE\;}} = \frac{1}{2}\; \times {I_f} \times \left( {\omega _1^2 - \omega _2^2} \right)\)

Calculation:

ω1 = 110 rad/sec, ω2 = 100 rad/sec, ΔKE = 1.05 kJ = 1050 J

1050 = \(\frac{1}{2}\; \times {I_f}\; \times \left( {{{110}^2} - {{100}^2}\;} \right)\)

\({I_f} = 1\;kg.{m^2}\)

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