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the moment of inertia of a wheel is 1000 kgm^2/. At a given instant, its angular velocity is 10 radian per seconnd. After the wheel rotates through an angle of 100 radians , the wheels angular velocity is 100 radian per second find torque applied on the wheel and the increase in rotational kinetic energy.
Asked by gcbharat986 | 11 Mar, 2019, 05:03: PM
If the wheel is rotating with uniform acceleration α, then its angular velocity at a given instant and that of a later instant are related as

ω2 = ωo2 + 2 α θ , where ωo and ω are initial and  later angular velocities, α is angular acceleration
and θ is angular displacement between two angular speeds.

hence we have 1002 = 102 + 2×α×100  or  α = 49.5 rad/s2

If I is moment of inertia, Torque = I α  = 1000×49.5 = 49,500 N-m

increase in rotational kinetic energy = (1/2) I (ω2 - ωo2 ) = (1/2)×1000×(1002 - 102) = 495×104 J
Answered by Thiyagarajan K | 11 Mar, 2019, 05:52: PM
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