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A wheel rotating at an angular speed of ...

A wheel rotating at an angular speed of 20 rad/s is brought to rest by a constant torque in 4.0 seconds. If the moment of inertia of the wheel about the axis of rotation is 0.20 `kg-m^2` find the work done by the torque in the first two seconds.

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`omega_1=20` rad/sec,`omega_2=0`, t=4 sec.So angular retardation`alpha=(omega_1-omega_2)/t=20/4=rad/sec^2` Now angular speed after 2 sec `omega_2=omega_1-alphat=20-5xx2=10rad/sec`Work done by torque in 2 sec 2 E2 = loss in kinetic`=1/2I(omega_(1)^(2)-omega_(2)^(2))=1/2(0.20)((0.20)-(10)^2)=1/2xx0.2xx300=30J.`
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