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The moment of inertia of a wheel is 1000...

The moment of inertia of a wheel is `1000 kg-m^(2)`. At a given instant, its angular velocity is `10 rad//s`. After the wheel rotates through an angle of `100` radians the wheel's angular velocity is `100 rad//s`. Find the torque applied on wheel.

Text Solution

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We know that `omega^(2) -omega_(0)^(2) = 2 alpha theta`
:. `alpha = (omega^(2) - omega_(0)^(2))/(2 theta) = ((100)^2-(10)^2)/(2 xx 100)`
=`49.5 rad//s^2`
`tau = I alpha = 1000 xx 49.5 xx 10^(4) N-m`.
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