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.

Explanation

The angular deceleration of the wheel durign the 4.0 seconds may be obtained by the equation
`omega=omega_0-alphat`
or` alpha=(omega_0-omega)/t=(2rad/s)/(4.0)=5.0rad/s^2`
The torque applied to produce this eceleration is `Gamma=Ialpha=(0.20 kg-m^2)(5.0rad/s^2)=1.0N-m`
The angle rotated in the first two seconds is
`theta=omega_0t-1/2alphat^2`
`=(20rad/s)(2s)-1/2(5.0rad/s^2)(4.0s^2)`
`=40rad-10rad=30rad`
The work done by the torque in the first 2 secons is therefore
`W=Gammatheta=(1.0N-m)(30rad)=30J.`