A wheel rotating at an angular speed of 20 rad/s ils brought to rest by a constant trouque in 4.0 secons. If the moiment 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

Here, `omega_(1) = 20 rad//s, omega_(2) = 0, t = 4 sec`.
`I = 0.20 kg m^(2), W = ?`
`alpha = (omega_(2) - omega_(1))/(t) = (0- 20)/(4) =- 5 rad//s^(2)`
Negative sign is for deceleration.
Torque, `tau = I alpha = 0.20(5) = 1.0 N-m`
Angle traced in first two seconds
`theta = omega_(1)t + (1)/(2) alpha t^(2) = 20 (2) + (1)/(2) (-5) (2)^(2)`
`= 30 radian`
`:.` Work done by the torque in 1st two seconds,
`W = tau theta = 1.0(30) = 30` joule