InterviewSolution
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InterviewSolution
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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. |
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Answer» 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 |
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