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A pulley with radius 2 m is acted upon tangentially by a force F = (20t − 5t²) N, where t is in seconds. If the pulley’s moment of inertia about its rotation axis is 10 kg m², then the number of revolutions completed before the pulley reverses its direction of rotation is

1. greater than 3 but less than 6
2. greater than 6 but less than 9
3. greater than 9
4. less than 3

Correct answer: greater than 3 but less than 6

Solution

Torque = F*r = (20t-5t^2)*2; alpha = torque/I = (40t-10t^2)/10 = 4t - t^2. omega = 2t^2 - t^3/3, which returns to zero at t=6 s. theta = integral omega dt = 2t^3/3 - t^4/12 = 144 - 108 = 36 rad at t=6. Revolutions = 36/(2pi) = 5.7, i.e. greater than 3 but less than 6.

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