# Winding Drum of Lift

In the figure (attached), the winding drum of a lift is driven by an electric motor via a gearbox having a reduction ratio of 20:1. The direct inertia on the motor shaft is 0.05 kg m2 and that on the winding shaft is 20 kg m2. The drum, which has a diameter of 600 mm, moves a lift by means of a cable wound around it. If the lift has a mass of 500 kg, find the torque required from the motor to raise the lift with an acceleration of 2 m/s2.

I have been given the answer of 101.91 Nm to this question, but I'm not sure how to obtain this. I hope you can help. Many thanks.

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#### Solution Summary

The answer is determined, after calculating the total torque of the motor shaft. Factors of inertia and angular speed had to first be taken into account, in order to yield the total torque equation, and ultimately the torque required for the specific acceleration of the motor shaft.