Question

A weight of 500 kg is being lifted up at a uniform speed of 1.5 M/S by a winch driven by a motor running at a speed of 1000 rpm. The moments of inertia pf the motor and winch are 0.3 and 0.3 kg-m respectively. Calculate the motor torque and the equivalent moment of inertia referred to the motor shaft. In the absence of weight, motor develops a torque of 100 N-m when running at 1000 rpm.

Answer 1

Since the weight is lifted up at uniform speed, that means the torque applied by weight on winch is equal to the torque applied by motor.

Lets calculate radius r of motor.

Given,

motor speed = 1000rpm = 50/3 revolution per sec

speed at which weight is lifted = 1.5 m/s

So 50/3 revolution of motor makes 1.5m .

So $2\pi r \times 50/3= 1.5$ , therefore r = 0.014m

So total motor torque = 100 + (5000 * 0.014) = 171.6 N-m