Question

Calculate the torque produced by an industrial motor (200 kw) when the angular speed is 4000 rev/min. A coil with a very small mass has a diameter of 0.500 m. it is attached to the arm of the aforementioned motor and the output power of the motor is used to lift a load attached to a rope wrapped around the coil. how heavy a load can the motor lift at constant speed

Answer 1

The torque produced by the industrial motor is 0.478 Nm.

Step-by-step explanation:

To calculate the torque produced by an industrial motor, we need to use the formula:

Torque = Power / Angular Velocity

Given that the power of the motor is 200 kW and the angular speed is 4000 rev/min, we need to convert the angular speed to radians per second. There are 2π radians in one revolution, so the angular speed in radians per second is:

Angular Speed = (4000 rev/min) * (2π radians/rev) * (1 min/60 s)

= 418.88 rad/s

Now we can substitute the values into the formula:

Torque = (200 kW) / (418.88 rad/s) = 0.478 Nm

The torque produced by the motor is 0.478 Nm.