Driving torque and load torque of a pmdc motor

Question

This question is a follow up to the question I had asked on electric torque of DC motors

I've done my research but couldn't find a satisfactory answer. I would appreciate if the answer was told with a fundamental concept to it.

Shown in the figure is a simple pulley system with a weight attached to it. The centre of the pulley (of radius, say R) is connected to the output shaft of the PMDC motor.

When I switch on the supply, the motor starts to rotate and produces a torque called the driving torque. The weight attached also produces it's own torque because it's a force acting at radial distance. Let this torque be the load torque.

My teacher says that as the pulley rotates at constant velocity, the torque produced by the motor = the torque due to the load (as there is no acceleration).

My question is this: if the torque due to the weight = the torque due to the motor, shouldn't the weight be stationary? How does it move up? Obviously the torque due to the motor has to be greater than the torque due to the weight to lift it off the ground right?

Moreover, how does the motor draw the right amount of current, thereby maintaining constant torque to keep the velocity constant? How does it not accelerate the weight by producing more torque than the load?

Answer 1

My question is this: if the torque due to the weight = the torque due to the motor, shouldn't the weight be stationary? How does it move up? Obviously the torque due to the motor has to be greater than the torque due to the weight to lift it off the ground right?

No, it doesn't necessarily be stationary.
Consider the translational example of a car driving on the road. To accelerate, it needs a needs a remaining force. But when the force of the motor that pulls the car forwards and the friction forces that work against it, cancel each other out, the car will move with constant velocity.
Likewise, when the torque due to the weight = the torque due to the motor, the disc will rotate with constant angular velocity and the weight will therefore have a constant velocity as well.

Moreover, how does the motor draw the right amount of current, thereby maintaining constant torque to keep the velocity constant? How does it not accelerate the weight by producing more torque than the load?

Your question is justified.
The motor does not know what torque it should 'produce' in order to cancel the torque caused by the gravitational force.
You need a controller that senses the position/velocity of the mass and will apply a voltage such that the motor torque cancels the torque caused by the gravitational force.

Answer 2

My question is this: if the torque due to the weight = the torque due to the motor, shouldn't the weight be stationary? How does it move up? Obviously the torque due to the motor has to be greater than the torque due to the weight to lift it off the ground right?

If the torques are equal then the weight is stationary.

For it to move up at a constant velocity you need to deploy a constant energy to achieve this. This is found from the mass x g x height formula for potential energy AND the rate at which you are using potential energy is power.

That power defines how quickly a weight is lifted through a certain height and is also the power of the motor's shaft (\$2\pi n T\$). That takes us to torque (T) and speed in revs per second (n).

Moreover, how does the motor draw the right amount of current, thereby maintaining constant torque to keep the velocity constant? How does it not accelerate the weight by producing more torque than the load?

A generalized motor may indeed produce enough torque to accelerate the weight too much. Sometimes folk use a control system to regulate acceleration and final velocity.