Before looking at motors, you need to define a few parameters, apart from the load mass and leadscrew pitch. First, what is the motion profile? I assume the device won't be in constant motion. Is there a defined sequence of movements? What distance and elapsed time for each move? What elapsed time between moves?
Next, the load. Is it affected by gravity? Are there other external forces acting on the load? What about friction? You should also know the diameter and material of the leadscrew (for inertia calculation).
With all of that, you can proceed to calculate peak torque, RMS torque, and speed required at the drive end of the lead screw. That will indicate what sort of motor you need.
Then you need to consider the ratio of the reflected load inertia to the motor's rotor inertia. A high ratio will necessitate de-tuning the position loop in order to get stability. Ideally, the ratio is 1:1 (hard to achieve), but under 3:1 is generally very good. Under 6:1 is generally acceptable. If the coupling is "stiff", 10:1 can be OK.
If the inertia ratio needs reduction, first look at a bigger motor. Otherwise, a gearbox may be required. Reflected inertia reduces as the square of the gear ratio. But speed requirements increase.
A constant current means, for an ideal motor, a constant torque. This is approximately true for real motors. It doesn't matter what you attach to the motor, or how fast it's turning.
What you seem to be missing is Newton's second law of motion. It states that force is the product of mass and acceleration:
$$ F = ma $$
The constant current you supply to the motor is one force. The weight opposes that force. The difference is the net force, \$F\$ in this equation, and \$m\$ is the mass of the weight, plus the mass of the rotor and the string and everything else the motor must move.
You set current to be sent to the motor so that the torque applied is 10 in-lbs without any load.
Not possible. There is nothing for the motor to "torque against". This is the mechanical equivalent of trying to develop 10 volts across a dead short. The motor will rapidly spin at its maximum speed, and the back-EMF will rise to the driving voltage such that your driving electronics are unable to supply enough voltage above the back-EMF to make enough current to have that much torque.
Let's just say you determine how much current is required for 10 in-lbs of torque, and you drive your motor with a constant-current supply set to that.
What happens when the torque from the weight/load is 5 in-lbs?
Assuming that the rotor and the string are massless and frictionless, the weight will be accelerated upwards by the net 5 in-lbs of torque (motor's 10 in-lbs, less 5 in-lbs from the weight). The rate of the acceleration is determined by the mass of the weight and Newton's law above.
As the speed of the motor changes (the weight is accelerating), the back-EMF also changes. Your constant-current supply to the motor will have to apply an increasing voltage to maintain the same current. Electrical power thus goes up, as does mechanical power.
What happens when the torque from the weight/load is 10 in-lbs?
Motor torque balances weight torque. However fast the weight is moving (if at all), it keeps doing that. Newton's first law applies.
What happens when the torque from the weight/load is 15 in-lbs?
The weight will accelerate downward, overpowering the motor. However, it won't be a free-fall. The motor cancels some of the force of the weight, resulting in a slower acceleration downwards.
If the weight overpowers the motor, then eventually it can get the motor to run backwards, relative to the way it would run if there were no load. When this happens, the back-EMF now adds (instead of subtracts) from the voltage you apply to the motor. At some point, your controller, which is attempting to maintain a constant current, must apply a negative voltage to maintain that current. In other words, the back-EMF is sufficient to create the necessary torque on its own: your controller must oppose it.
This is perfectly symmetrical with the first case, where the motor was overpowering the weight. In that case, electrical and mechanical power went up (without bound, if you let them). In this case, electrical and mechanical power go down (negative, if you let them). Energy is conserved because you are changing the gravitational potential of the weight.
The need to resist the back-EMF usually means storing electrical energy in a capacitor or battery, or using it to heat a resistor. If you can't do this fast enough, then the motor will create more torque than your desired 10 in-lbs, and you have hit the limits of your "constant current" driver.
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Best Answer
Some motor basics.
A motor's stall current is determined by the resistance of the coil. (quoted at the nominal operating voltage). If the shaft is clamped so it cannot rotate then the 'motor' will act like any other resistor and follow Ohm's law - so yes, the stall current will increase with increasing applied voltage.
Once the rotor starts to rotate it will induce a back emf so the voltage 'seen' across the rotor will be reduced. The net current through the motor will be reduced.
Increasing the applied voltage will increase the speed (how much depends on the motor). With increasing speed comes increasing loss (friction, windage etc.) so there will be a corresponding increase in current.
Applying a load (torque) to the output shaft slows the rotation speed and reduces the back emf. This in turn will increase the motor current. Applying too much torque may reduce the shaft speed to 0 and this is once again back to a stall.