This is a continuation of me trying to understand torque and stepper motors in my other question. I'm trying to understand the torque a motor would be required to generate to lift a small weight, and the formulas involved.

The first part of my question is to verify if I am calculating this correctly:

Let's say I have a 450 g weight (roughly half a pound) then the force of gravity pulling it down is:

$\begin{align} F &= ma \\ &= 0.450 \:\mathrm{kg} * 9.8 \:\mathrm{m}/\mathrm{s}^2 \\ &= 4.41 \:\mathrm{N} \\ \end{align}$

If I have a stepper motor with a spindle for my string that pulls up my motor with a radius of 5 cm. I think my torque needed would be:

$\begin{align} T &= Fr \\ &= F * 0.005 \\ &= 0.022 \:\mathrm{Nm} \\ \end{align}$

So now if I want to move that weight I need to find a stepper motor that can output more than 0.022 Nm of torque, right?

The follow-on to my question is that if I want to see how fast I can move it then I need to look at a Torque speed curve, right?

My confusion is this: do I have to ensure that I'm moving slow enough to get the torque I need, or does that curve say if you need this torque you won't be able to go above this speed because the motor won't let you?

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    $\begingroup$ "450g weight" is nonsensical. That should be "450 g mass". $\endgroup$ – Olin Lathrop May 14 '15 at 12:28
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    $\begingroup$ It's not nonsensical at all, we know exactly what they mean. It's just not formally correct. $\endgroup$ – Ethan48 May 14 '15 at 14:33
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    $\begingroup$ @OlinLathrop Calibration weight. $\endgroup$ – Air May 14 '15 at 17:03
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    $\begingroup$ @OlinLathrop - Tone down the rhetoric, please. While you are correct regarding the terminology, you are incorrect regarding the importance of terminology in this particular case. I agree that usage of correct units is important, but the hyperbole regarding units is unnecessary. $\endgroup$ – user16 May 14 '15 at 17:13
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    $\begingroup$ @OlinLathrop you are right I should have said mass and not weight that was laziness on my part. Although I thought the broken windows philosophy was dis-proven :) $\endgroup$ – confused May 14 '15 at 18:08

You have the right concept, but slipped a decimal point. 5 cm = 0.05 m. The gravitational force on your 450 g mass is 4.4 N as you say, so the torque just to keep up with gravity is (4.4 N)(0.05 m) = 0.22 Nm.

However, that is the absolute minimum torque just to keep the system in steady state. It leaves nothing for actually accellerating the mass and for overcoming the inevitable friction.

To get the real torque required, you have to specify how fast you want to be able to accellerate this mass upwards. For example, let's say you need at least 3 m/s². Solving Newton's law of F = ma:

(0.450 kg)(3 m/s²) = 1.35 N

That, in addition to the 4.4 N just to balance gravity means you need 5.8 N upwards force. At 0.05 m radius, that comes out to a torque of 0.28 Nm. There will be some friction and you want some margin, so in this example a 0.5 Nm motor would do it.

Note also that torque isn't the only criterion for a motor. Power is another important one. For that you have to decide what's the fastest speed you want to be able to pull the mass upwards at. Let's say 2 m/s for sake of example. From above, we know the highest upwards force is 5.8 N.

(5.8 N)(2 m/s) = 11.6 Nm/s = 11.6 W

After accounting for some losses due to friction and leaving a little margin, the motor should be rated for about 15 W minimum.

  • $\begingroup$ I have a comment sir. Your explanation was really good and clear but I have a question: let's say I want to spin the motor very fast so I can pull up the weight fast. Let's say I want to pull it at 10m/s with constant speed. If I provide the initial 10 m/s speed with a push from my hand so the motor reaches that speed, will the motor only have to provide 0.22Nm plus a little more to overcome gravity and friction in order to keep the 10 m/s speed? The same applies to 100 m/s? If I, somehow, give the initial push to the motor achieve 100 m/s, will the motor only need to provide 0.22Nm? $\endgroup$ – Samul Apr 19 '16 at 15:13
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    $\begingroup$ @Samul: Yes, you seem to have the right concept regarding torque. Note that the power required to sustain the upward speed still has to be there. $\endgroup$ – Olin Lathrop Apr 19 '16 at 18:35
  • $\begingroup$ thank you so much! I couldnt believe what I was saying was correct... so if I want to keep a speed of 100000m/s I just need to accelerate till that speed and keep a realllly small torque to keep lifting the weight? That's amazing! Of course I think that friction may rise exponentially so maybe at a high speed I will have to provide a huge torque in order to fight the friction, am I correct? This is my last question, I promise :) $\endgroup$ – Samul Apr 19 '16 at 20:47
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    $\begingroup$ @Samul: At any steady speed, you only need enough torque to counter gravity and friction. However, there are two gotchas with this. First some part of the friction will be proportional to speed (viscous friction). Air drag is even worse, proportional to the square of the speed. Second, the power required is the speed times the torque. In a ideal system with no friction, the torque to maintain 10 m/s and 1000 m/s is the same, but the power required to produce that torque at the higher speed is 100 times more. If you needed 15 W at 10 m/s, then you'll need 1.5 kW at 1000 m/s. $\endgroup$ – Olin Lathrop Apr 19 '16 at 21:46
  • $\begingroup$ thank you so much!!! You were very very very clear! I finally could understand it. It may look obvius at last but my misconception abouth phisics made this a little confused, but now you made it very clear! $\endgroup$ – Samul Apr 20 '16 at 13:15

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