# How to calculate the required torque in a static equilibrium system

I am working on a personal project where I need to select a motor to rotate a system that is in static equilibrium. The motor torque specification is given in kg.cm.

This is what I have: Bar : 2 meters in length, weighing 1kg (for simplicity) Axle: 10mm (5mm radius) with a bearing to reduce friction. F : F1 and F2 both weigh 10kg

(F1)       (F2)
10kg       10kg
-------o-------
^
T1               -T2

I am not entirely sure what all the parameters are to arrive at a formula to use. Also, what formula can I use to calculate the torque required to rotate the system at a constant speed of 3rpm ?

Thank you in advance.

I would disregard friction and drag.

let's say an estimated acceleration to get to 3rpm is 10 seconds.

$$I=(1/12ml^2)_{bar}+(2*10*1^2)_{10kgmass}=\\1/12*1*4+20=1/3kg+20kg=20.33kg.m$$

This line added after OP's comment

Using Newton's second law $$F*dt= \Delta mv\quad or\ \tau dt=\Delta L=\omega I$$

$$\tau dt=I\omega=20.33*3/60=61/60\ \approx 1 kgm^2/s$$

plugging 10s for dt we get the torque.

$$10\tau= 1, \ \tau=1/10*9.8 \approx 1Nm,\ or\\ \tau= 10kg.cm$$

Now by changing dt to up or down you get the torque you need.

• Thank you very much for the detailed answer. If you don't mind, briefly explain to me how you went about the calculation. I did look at Inertia, but by that time I was already very confused. Dec 7, 2020 at 1:41
• @TinoFourie, we use Newton's 2nd law but in rotational acceleration. T=Ialpha as F=ma. then we multiply both sides by dt to get alphadt= angular velocity and Tdt = alphadtI= rotation velocityI= omega*I. we calculated the I and we plug angular acceleration. later I will add steps to my answer to clarify this. Dec 7, 2020 at 3:16
• I thank you for your time, patience, and explanation. Dec 8, 2020 at 11:08

If:

• 3rpm is the constant velocity that your system will rotate
• you are not worried about how fast you accelerate to 3rpm
• the system loading is symmetric (balanced)

then you'd only need to worry about friction losses and maybe cogging torque. However, those two are really small generally.

If you wanted to get within an order of magnitude(or two), I would suggest assuming an angular acceleration $$\alpha$$ which would get you from zero to the constant velocity in a reasonable for your project time, and then multiply that by the moment of area ($$I= \frac{1}{12}m_{bar}L_{bar}^2$$).

However, probably you'd be better off measuring the torque required to turn the axle. The friction of the bearing is dependent on many parameters. One of them is the actual installation and alignment, which can have detrimental effects (you can get the best bearing and install them the wrong way and you get something that doesn't work).

• Thank you for your reply and in depth explanation. I assume "cogging" refers to "start-up" torque, which I believe should be factored in as it will probably require the most torque (in my humble and uneducated opinion). I would agree on the point where measuring the torque required to turn the axle, however, I have no clue as to what formula to use or even how to go about arriving at an answer to this. Dec 5, 2020 at 12:34
• what you'd need to measure the torque needed to maintain constant velocity of the shaft is a force or (even better) torque gauge. If you use a force gauge, you'd need to measure also the distance (force times distance). Any torque more that that should be ok. Dec 5, 2020 at 20:19