hi, im stuck in a part of a project, im not the strongest in physics and here im trying to build a system but im struggling to calculate the torque needed to rotate a load via belt attached to a pulley and that pulley attched to a motor. an example is here in the image of what im trying to build , now the speed is unrelevent aslong as its rotating even if its very slow. thanks in advance.

 

Let us start with the basics. In this case the basics are units of measurement. To get your arms around this problem we need:

1. Angular Velocity
2. Angular Acceleration
3. Moment of Inertia

You may have some intuition about angular velocity and angular acceleration, so let us put those aside for a moment. The thing you want to become familiar with is Moment of Inertia. The idea is that your rotating disc is made up of a whole bunch of little masses. Each of those masses resides in the rigid object at some distance from the center of rotation. Masses close to the center of rotation contribute a small increment to the moment of inertia, while masses that are further away from the center of rotation contribute a much larger increment to the moment of inertia. You can search online for formulas to calculate the moment of inertia for a disc but you need to be careful about the units.

You can use the English system or the Metric system, but whatever you do you must be consistent.

In the Metric system the units are:

\(\text kg\cdot m^2\)

In the English system the units are:

\(\text lb\cdot \text ft \math ^2\)

In the Metric system we have units that are intuitively reasonable we have mass times distance squared. In the English system we have pounds, which is a unit of force, not mass, times distance squared. What is up with that? More on this later.

In the case of your rotating disc, the moment of inertia about the z-axis, which is the axis of rotation is given by:

\(I_z=\frac{mr^2}{2}\)

Where Iz is the moment of Inertia about the z-axis, and m is the mass, and r is the radius. Notice that the thickness does not enter into the calculation -- only the total mass and the radius.

Now is where we get back to units. In the English system the unit of mass is the slug. The conversion from slugs to pounds on the surface of the Earth is the acceleration of gravity or 32.4 ft/sec/sec. So I weigh about 240 lbs., or you could say I have a mass of ≈ 7.40 slugs. To convert from slugs/ft/ft to pounds/ft/ft we multiply the mass in slugs by 32.4. That's the trickery used by members of the cognoscenti. This only works on the surface of the Earth you understand. If you're on Mars you have to use slugs, or the Martian acceleration of gravity, ≈ 12.316 ft/sec/sec, to get lb(Mars).

So tell us:

1. What is the mass of the large disk, in kilograms and slugs?
2. What is the radius of the large disc in feet and in meters?
3. What is the moment of Inertia in both English and Metric units?

 

The moment of inertia needs to be considered only when you are accelerating or decelerating the rotating load. At constant speed you can ignore it, since torque is then relevant merely to overcoming friction.

 

The moment of inertia needs to be considered only when you are accelerating or decelerating the rotating load. At constant speed you can ignore it, since torque is then relevant merely to overcoming friction.

Yes @Alec_t, but he does need to get the thing started. I thought he wanted to learn the Physics, but maybe I was wrong. If he has enough torque to get it started, then he certainly has enough to keep it running. So tell me Doktor -- how much does he need?

 

I think the OP just wants to spin a turntable.

Getting too deep into inertia calculations here, my eyes are bleeding already.
in simple terms, it's really a matter of how much friction is in the system.
There is no way to calculate this friction- you need to MEASURE it, it's a function of the design of the turntable and bearings - real world stuff.

Build the basic turntable, attach a string to the edge- use a spring scale to measure how much pulling force it takes to rotate the turntable.
Measure the radius from the center where the string attaches to the edge, your torque is equal to: ( force * radius )

Now divide this torque by the pulley ratio to get the motor torque required, then double that result for a safety margin.

 

Hi,

Rotational inertia is not a consideration unless you want to get it up to speed in a certain time or of course if you want to learn how this works in general. Let me try to illustrate with a simple example.

We have a 10 foot diameter round platform 1 foot thick made out of pure lead mounted on a single frictionless pivot point at the very center of the disc so if the disc is spun around it turns on that pivot point round and round like a record on a record player. Because it is made out of lead, it has a huge huge weight. How much force applied tangentially along the outside diameter does it take to get it to start moving (rotating), and and how much force does it take to get up to a speed of 100 rpm ?
Answer for both: almost no force at all, you could start it rotating with your little finger.
Reason for both: there is no friction, and inertia alone can not stop anything from moving when a force is applied, it only makes it take longer to move with a certain limited force. That is until relativistic effects set in, which at these normally low speeds will not kick in.

What does stop something from moving with a force that is too small is static friction, sometimes called 'sticking' friction. In order to overcome the sticking friction a certain minimum force is required.
Second, once moving, the 'sliding' friction comes into play, and that is the friction that must be overcome in order to keep the object moving. These two apply to either rotational or translational movements.

What this means is if the static friction is low (good bearings) then not that much force is required to start the object moving, and if the dynamic friction is also low then it does not take much force to keep it moving. This works independently of size or weight, but for heavy objects you need bearings that are good enough to support the object properly and still have low friction. Also, if you start with a light weight object and then later intend to add objects to the platform, then you have to take that into consideration when choosing the bearings.

A rough sticking friction measurement can be done with a rope and spring scale. Sliding friction can be done the same way if you have enough room to keep pulling on the spring scale as you get the measurement. The minimum required torque can be calculated from those measurements, and since sticking friction is usually larger than sliding friction that measurement will usually dictate what you need to get going.
Other than that there is also another friction which we might call 'catch' friction, where the surface slides freely sometimes and seems to have abrupt changes in that friction as it moves over time. This usually does not come into play unless the surfaces are very rough.
There's also the long term sticking friction, which may come into play if the device sits for long periods of time and may build up junk or dust or even rust which will prevent it from moving with the previously measured torque. In that case you have to take special measures to get it going again which could mean just a quick push with the hand or require special oils and a much higher than normal force.

 

So tell me Doktor -- how much does he need?

To get the turntable (and its drive system) moving in a frictionless system the required torque (T) = moment of inertia (I) times chosen angular acceleration (a). If you know I and choose a then you can calculate T. Then add the measured torque required to overcome friction.

 

So in effect you now agree that the TS needs to have at least an approximate estimate for the Moment of Inertia. Thanks for the leadin to my second post.

 

So in effect you now agree that the TS needs to have at least an approximate estimate for the Moment of Inertia.

Always did. (First sentence, post #3).

 

So once we get the mass and the dimensions of the discs we can start talking about angular velocity and angular acceleration. Holding my breath.

 

thank you very much guys really appreciate your efforts, the system is designed so that i put some load on it, its meant to track a light signal so it will move very slowly, from reading the comments i realized i defiantly need angular acceleration as i need to move from rest, in terms of friction im using a ball bearing or its known as a wheel hub sometimes so it will be very little friction. the mass on the turn table in total is 12kg, the diameter of the turntable is about 75 to a 100cm.

but honestly guys thank you very much, mechanics isn't my area, and thats why im seeking your help, sorry for the late replay!!

 

I built a similar arrangement and placed a single steel ball bearing slightly smaller than the shaft dia in order to support the shaft and allow it to turn on.
Also if you use an angular thrust bearing there is very much less friction than a sealed bearing.
Max.

 

i'm thinking as it will move and stop continuously my acceleration should be slow and therefore it would reduce the torque right? , one more thing what about the tension of that rope? doesnt it go with the calculation of torque, i.e how much strength that belt is meant to have ? many thanks.

 

thank you very much guys really appreciate your efforts, the system is designed so that i put some load on it, its meant to track a light signal so it will move very slowly, from reading the comments i realized i defiantly need angular acceleration as i need to move from rest, in terms of friction im using a ball bearing or its known as a wheel hub sometimes so it will be very little friction. the mass on the turn table in total is 12kg, the diameter of the turntable is about 75 to a 100cm.

but honestly guys thank you very much, mechanics isn't my area, and thats why im seeking your help, sorry for the late replay!!

You said the mass on the turntable is 12 Kg. Did you mean the mass of the turntable is 12 Kg. Anything you put on the turntable will add to the moment of inertia and be very likely to fall off at any significant angular velocity.

In any case, using the numbers you given us:

\(I_z = \frac{12\; \text Kg\cdot (0.5\;\text m. )^2}{2} \;=\;1.5\; \text Kg\cdot m^2\)

I don't know how fast you want to go but let's try some numbers. Suppose you want to accelerate at .25 radians/sec/sec then

\(T = (1.5\; Kg\cdot m^2)(0.25\; \text radians/sec/sec)=0.375\;\text N\cdot\text m\)

As has already been mentioned you will need a bit more at first to overcome friction. After 1 second the disc will be going .25 radians/sec or about 2.4 rpm. After 2 seconds the disc will be going 0.5 radians/sec or about 4.75 rpm. Is that about the speed range you wer looking for? Naturally once you get to the desired speed you can reduce the torque to keep going at a constant speed.

As for the belt it needs to be tight enough so that it does not slip when you accelerate the disc.

 

So once we get the mass and the dimensions of the discs we can start talking about angular velocity and angular acceleration. Holding my breath.

that's what i done so far iz= mr^2/2= 12kgx0.5m^2/2=1.5kg.m^2
since my system acceleration is really slow so its almost negligible and therefore im using a low speed motor, i assume that my angular acceleration is 5RPM which is 0.5 rads/s^2
this means torque = 1.5kg.m^2 x 0.5 rads/s^2 =0.78 n.m
(i couldn't justify my angular acceleration as im building a tracker system of the sun, so im assuming its really slow and therefore i use 5rpm value.) ps i havant included anything about my tension of the belt ! what do you think sir?
thanks in advance.

 

You said the mass on the turntable is 12 Kg. Did you mean the mass of the turntable is 12 Kg. Anything you put on the turntable will add to the moment of inertia and be very likely to fall off at any significant angular velocity.

In any case, using the numbers you given us:

\(I_z = \frac{12\; \text Kg\cdot (0.5\;\text m. )^2}{2} \;=\;1.5\; \text Kg\cdot m^2\)

I don't know how fast you want to go but let's try some numbers. Suppose you want to accelerate at .25 radians/sec/sec then

\(T = (1.5\; Kg\cdot m^2)(0.25\; \text radians/sec/sec)=0.375\;\text N\cdot\text m\)

After 1 second the disc will be going .25 radians/sec or about 2.4 rpm. After 2 seconds the disc will be going 0.5 radians/sec or about 4.75 rpm. Is that about the speed range you wer looking for?

that's exactly what i done just with different speed because i don't know how to get that angular acceleration as my device should track the sun ideally. and one thing is the belt tension isn't being included in the calculation is that fine ?

thank you very much sir for your effort.

 

that's what i done so far iz= mr^2/2= 12kgx0.5m^2/2=1.5kg.m^2
since my system acceleration is really slow so its almost negligible and therefore im using a low speed motor, i assume that my angular acceleration is 5RPM which is 0.5 rads/s^2
this means torque = 1.5kg.m^2 x 0.5 rads/s^2 =0.78 n.m
(i couldn't justify my angular acceleration as im building a tracker system of the sun, so im assuming its really slow and therefore i use 5rpm value.) ps i havant included anything about my tension of the belt ! what do you think sir?
thanks in advance.

Wait a minute

\(1.5 \times 0.5 \;=\; 0.75\text , Not 0.78 as you mentioned in your text\)

But you are on the right track

 

The method in Post #5 should give you any breakaway torque figure and the fact that it is turning slowly, no appreciable speed or acceleration, should put you in the ball park.
Max.

 

The method in Post #5 should give you any breakaway torque figure and the fact that it is turning slowly, no appreciable speed or acceleration, should put you in the ball park.
Max.

Wouldn't something like this, a sun tracker, be almost all "break away" torque? The movement should be very intermittent. I also thought this kind of thing used a stepper motor not a regular PM type motor.

 

Wouldn't something like this, a sun tracker, be almost all "break away" torque?

Yes.

I also thought this kind of thing used a stepper motor not a regular PM type motor.

Either would work, if suitably rated.