Calculate torque of an 80kg object rotating on its central axis

[Yafimski]

Hello,

I have a setup whereby:
1) A circular plate with fillet under-edge, is sitting like a bowl on some ball bearings. Imagine if you will a bowl sitting in a way that it can rotate and move a bit up and down etc. without much friction.
2) The plate needs to rotate at about 4 or 5 rpm.
3) The plate is 80kg.
4) The radius of the plate is 0.15m
5) A motor in the center spins it.

What I am trying to find out is the strength of a motor needed in order to rotate those 80kg continuously. Right now I am using a NEMA 17 planetary gear motor, which can only move a few kg and is hardly enough for what I need.

I calculated the following:

Angular velocity (w) = 0.166 rad/s
Angular acceleration (a) = 0.55 rad/s2
Linear velocity (at edge) (v) = 0.025 m/s
Rotational Inertia (i) = 0.9 kg*m2

So my questions are:

1) Are my calculations correct so far?
2) Given the formula of torque = i * a, does this mean just to multiply (0.9kg*m2) * (0.55rad/s2)?
In that case, I'm not sure I understand the units here... what is m2*rad/s2??
3) Eventually what I want to know as a result is the strength of the motor, meaning it's torque, in order for me to get a proper motor for this application.Thank you

 

[Dr.D]

In your description, you make no mention of any need to accelerate the system, but then you calculate an angular acceleration. How? What are you saying is happening?

In answer to your other question, a kg-m^2/s^2 = (kg-m/s^2) * m = N-m, a torque unit.

It really looks like you have neglected rolling friction which is the biggest consumer of power at steady speed.

 

[Yafimski]

@Dr.D

Thanks! Okay so basically the torque is 0.5Nm, but this doesn't make sense when considering an 80kg weight.

The plate is turning via a motor shaft, and it's starting from 0 and reaching the Angular velocity (w) = 0.166 rad/s.
From what I understand, angular acceleration is the change of angular velocity over time, so I did (0.166rad/s) / (3s) which is an estimate of the time taking the system to "reach constant motion".

Could you help me understand how to make sense of this 0.5Nm figure I arrived at...?

 

[Dr.D]

If I understand you correctly, you are saying that you assumed a uniform acceleration over a 3 sec time interval to reach a speed w = 0.166 rad/s. The uniform acceleration is then alpha,
alpha = delta(w)/delta(t) = (0.166-0)/(3-0) = 0.0553333 rad/s^2

Then it looks like you applied T = I*alpha to get the torque,
T = (0.9) *(0.0553333) = 0.0498 N-m
(Looks like I disagree with you by a factor of 10 here.)

The big oversight seems to me to be that you neglected friction. You only calculated the torque required to accelerate the system, but with your model, it should spin freely forever once it reaches operating speed, without the need for any further input torque. It just will not work that way.

The assumption of constant acceleration is also somewhat dubious. Very few systems are capable of doing even close to this.

Also, where did the I = 0.9 kg-m^2 come from? Are you confident about this figure?

 

[Yafimski]

Thanks for the help so far Dr.D.

Yes you're right, I was wrong by a factor of 10 :/

I'm really clueless about friction laws. Which specific kinds of forces should I take into account here?

The 0.9 comes from the equation for Rotational Inertia, which if I got it right is:
I = 1/2*m*r2
so:
I = 1/2 * 80kg * (0.15m)2 = 0.9 Kg-m2

I know there is also a version of that formula which is I = m*r2, meaning 0.9 turns into 1.8, but I don't think that gets me very far...

Thanks :)

 

[Dr.D]

If your 80 kg object is a simple disk, then this is the correct MMOI formula. You described it as a bowl, and that suggests something other than a simple disk, but I have no clear idea of what your actual shape is.

As to friction, I cannot tell you very much without a much better understanding of your system. You have not given us a figure or any real details, so I'd simply be guessing in the dark.

Does this system start and stop regularly/frequently, or does it start up and run for a long time?

 

[Yafimski]

Hi,
Yes sorry - I'm attaching an image of the basic setup. The reason it is a "bowl" shape is because it's resting on inclined ball bearings, so basically the sides of the disk are curved to accommodate that function of 'resting' on the ball bearings. This is where friction comes in in a way.

The bowl rotates and should move around 80kg which are directly placed on top of it.

The system should optimally start and run for quite some time.

Does this help..?

 

[Dr.D]

Well, it helps in th e sense that nothing done above makes much sense now.

Most critically, just how that 80 kg is distributed will make all the difference in the world as to the value of the MMOI. Imagine two cases for a moment:
(1) the mass is all piled in a tall column with a very small base; it is essentially a slender rod located at the center of the bowl.
(2) the mass is all piled in a ring around the edge of the disk, with nothing in the center.
The MMOI for these two cases is radically different, and your 0.9 kg-m^2 value is utterly meaningless.

Now, regarding friction, I suggest you do some internet research on rolling element bearing friction. There is more too it than I can go into here.

In view of all the foregoing, don't you think it might be well to hire someone who knows how to do such things, rather than fumble through it yourself?

 

[JBA]

Once the plate has reached a steady rotational speed then its inertia is irrelevant to the power and torque required to rotate it. The only effect of its inertia once it is rotating is to reduce the amplitude of momentary rotational speed oscillations due to the drive system or outside forces. Only external forces, including the rolling bearing friction discussed above, are relevant.

Apart from the friction issue, is this an absolutely smooth round plate or are there any projections or irregularities on the plate's surfaces; or, are there any other external forces or loads being applied to the bowl while it is rotating; and, what is the plate's diameter? While the rotation speed in rpm is low, if the plate diameter is very large or has irregular surface features then there could also be a some, if small, amount of aerodynamic surface drag as well.

You state:

it can rotate and move a bit up and down etc. without much friction.

Can you clarify this statement?

Edit: now that I have seen your last post and diagram, if there is is any wobble in the bowl because the bearings are running on a curved surface or the load placed upon the plate is irregular or off center then, as stated above, this adds a whole new range of factors to the issue.

 

[JBA]

Please see my last post edit

 

[Yafimski]

Hi,
@Dr.D - the plate is 15cm in radius, and the weight on it would be a solid object, a cylinder, with 80kg in mass.
I'm broke, can't hire anyone :)

@JBA - The plate underside is indeed not completely smooth. This is part of what causes friction in the tests I made. The problem I am presenting is a simplified version of the mechanism in all of its detail because I am trying to figure out the simplest case possible.

In the case of the diagram I attached, given a bare minimum or no friction at all, what would be the strength of motor to move 80kg. That's all I'm trying to figure out. So the plate will ideally be rotating with a load on top of it, the diameter as mentioned, is 30cm.

In reality, the plate can 'move up and down' because it is in theory capable of being pushed gently to the sides, moving 1-2 cm and therefore moving 'up' the ball bearings that hold it (the ball bearings are on an inclined plane as in the diagram).

 

[Yafimski]

Right now the motor I am using is a NEMA 17 planetary gear, but it is hardly enough, it can barely move if you put 5kg on it.

 

[JBA]

Can you run a test to determine the maximum weight that can be applied and still have the plate rotate at your required rpm?

Also, can your plate be modified so to insure that your cylinder is absolutely and reliably centered on the rotating plate and is the cylinder uniformly round and of uniform density; and, its centerline perfectly perpendicular to its bottom face?

 

[Yafimski]

Yes I've run some tests, it's around 5kg max, maybe a bit less.

Regarding the centered cylinder - that's the theory that I am trying to figure out - can we assume that it is perfectly uniform etc and centered and calculate the motor for that?

 

[Nidum]

Fit a radial bar temporarily and use a spring balance on the outer end to find how much force is needed to turn the plate at about the right speed and under the least favourable conditions of loading .

 

[Dr.D]

There seems to be little point to trying to size a motor for the best case (no friction) when you already know that friction is a major factor.

For the ideal system (no friction, no vertical motion) after the system is up to speed, no further torque is required to maintain speed because there is nothing else doing work on the system. But this is a meaningless case because you have said it does not really apply.

 

[JBA]

From my point of view, you have already determined an approximation for your required motor size. What is the power rating of your current motor?

 

[Yafimski]

I've spoken with a friend who said I should take the friction coefficient between metal-plastic and multiply that times gravity and then times the Kg load I want.
Given an extreme example of a high load:

0.3 * (9.81 m/s2) * 120 kg = 353.16 Kg-m / s2
He said I should then multiply that by the radius of the plate, which is 0.15m, so I get 353.16N * 0.15m, which is 53Nm. This is starting to make more sense.

Let's assume that the friction is even higher, 0.5 if we're really grinding stuff, so that means 88Nm.
The motor I have now has the following specification:

NEMA 17 Planetary Geared Stepper Motor
Weight = 0.5Kg
Rated Voltage = 12V DC
Rated Current = 0.4A DC
Rotor Inertia = 35gcm2
Holding Torque before gearbox = 26Ncm

Gearbox:
Gear ratio = 26 (103/121)
Max. Permissible Torque = 3Nm
Moment Permissible Torque = 5Nm
Shaft maximum Axial load = 50N
Shaft maximum Radial load = 100N

I'm not sure how my 53 to 88 Nm result fits into this. I think it means I need a motor at least x10 stronger.

 

[Dr.D]

The OP said, "I've spoken with a friend who said I should take the friction coefficient between metal-plastic and multiply that times gravity and then times the Kg load I want."

This is a pretty poor approach to modeling rolling friction, but whatever ...

This should result in a gross over-design specification, but it only your money.

You are substituting the Coulomb friction model for rolling friction, not a common approach.

 

[Yafimski]

@Dr.D,

I am totally aware that I have near-zero understanding of friction. Now that you say that I should be using Coulomb friction, I'll look into it. Obviously I'm here to learn more and understand things better to get a better result, so if you can explain about these things in depth, please do.

 

[Dr.D]

No, I did not say you should be using the Coulomb model. I said that is the implication of your friend's approach, but that it makes little sense. Stick with rolling friction if there are ball bearings involved.

 

[Yafimski]

Ah okay. Does rolling friction mean the following formula I found? :

Fr = μrN

where:

• Fr is the resistive force of rolling friction
• μr is the coefficient of rolling friction for the two surfaces (Greek letter "mu" sub r)
• N is the normal force pushing the wheel to the surface

In this case, I know N and Ur, but not Fr. Which force is that?

 

[JBA]

I have done a bit of investigating and I suggest you use the following two web addresses for SKF ball bearings for calculating your bearing friction. The first one will allow you to calculate your "Dynamic Bearing Load"; and, with that information the second one will allow you to calculate your bearing friction torque.

http://www.skf.com/us/products/bear...ds/equivalent-dynamic-bearing-load/index.html

http://www.skf.com/us/products/bear...ction/estimating-frictional-moment/index.html

 

[Yafimski]

@JBA
Thank you! I will delve into it in the next 24 hours and see what I can come up with, and report back on progress :)

Cheers

 

[Yafimski]

@JBA

I've tried to understand what the first link says. I'm not sure I'm able to... it's really unclear to me what I should focus on that page..? What is relevant there? I can't understand what they are talking about... :/

 

[JBA]

All the first page is addressing is combining the radial and axial loads geometrically to find the combined vector of the two loads.

In your case, you should do is take your total vertical load (90 kg + plate weight) and the angle of the cone surface on which your balls are running to establish the "equivalent dynamic load" to use in the second link calculation.

For example: If your outer surface on which the balls are running is 45° to the horizontal, then the "equivalent dynamic load" = (90 kg + plate weight) / sin 45°

 

[Yafimski]

Okay so I calculated it and the 'frictional moment' is around 2.5Nmm, which is 0.0025Nm.

I'm a bit lost. This is quite a negligible amount isn't it? How can this help me understand the total torque needed for the system...?

 

[JBA]

Please show me your calculations; and, what dimension did you use for the "bearing bore diameter" (it should be the inside diameter of the circle of balls in your case).

 

[Yafimski]

Yes sorry:
I think in the previous answer I made a x10 mistake again, but here is my new trial:

Let's say the plate+weight = 90kg, the ball bearing angle is 45.
So I calculated --> 90kg / 0.707 = 127.3 kg. That's the "equivalent dynamic load".

If I plug that into the second link you provided, I get:

M = 0.5 * 0.0024 (this is the ball bearing coefficient which they put in the table) * 1273N * 8mm = 12.22 Nmm --> which is equal to 0.01222 Nm

 

[JBA]

Are you saying that the inside diameter of your circle of bearing is only 8mm, that seems a very small diameter to me because I am under the impression that the bearings are located near the outer rim of your plate.

 

[Yafimski]

From what I understand they meant "bearing bore diameter [mm]" to mean the size of the ball bearing right?
I have a collection of them positioned at 45deg, but each one has a diameter of 8mm, yes.

The entire plate is 300mm in diameter if that's what you mean.

 

[JBA]

The correct d diameter is the diameter of the circle where your ball bearings run on the bottom of the plate.

 

[Yafimski]

In that case it comes to about 430Nmm which is 0.43Nm

But that still doesn't make sense because the motor I'm using now is more powerful than that...

 

[JBA]

Are your balls running dry or lubricated?

 

[JBA]

What is the gear ratio of your planetary gearbox?