Torque due to moment of inertia (?)

  • Thread starter Mech_LS24
  • Start date
  • #1
127
16
Summary:
Just a want to have an indication how much torque I need for rotating a plateau with 7.5 kg. See sketch for clearer view.
Hi!

I would like to calculate (roughly) how much torque is needed bringing the blue plateau in movement. Assume the blue plateau is loaded with 7.5 kg. The radius of the blue circle is 100 mm.

1626097074669.png
 

Answers and Replies

  • #2
anuttarasammyak
Gold Member
1,030
472
Any amount of torque would rotate the disk. So you should set some goal of quantity.
 
  • #3
127
16
Thanks for your reply @anuttarasammyak .

Not any amount would rotate the disk, the disk is loaded with a weight of 7.5 kg. This would give the disk a sort of moment of inertia isn't is? Or do I miss the mark here?

NB. the load is applied in the Z-direction, so actually 'into' the screen.
 
  • #4
berkeman
Mentor
60,390
10,698
Not any amount would rotate the disk, the disk is loaded with a weight of 7.5 kg. This would give the disk a sort of moment of inertia isn't is? Or do I miss the mark here?
If the bearings are frictionless, then the rotation will happen with any torque, just more slowly for higher moments of inertia. The relevant equation is ##\tau = I \alpha##

However, if there is bearing friction, you need to overcome the static bearing friction to start the disk rotating. Have you measured the friction of the loaded bearings? Are you using the right type of bearings for the way you are loading the axle?
 
  • Like
  • Informative
Likes crudux_cruo, Lnewqban and Mech_LS24
  • #5
Baluncore
Science Advisor
9,700
4,138
Does the plate rest on a friction free mechanical bearing that keeps it centred ?
or does it rest with friction, like on a bed of sand ?
 
  • #6
127
16
If the bearings are frictionless, then the rotation will happen with any torque, just more slowly for higher moments of inertia. The relevant equation is
If I take an angular acceleration of 3 rad/s^2, the torque needed is:
I= 0.5m*r^2 = 0.5*7.5*100^2 = 50000 kg*mm^2
α = 3 rad/s^2

τ= I*α = 0.05 (kg*m^2) * 3 (rad/s^2) = 0.15 Nm

Ok, this is frictionless.
Can I calculate the angular acceleration to RPM? I can't imagine 3 rad/s^2 yet.

However, if there is bearing friction, you need to overcome the static bearing friction to start the disk rotating. Have you measured the friction of the loaded bearings? Are you using the right type of bearings for the way you are loading the axle?
I have measured the friction of the bearings yet, this was just a concept. I wanted to know what amount of torque I roughly need for the motor. Maybe as a nice learning process for me I can calculate the static bearing friction for a plain bearing and angular contact bearings?

Using the following reference: Bearing friction coefficients
Angular contact bearings: .0020

Now calculate the torque:
1626156866161.png

P = 7.5kg * 10 = 75 N
μ= 0.0020
dm = 10 mm = 0.01 m (assumption)

Torque = (75N*0.002*0.01m)/2
Torque = 7.5*10^-4 Nm

Do I need to add the torque values for the total torque needed?
 
  • #7
127
16
Does the plate rest on a friction free mechanical bearing that keeps it centred ?
or does it rest with friction, like on a bed of sand ?
Yep it does rest with friction. I have tried to calculate the friction in post #6. Hopefully on the correct way :)
 
  • #8
127
16
Anyone who could support with post #6? Thanks
 
  • #9
Baluncore
Science Advisor
9,700
4,138
Anyone who could support with post #6? Thanks
How many alternative situations are you considering? What is, or is not, friction free? What stops things moving sideways? Where are the bearings? What sort of bearings are used?

Just because you put numbers in an equation does not mean the equation is applicable to the situation you had in mind. You must describe the situation and ask the question clearly.

We are not mind readers and won't waste valuable time trying to guess what you do not tell us. Instead we just walk away to escape the confusion, and solve some clearly specified problem elsewhere.
 
  • #10
127
16
Thanks for your reply @Baluncore .

Just a want to have an indication how much torque I need for rotating a plateau with 7.5 kg
As mentioned in post #1 I want to have an indication for the needed torque. With this torque I assumed I could select a proper motor.

But (again) as former student and just graduate, I am at the beginning of learning 'new' things. I appreciate you mentioned me on that friction. But you can't expect any magic from me. I shouldn't open this topic if I already knew it. By asking: "does it rest with friction", I can not come with all the things needed for that, because I just doesn't be familiar with that yet.

Now let's try to figure out that friction. I will try to make some sketches and come back to you.
 
  • #12
Baluncore
Science Advisor
9,700
4,138
I assume that by blue “plateau” you mean a plate, disk, or circle.
I assume that the small white circle marks the fixed centre of disk rotation.
What is the significance of the blue disk 100 mm radius specification?

I assume the black rectangle on the right hand side of the post #1 diagram is a bar that rotates about the fixed white circle.
I assume that the two grey lines are fixed length links that connect the circle to the bar.
I assume that the small red circles mark moving connection pins.
I observe that since the diameter of the blue circle is less than the length of the black bar, the system will be rigid and so cannot rotate.
What is the purpose of the black bar and two grey links?

How much does the blue disk (and shaft)? and outer ball race, and half the links weigh? When you say the blue disk is loaded with 7.5 kg, how is that load arranged and placed? I notice the added load has now risen to 10 kg. Notice that the moment of inertia is dependent on how the mass is distributed over the radius. But if you want to know the torque needed to break the friction then load distribution is not important.

The diameter of the ball centres in the bearing is needed from the data sheet.
What is the part number, 72XX, 73XX, of the angular contact ball bearing ?
 
  • #13
127
16
Thanks @Baluncore .

A description of the concept and hopefully it gives a more detailed view for your assumptions, as well I sketched the situation which is visualized below the description:

  • The blue plateau is as disk indeed, this disk carries the load
  • The white points are rotation pivots
  • I specified the radius of 100 mm for the disk, but haven't looked at details here. I can think of the amount of torque in the rods increase by a larger radius. So, if the radius increase, the forces in the rods decreases.
  • The black rectangle is an kind of a transmission for transfer the torque from the motor to the arm which brings the system into 'movement'
  • The red circles are pins which connects the rods from the disk to the motor and vice versa.
  • Good point mentioning the blue disk should be larger as the radius of the black bar. Changed that.
  • The disk is made for example from aluminium, the rods (arm) are most likely made of a steel. So I assume the weight of those parts are in total 1.5 kg. Added to the 10kg makes it 11.5 kg in total.
  • The load is probably loaded equally over the disk, I didn't find a nice way to draw that.
  • The bearing is an 7200 BECBP angular contact bearing. I am not a bearing experts so all recommendations are welcome :)
  • The bearing is locked with a bolt. A disk is pushed through the bolt into the fixed pin.

Please let me know if the description is understandable for you.

Motor_torque_calc_concept.jpg
 
  • #14
Baluncore
Science Advisor
9,700
4,138
Without specifying the distribution of the mass we cannot compute inertia. As I understand it, you only want to know the force needed to initiate movement of the disk. You do not require the rate of disk acceleration.
You need the coefficient of static friction, but the coefficient of dynamic friction is usually specified. I think we must assume the coefficient of static friction will be about twice that specified for dynamic friction. https://en.wikipedia.org/wiki/Friction#Static_friction

The effective radius of the 7200 bearing is (d+D)/4 = 10 mm radius.

Only half the weight of the rods is carried by the 7200 bearing, the other end is carried by the motor bar. You need to itemise the materials list so as to proportionately allocate mass to load the motor bar or the disk.

Since the motor and bar must rotate before the disk, you will also need to know the motor bearing and drive train friction. The motor and bar will transmit a force to the disk through the two rods and four pins. Friction in those rod pin-joints will almost certainly be greater than friction in the 7200 bearing.
 
  • #15
127
16
I would like to know the size of motor is needed for a concept like this. Probably I have been at the beginning of this concept been too complacent.

The effective radius of the 7200 bearing is (d+D)/4 = 10 mm radius
What is the effective radius? The formula is based on? I got web hits with bolted joints when I search effective radius bearings o_O

Hmm okay regarding the materials list, I take this:
  • Disk: EN-AW6082, t= 5mm, d= 200mm, density = 2700 kg/m^3, weight = 0.42 kg
  • Motor Bar: POM-C, 100x30x8 (L*W*H), density = 1.41 g/cm^3, weight = 0.034 kg
  • Rod: 1.4301 (Stainless steel), d= 10 mm, length = 300 mm, density = 8 kg/m^3, weight = 0.19 kg
  • Bearing 7200 mass = 0.03 kg
Next....?
Static friction (Wikipedia):
{\displaystyle F_{\text{max}}=\mu _{\mathrm {s} }F_{\text{n}}}
{\displaystyle F_{\text{max}}=\mu _{\mathrm {s} }F_{\text{n}}}

Us = 0.75 (assuming steel on steel)
Fn = 100 N (load on the disk) + ((0.42+0.034+(0.19*2)+0.03)*9.81) = 108.5 N

Fmax = 0.75*108.5 = 81.4 N

Since the motor and bar must rotate before the disk, you will also need to know the motor bearing and drive train friction. The motor and bar will transmit a force to the disk through the two rods and four pins. Friction in those rod pin-joints will almost certainly be greater than friction in the 7200 bearing.
Shouldn't I first define the motor? Then check the specifications on motor bearing and drive train friction? The pin-joint will probably be with spherical plain bearings, I found the following formula on the SKF site:
M = 0,5 μ P dm
 
  • #16
Baluncore
Science Advisor
9,700
4,138
What is the effective radius? The formula is based on?
The inner and outer race thickness are close to equal. So; re = (d+D)/4; is the average of inner and outer diameters, converted to radius, which is the length of the torque arm.
Because mean diameter, dm = 2 * effective radius of the ball race.
M = 0,5 · μ · P · dm ; can be reduced to; M = μ · P · re.

Shouldn't I first define the motor?
Specify the type of motor, and select an initial choice. Use the specifications to evaluate that choice in your application.
If you specify a DC servo motor, then the initial torque will be proportional to the initial current, which will simply be; motor voltage divided by motor resistance.
 
  • #17
127
16
The inner and outer race thickness are close to equal. So; re = (d+D)/4; is the average of inner and outer diameters, converted to radius, which is the length of the torque arm.
The term 'race' thickness isn't known by me, but I got the message (sketch below).
I was thinking about an TowerPro servo motor, I have used one in the past at school, because it was easy to program with Arduino. I think the motor isn't able to 'move' the load.

I am a bit confused now what I need to consider, the following is the case now(?):
  • Total mass that should be consider is 108.5 N (see Fn in post #15)
  • Bearing is a 7200 angular contact bearing
  • Re for 7200 bearing is 10mm
  • Static friction for the 7200 bearing is 81.4 N (see Fmax post #15)
  • Type of motor is a TowerPro servo motor
  • Output torque of the TowerPro motor is 1.6 kg*cm = 0.16 Nm (I can't find the motor resistance in the specifications..)
  • I take four rod ends to connect the rods from the disk to the servo motor. The rod ends are SI 8 C with dk = re = 13 mm. The frictional coefficient is 0.2 (steel/steel, SKF data).
Given the above I can determine following:
  • The torque for four rod ends (chosen: SI 8 C) are M = μ · P · re (SKF-formula). μ = 0.2 (SKF table, Steel/Steel). So, M = 0.2*108.5(N)*13(mm)*4 = 1.13 Nm

Open questions for me are:
  • The static friction for the 7200 bearing is 81.4 N, how can this be converted into torque? So I can add the static friction with the torque of the rod ends. Yes I know the force times an arm, but which arm should be taken into account?
  • The rod ends comes already with a torque of 1.13 Nm, the TowerPro servo motor can be wiped out
I added a sketch with the updated situation, I try to give you a better view of the situation.

Like to hear.

1626895385973.png



Average_inner_outer.jpg
 
  • #18
Baluncore
Science Advisor
9,700
4,138
The term 'race' thickness isn't known by me, but I got the message (sketch below).
The OD of the outer race is the ID of the housing. The ID of the inner race is the OD of the shaft. If you look at a ball bearing in section, (in the SKF data sheet), you will see that the inner and outer races have similar thickness dimensions, as the balls and the two races must fit in the space between the shaft and housing.

The balls all travel in the same circular path. Friction in a ball bearing is not just lubricated rolling, it is also contact between the balls and ball-cage that rotates with the balls at about half the speed of the shaft. The radius to the path of the ball centres is needed to compute torque due to friction from all sources.
https://en.wikipedia.org/wiki/Race_(bearing)
https://en.wikipedia.org/wiki/Ball_bearing

I notice again in your diagram, the rods are not parallel.
They need to be parallel and the same length.
 
  • #19
127
16
The OD of the outer race is the ID of the housing. The ID of the inner race is the OD of the shaft. If you look at a ball bearing in section, (in the SKF data sheet), you will see that the inner and outer races have similar thickness dimensions, as the balls and the two races must fit in the space between the shaft and housing.
Very clear, get it. Thanks. Cool process centerless grinding!

The radius to the path of the ball centres is needed to compute torque due to friction from all sources.
Thus, I can take the 81.4 N as force and the radius to the path of the ball centre as arm? So:
M = F*arm length
M = 81.4(N)*10(mm) = 814 Nmm = 0.81 Nm

Now I can filter motor with an amount of torque of at least 0.81(Nm) + 1.13 (Nm) = 1.95 Nm. If I add a safety factor of 1.5 makes it, 2.92 Nm.

Correct?

friction_bearing_motor_calc.jpg
 
  • #20
Tom.G
Science Advisor
4,006
2,717
1626895385973-png.png


I notice again in your diagram, the rods are not parallel.
They need to be parallel...
Parallel? What am I missing? Somehow I don't see that requirement.

The linkage is configured as a trapezoid, or a 4-bar linkage, which if freestanding is subject to collapse. It is also similiar to the configuration of a belt drive with speed reduction (or torque increase).

Cheers,
Tom
 
  • #21
Baluncore
Science Advisor
9,700
4,138
The linkage is configured as a trapezoid, or a 4-bar linkage, which if freestanding is subject to collapse.
Don't forget that the centres of the disk and the bar are fixed. The double-rod system is then overconstrained if the rods are not parallel. Think of the parallel connecting rods, in quadrature, on a railway engine.
It is also similiar to the configuration of a belt drive with speed reduction (or torque increase).
The belt passes around a pully so the length is fixed and the direction is not important. A link attached to a radius arm does change direction as it moves.
 
  • #22
Tom.G
Science Advisor
4,006
2,717
Hmm... To convince myself that the rods must be parallel, I just built a model.

Result was it worked fine with non-parallel links!

It's a bit late for me to to post photos tonight but the model is easy to build.

I tore strips from a 8.5x11 in sheet of paper for the 'hardware', some bent paper clips for the joints, and some straight pins to hold the points of shaft mounting to the carpet. The two links were stacked together and holes poked for the pivot points.

The disk and motor arm had pivot holes at random spacing (see below) and their center points found by folding to align the pivot points. The fold line of course was the center and a straight-edge between the pivot points and a pencil marked the 'shaft' attachment position. (where the straight pins held them to the carpet)

Link spacing on large disk: 7.25in. (18.5cm)
Link length: 7.5in (19cm)
Link spacing on short (motor) arm: 4.25in. (11cm)

Cheers,
Tom
 
  • #23
Baluncore
Science Advisor
9,700
4,138
Result was it worked fine with non-parallel links!
I can assure you that the rods will be under tension or compression for angular displacements of the short motor bar. For small angles you will not notice it with a pins-in-paper model because it is only a few percent. But a few percent in a mechanism can be destructive, it would certainly build up sufficient force to significantly increase friction in the bearings and pin joints. The motor would not be happy.

I will see if I can give you a numerical example based on your model dimensions.
 
  • #24
Baluncore
Science Advisor
9,700
4,138
Code:
' fixed rod = 190 mm
' bar    second
' deg  rod length
'   0  190.000000
'   5  190.033945
'  10  190.138177
'  15  190.317948
'  20  190.581001
'  25  190.938046
'  30  191.403370
'  35  191.995625
'  40  192.738782
'  45  193.663238
'  50  194.806986
'  55  196.216652
'  60  197.948002
'  65  200.065281
'  70  202.638443
'  75  205.737193
'  80  209.421051
'  85  213.725686
'  90  218.647688
 
  • #25
127
16
Interesting discussion.

Code:
' fixed rod = 190 mm
' bar    second
' deg  rod length
'   0  190.000000
'   5  190.033945
'  10  190.138177
'  15  190.317948
'  20  190.581001
'  25  190.938046
'  30  191.403370
'  35  191.995625
'  40  192.738782
'  45  193.663238
'  50  194.806986
'  55  196.216652
'  60  197.948002
'  65  200.065281
'  70  202.638443
'  75  205.737193
'  80  209.421051
'  85  213.725686
'  90  218.647688
How can I imagine this code? The rod length is increasing by stepping up the degrees, but the rod has a fixed length and thus can't change in length, right?

Thus, I can take the 81.4 N as force and the radius to the path of the ball centre as arm? So:
M = F*arm length
M = 81.4(N)*10(mm) = 814 Nmm = 0.81 Nm

Now I can filter motor with an amount of torque of at least 0.81(Nm) + 1.13 (Nm) = 1.95 Nm. If I add a safety factor of 1.5 makes it, 2.92 Nm.

Correct?
Is the approach above correct? As the discussion has skipped this a bit.
 

Related Threads on Torque due to moment of inertia (?)

Replies
1
Views
9K
  • Last Post
Replies
2
Views
745
  • Last Post
Replies
7
Views
7K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
2
Views
4K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
4
Views
8K
Top