Momentum of this rotating gear?

[risecolt]

I assume that the equation that applies for this problem is M = mg * r.
The large gear is rigid while the small gear rotates about the axis which passes through the center of the large gear by rotating a motor. What is the momentum which is required to turn the motor?

I'm confused because I'm used to calculating the momentum of an object rotating about an axis, when the point of rotation is at the same location as the axis. But in my problem, the center of rotation and the center what is making it rotate is different. The gears are weigthless.

http://cognitivenetwork.yolasite.com/resources/Gear.png

 

[etudiant]

If the gears are weightless, they will have 0 momentum.
The problem needs a bit more exact specification.

 

[risecolt]

If the gears are weightless, they will have 0 momentum.
The problem needs a bit more exact specification.

The gears are weightless, but not the motor which makes it rotate. The gear carries the motor around the large gear, as it rotates about it.

 

[tiny-tim]

welcome to pf!

hi risecolt! welcome to pf!

The gears are weightless, but not the motor which makes it rotate. The gear carries the motor around the large gear, as it rotates about it.

yes, but if there's no resistance, no force is needed

doesn't the question mention any resistance?

(the mass of the motor isn't resistance, it's just mass)

 

[etudiant]

The problem is not well defined.
Unless the small gear and motor are constrained, they will simply slip off the large gear rather than rotating around it. One might get some momentum estimate from the mass of the motor and the rpm of the small motor, which would give a tip speed of the gear and hence an approximate forward speed of the motor/gear assembly.

 

[risecolt]

The problem is not well defined.
Unless the small gear and motor are constrained, they will simply slip off the large gear rather than rotating around it. One might get some momentum estimate from the mass of the motor and the rpm of the small motor, which would give a tip speed of the gear and hence an approximate forward speed of the motor/gear assembly.

There is a force, that force is equal to gravity multiplied by the mass of the motor.
That force is located at the center of the motor.
The arm is equal to the radius of the small gear.
Are you saying that I can't use the formula M = F x r, because the momentum required depends on the friction coefficient between the contact surfaces of the two gears?

Those two gears are held in place by a mechanical structure. Just imagine that they do not simply slip off.

 

[sophiecentaur]

I don't see where gravity comes into this. It all seems to be in a horizontal plane.??
You would need to draw in, explicitly, any framework that is holding this arrangement together - else the problem is not defined. What is fixed and what is allowed to move?

 

[risecolt]

I don't see where gravity comes into this. It all seems to be in a horizontal plane.??
You would need to draw in, explicitly, any framework that is holding this arrangement together - else the problem is not defined. What is fixed and what is allowed to move?

Yes, it's all on the horizontal plane, or more technically the top plane.
You don't see how the gravity would pull the mass of the motor downwards?!
The large gear in the center is rigid. The small gear is attached to the motor.
The motor turns, and the small gear rolls around the rigid gear in the center.
But the gears are still held in place, from falling apart. Meaning I have eliminated one of their three degrees of freedom. The plane it moves in is in the x and z coordinate, while it is locked from moving in the y coordinate.

 

[etudiant]

The mass of this lashup would be centered at the axis of the small wheel, at distance
r1+r2 from the large wheel axis.
Assume the gears are held in contact by some frame linking the two axes, then the angular momentum of the small gear/motor combo will be set by the rotation speed of the small gear translated into angular velocity around the large gear.

 

[sophiecentaur]

Yes, it's all on the horizontal plane, or more technically the top plane.
You don't see how the gravity would pull the mass of the motor downwards?!
The large gear in the center is rigid. The small gear is attached to the motor.
The motor turns, and the small gear rolls around the rigid gear in the center.
But the gears are still held in place, from falling apart. Meaning I have eliminated one of their three degrees of freedom. The plane it moves in is in the x and z coordinate, while it is locked from moving in the y coordinate.

I see that but what has that got to do with Momentum, which involved Mass and not Weight? I thought this was an 'ideal' system, with no friction - and it would only be friction that could be affected by gravity here.

 

[risecolt]

I see that but what has that got to do with Momentum, which involved Mass and not Weight? I thought this was an 'ideal' system, with no friction - and it would only be friction that could be affected by gravity here.

This is starting to make sense to me. So that means that I should set up the equation:

Angular Momentum = Force x arm = angular velocity x moment of inertia?

F x r = w x I

 

[risecolt]

I see that but what has that got to do with Momentum, which involved Mass and not Weight? I thought this was an 'ideal' system, with no friction - and it would only be friction that could be affected by gravity here.

The angular momentum is what determines the torque required to make the motor turn the mass of the motor about the axis of the large gear. The weight is determined by the mass multiplied by the gravity generated by the Earth's magnetic field. You can exclude friction in this example.

I was thinking if I could perhaps use the following equation: M = angular velocity x moment of inertia. So that if the maximum torque which the motor can generate by the motor is known, and if I could calculate the moment of inertia, I could calculate the fastest speed the gear could rotate around the axis?

 

[risecolt]

The mass of this lashup would be centered at the axis of the small wheel, at distance
r1+r2 from the large wheel axis.
Assume the gears are held in contact by some frame linking the two axes, then the angular momentum of the small gear/motor combo will be set by the rotation speed of the small gear translated into angular velocity around the large gear.

I think I have finally found the solution because of your help.
I need to know what the angular velocity and the moment of inertia is, because calculating the force acting on the gear/motor alone is not enough, because torque increases by the increase of radius or speed. So if I consider the gear/motor as a cylinder to simplify things, I = 0.5 * (r1^2 + (r1+r2)^2), where r1 is the radius of the small gear, while the length at which what is rotating about the axis, is dislocated from its center, is the radius of the small gear + the radius of the large gear. So in an ideal system, if the max torque is known, the maximum allowed angular speed would be determined by the moment of inertia.

 

[sophiecentaur]

BTW, are you assuming that the motor has no moment of inertia but just consists of an 'ideal' point mass?

 

[risecolt]

BTW, are you assuming that the motor has no moment of inertia but just consists of an 'ideal' point mass?

Of course the motor has a moment of inertia, everything has. Only friction should be excluded, just like air resistance, because they can be very difficult to calculate. What is ideal about the mass is that it is centered in the motor.

 

[risecolt]

BTW, are you assuming that the motor has no moment of inertia but just consists of an 'ideal' point mass?

Would you be so kind and see my last post, it is more relevant.

https://www.physicsforums.com/showthread.php?t=679731

I had made a mistake previously. Torque is the change of rate of angular moment.
So if you know the angular moment, divide it by s (second).

 

[sophiecentaur]

@risecolt
This has been muddled from the start. To get an answer, you have to start with a meaningful question
In the OP, you wrote:

What is the momentum which is required to turn the motor?

This means nothing. In a frictionless system the torque (!) needed is infinitessimal.

The weight is determined by the mass multiplied by the gravity generated by the Earth's magnetic field.

You are still bringing "weight" into the argument (I assume you mean 'gravitational field'). Weight is irrelevant in the horizontal plane. It's all about Mass and Moment of Inertia.

There is no 'maximum speed' involved unless the motor itself is speed limited. The total angular momentum starts at zero and will remain at zero as the system is not fixed to the Earth (rotationally) - I have to assume. The casing of the motor will rotate one way and the whole motor will travel in a circle around the central pivot. The relevant thing to find out would be the relative angular acceleration of the motor (around its own axis to the angular acceleration of the motor about the central pivot. The relationship between the two angular rotation speeds will be given by the gear ratio and you can easily set up an equation involving this and equating angular momentum to zero.
It's just a more complicated form of the equivalent linear situation of rocket propulsion (reaction drive).

 

[risecolt]

@risecolt
This has been muddled from the start. To get an answer, you have to start with a meaningful question
In the OP, you wrote:

This means nothing. In a frictionless system the torque (!) needed is infinitessimal.


You are still bringing "weight" into the argument (I assume you mean 'gravitational field'). Weight is irrelevant in the horizontal plane. It's all about Mass and Moment of Inertia.

There is no 'maximum speed' involved unless the motor itself is speed limited. The total angular momentum starts at zero and will remain at zero as the system is not fixed to the Earth (rotationally) - I have to assume. The casing of the motor will rotate one way and the whole motor will travel in a circle around the central pivot. The relevant thing to find out would be the relative angular acceleration of the motor (around its own axis to the angular acceleration of the motor about the central pivot. The relationship between the two angular rotation speeds will be given by the gear ratio and you can easily set up an equation involving this and equating angular momentum to zero.
It's just a more complicated form of the equivalent linear situation of rocket propulsion (reaction drive).

Thank you for your reply. I have previously suggested in a new thread that the torque is the change of rate in angular momentum. So if I divided the angular momentum by a second, I get the torque. But now I realize, it is more practical to just use Torque = angular acceleration * moment of inertia. So far so good. I know that the gear ratio is 0,55555555555555555555555555555556. The larger gear is 1,8 times larger.

However, when I've calculated this torque, it would be located on the (vertical) y-axis of the motor. But is this torque simply transferred from the teeth of the small gear, to the motor, or do I still have to consider that for the small gear, there is rotation and (radial) translation (about the axis of the large gear).

Should following be the solution to my problem?

$\tau$ = $\alpha$ * $I$

By the way, I do realize, gravity is not taken into account, and I do find it a bit strange.
So I imagine myself that the system was confined within empty space with no gravity, I do agree only its own mass acts on the system, where moment of inertia is relevant.

 

[jbriggs444]

Thank you for your reply. I have previously suggested in a new thread that the torque is the change of rate in angular momentum. So if I divided the angular momentum by a second, I get the torque.

If you divide angular momentum by time then you get torque.

If you divide angular momentum by one second you get a number of no particular relevance.
[You get the torque that would be required to achieve the given angular momentum if that torque were applied for a duration of one second]

You may have misread what risecolt wrote. If you have angular momentum expressed in kilogram meter² per second and you divide by time expressed in seconds then you get torque expressed in kilogram meters² per second². The "second" enters in as a unit of measurement, not as a physical parameter of the problem.

But now I realize, it is more practical to just use Torque = angular acceleration * moment of inertia. So far so good. I know that the gear ratio is 0,55555555555555555555555555555556. The larger gear is 1,8 times larger.

That's a lot of 5's. Call the gear ratio 5 to 9 and save yourself some typing.

 

[risecolt]

If you divide angular momentum by time then you get torque.

If you divide angular momentum by one second you get a number of no particular relevance.
[You get the torque that would be required to achieve the given angular momentum if that torque were applied for a duration of one second]

You may have misread what risecolt wrote. If you have angular momentum expressed in kilogram meter² per second and you divide by time expressed in seconds then you get torque expressed in kilogram meters² per second². The "second" enters in as a unit of measurement, not as a physical parameter of the problem.



That's a lot of 5's. Call the gear ratio 5 to 9 and save yourself some typing.

Thank you for replying. I am getting to the finish line now. So far I understand and accept that I must use the equation: torque = angular acceleration * moment of inertia.
This calculated torque would cross the axis of the center of the motor.
I know what the gear ratio is, but (1) do I use the gear ratio to calculate the angular acceleration of the small gear, and the torque remains the same, it's just transferred to the pin of the motor, or (2) I use the gear ratio to calculate the torque?

 

[risecolt]

If you divide angular momentum by time then you get torque.

If you divide angular momentum by one second you get a number of no particular relevance.
[You get the torque that would be required to achieve the given angular momentum if that torque were applied for a duration of one second]

You may have misread what risecolt wrote. If you have angular momentum expressed in kilogram meter² per second and you divide by time expressed in seconds then you get torque expressed in kilogram meters² per second². The "second" enters in as a unit of measurement, not as a physical parameter of the problem.



That's a lot of 5's. Call the gear ratio 5 to 9 and save yourself some typing.

Let's say the gear ratio is 0,5. The radius of the gear is 25mm, and the torque has been calculated to be 1Nm. What do I need to do to place the torque to the edge of the gear, instead of being centered on the gear?

 

[sophiecentaur]

Let's say the gear ratio is 0,5. The radius of the gear is 25mm, and the torque has been calculated to be 1Nm. What do I need to do to place the torque to the edge of the gear, instead of being centered on the gear?

Nm is the same at the centre and at the edge. This would give you the force on the gear teeth as a function of the diameter.

But I still don't really know what your question is. Some of it is there but it isn't a complete or answerwable question, I think. You didn't say I was wrong in supposing that the outer case of the motor is free to rotate so the MI of the motor casing will tell you the angular acceleration of the spinning case. This will have a bearing on the acceleration of the motor around the central pillar - reaction and action being equal and opposite. The angular velocity around the central pillar will depend upon how fast the motor spins.
As I see it, there is an equal and opposite torque on the inner and outer of the motor so you can equate them.

If ω₁ is the angular velocity of the motor casing, M is the mass of the motor, I is the MI of the motor casing. R is the distance from the centre of the motor to the central pivot and ω₂ is the angular velocity around the pivot then:
ω₁MR²=ω₂I

You could improve on that by including the MI of the motor armature I', in which case the equation becomes

ω₁(MR²+I')=ω₂I

This doesn't include anything about the relative diameters of the wheels but I don't think that actually make a difference in an ideal case. This is because your question only involves torque and not power. If power were taken into account then the time taken to reach a given speed would be relevant and then the gearing would, in fact, govern how much torque was available for a given power.

Angular momentum change (and hence the angular velocity after a given time) would be given by
Δω₂ = τ/I

Isn't that what you wanted to know?

You needn't worry about the gravity thing. If you do calculations about linear motion on a smooth horizontal plane, you wouldn't include g, would you? You can resolve force and acceleration into independent horizontal and vertical components.

 

[risecolt]

Nm is the same at the centre and at the edge. This would give you the force on the gear teeth as a function of the diameter.

But I still don't really know what your question is. Some of it is there but it isn't a complete or answerwable question, I think. You didn't say I was wrong in supposing that the outer case of the motor is free to rotate so the MI of the motor casing will tell you the angular acceleration of the spinning case. This will have a bearing on the acceleration of the motor around the central pillar - reaction and action being equal and opposite. The angular velocity around the central pillar will depend upon how fast the motor spins.
As I see it, there is an equal and opposite torque on the inner and outer of the motor so you can equate them.

If ω₁ is the angular velocity of the motor casing, M is the mass of the motor, I is the MI of the motor casing. R is the distance from the centre of the motor to the central pivot and ω₂ is the angular velocity around the pivot then:
ω₁MR²=ω₂I

You could improve on that by including the MI of the motor armature I', in which case the equation becomes

ω₁(MR²+I')=ω₂I

This doesn't include anything about the relative diameters of the wheels but I don't think that actually make a difference in an ideal case. This is because your question only involves torque and not power. If power were taken into account then the time taken to reach a given speed would be relevant and then the gearing would, in fact, govern how much torque was available for a given power.

Angular momentum change (and hence the angular velocity after a given time) would be given by
Δω₂ = τ/I

Isn't that what you wanted to know?

You needn't worry about the gravity thing. If you do calculations about linear motion on a smooth horizontal plane, you wouldn't include g, would you? You can resolve force and acceleration into independent horizontal and vertical components.

Ok, so if the torque calculated from moment of inertia multiplied by the angular acceleration, is the result of the motor's translation about the axis. Let's say it was the large gear that was rotating the motor about its own axis, it would use the same torque as if the small gear was rotating about the axis of the large gear? You mean that the torque will be the same, because all that matters is how far the motor is from the center of rotation and how much the angular acceleration is?

 

[sophiecentaur]

Ok, so if the torque calculated from moment of inertia multiplied by the angular acceleration, is the result of the motor's translation about the axis. Let's say it was the large gear that was rotating the motor about its own axis, it would use the same torque as if the small gear was rotating about the axis of the large gear? You mean that the torque will be the same, because all that matters is how far the motor is from the center of rotation and how much the angular acceleration is?

That's what the sums tell us. It's not intuitive, I agree, but, as I said, torque is only a part of the situation. Imagine that the gear wheel on the motor was very tiny. The motor case would spin up very fast. If the gear wheel on the central shaft, the motor case would only rotate once per revolution around the pivot. The sums say that the rotation (i.e. angular acceleration) about the pivot would be the same. But the power from the motor would have to be much higher to provide that torque with a small wheel on the motor because Power = ωτ

 

[risecolt]

That's what the sums tell us. It's not intuitive, I agree, but, as I said, torque is only a part of the situation. Imagine that the gear wheel on the motor was very tiny. The motor case would spin up very fast. If the gear wheel on the central shaft, the motor case would only rotate once per revolution around the pivot. The sums say that the rotation (i.e. angular acceleration) about the pivot would be the same. But the power from the motor would have to be much higher to provide that torque with a small wheel on the motor because Power = ωτ

That's interesting, and it sounds reasonable. If the gear wheel on the motor was very tiny, but the acceleration remained the same, the RPM of the gear wheel would increase, but it wouldn't go farther.

So the current output (to the motor) would have to increase. So the conclusion is that the torque would be limited by the angular acceleration and opposite, with respect to the moment of inertia and the size of the gear would be limited by the current output limit to the motor.

 

[risecolt]

That's what the sums tell us. It's not intuitive, I agree, but, as I said, torque is only a part of the situation. Imagine that the gear wheel on the motor was very tiny. The motor case would spin up very fast. If the gear wheel on the central shaft, the motor case would only rotate once per revolution around the pivot. The sums say that the rotation (i.e. angular acceleration) about the pivot would be the same. But the power from the motor would have to be much higher to provide that torque with a small wheel on the motor because Power = ωτ

Would the (sum of) moment of inertia for (1) the small gear and (2) the cube shaped motor be:
(1) I = 1/2 * m(gear) r^2 + mR^2
(2) I = 1/6 * m(motor) r^2 + mR^2

 

[sophiecentaur]

Would the (sum of) moment of inertia for (1) the small gear and (2) the cube shaped motor be:
(1) I = 1/2 * m(gear) r^2 + mR^2
(2) I = 1/6 * m(motor) r^2 + mR^2

The MI of the motor parts are not predictable. For a start, they don't have uniform density and what would be the radius of the armature? You are straying into practicalities now. Dodgy!

 

[risecolt]

The MI of the motor parts are not predictable. For a start, they don't have uniform density and what would be the radius of the armature? You are straying into practicalities now. Dodgy!

That didn't answer my question.

 

[risecolt]

The MI of the motor parts are not predictable. For a start, they don't have uniform density and what would be the radius of the armature? You are straying into practicalities now. Dodgy!

I think I have come to a conclusion. In order for me to calculate whether the motor has enough holding torque rating, will be determined by its nominal speed, nominal torque etc.

So time for motor to reach nominal speed = (moment of inertia * nominal angular speed) / (nominal torque - torque load).

The nominal torque = nominal power * 9550 / nominal speed
torque load = nominal speed torque, and I don't understand how torque load is different from torque load.

I suppose I can find these ratings on datasheets for the motors?

Look at page 17
http://www05.abb.com/global/scot/scot201.nsf/veritydisplay/a3ef20fdc69ccc9ac12578800040ca95/$file/ABB_Technical%20guide%20No.7_REVC.pdf

But I don't know if this applies for stepper motors, which I am using.
If there is a rating for nominal speed and nominal time for it to reach that speed, and I neglect the friction, then I can use that time to see if it exceeds the maximum holding torque rating of the motor and use the equation: Torque =moment of inertia * nominal speed / time.

What do you think?

 

[risecolt]

@risecolt
This has been muddled from the start. To get an answer, you have to start with a meaningful question
In the OP, you wrote:

This means nothing. In a frictionless system the torque (!) needed is infinitessimal.


You are still bringing "weight" into the argument (I assume you mean 'gravitational field'). Weight is irrelevant in the horizontal plane. It's all about Mass and Moment of Inertia.

There is no 'maximum speed' involved unless the motor itself is speed limited. The total angular momentum starts at zero and will remain at zero as the system is not fixed to the Earth (rotationally) - I have to assume. The casing of the motor will rotate one way and the whole motor will travel in a circle around the central pivot. The relevant thing to find out would be the relative angular acceleration of the motor (around its own axis to the angular acceleration of the motor about the central pivot. The relationship between the two angular rotation speeds will be given by the gear ratio and you can easily set up an equation involving this and equating angular momentum to zero.
It's just a more complicated form of the equivalent linear situation of rocket propulsion (reaction drive).

My solution. I think I have got it now. There seem to be ratings for nominal acceleration on stepper motors. One that I found had a nominal acceleration of 280 steps/sec, which is equal to 504 degrees / sec, which is equal to 8,79 rad / sec. Then I just need to find out whether the moment of inertia exceeds the holding torque by using Torque = nominal angular acceleration * moment of inertia. Got any comments?

 

[sophiecentaur]

My solution. I think I have got it now. There seem to be ratings for nominal acceleration on stepper motors. One that I found had a nominal acceleration of 280 steps/sec, which is equal to 504 degrees / sec, which is equal to 8,79 rad / sec. Then I just need to find out whether the moment of inertia exceeds the holding torque by using Torque = nominal angular acceleration * moment of inertia. Got any comments?

I really really don't understand any of this, now. How are you going to deliver power to this free moving motor? I have to repeat my original question - What is fixed to what, what is allowed to rotate and what is fixed to the earth?
You clearly have a picture of this in your head which has more details in it than that basic figure you have provided.

And god only knows what you mean by this. How can a torque be compared with a Moment of Inertia? They have totally different dimensions. And what is "holding torque"? We have suddenly migrated from ideal, theoretical to nuts and bolts (totally practical). V. confusing.

 

[risecolt]

I really really don't understand any of this, now. How are you going to deliver power to this free moving motor? I have to repeat my original question - What is fixed to what, what is allowed to rotate and what is fixed to the earth?
You clearly have a picture of this in your head which has more details in it than that basic figure you have provided.

And god only knows what you mean by this. How can a torque be compared with a Moment of Inertia? They have totally different dimensions. And what is "holding torque"? We have suddenly migrated from ideal, theoretical to nuts and bolts (totally practical). V. confusing.

lol. Are you ****ing kidding me? Torque = angular acceleration * moment of inertia.
A motor has its own lowest acceleration rating. The big gear is rigid, the small gear is not rigid.
The small gear rotates about the big gear, because the motor spins, the small gear thus also have to rotate about itself, as it is in contact with the large gear. That is all the information you need to get into your head.

 

[sophiecentaur]

lol. Are you ****ing kidding me? Torque = angular acceleration * moment of inertia.
A motor has its own lowest acceleration rating. The big gear is rigid, the small gear is not rigid.
The small gear rotates about the big gear, because the motor spins, the small gear thus also have to rotate about itself, as it is in contact with the large gear. That is all the information you need to get into your head.

I could say the same thing.

You have still not described what happens to the body of the motor. You 'agreed' that it is free to rotate (or did not read my question). OK for a thought experiment but now you tell me you have a real motor in mind, with real characteristics. SOOOOOO it needs power from somewhere. Where? A long flexible power lead?
If you cannot actually draw out your apparatus with more than just 'sky hooks then you have to specify, fully, what you have in your head. I suspect that there is something in your imagined set-up that you may not have fully considered (or, at least, keeping secret) - which is why I was after a proper diagram.

You have done the equivalent of drawing an electronic circuit without including power supply voltages and a ground and expecting someone to say how it will function.

 

[etudiant]

lol. Are you ****ing kidding me? Torque = angular acceleration * moment of inertia.
A motor has its own lowest acceleration rating. The big gear is rigid, the small gear is not rigid.
The small gear rotates about the big gear, because the motor spins, the small gear thus also have to rotate about itself, as it is in contact with the large gear. That is all the information you need to get into your head.

It is rude to talk down to people who are trying to help answer a poorly posed question such as yours.
In a class, I'd tell you to sit in the corner. Dunce cap for the next offense.

 

[sophiecentaur]

It is rude to talk down to people who are trying to help answer a poorly posed question such as yours.
In a class, I'd tell you to sit in the corner. Dunce cap for the next offense.

Cheers, but it's water off a ducks back for me. I spent a long time being paid for dealing with hormonal teenagers so I tend not to take offence!