Angular momentum conserved with multiple gears?

In summary, we have a motor attached to the Earth with gear A, that drives identically sized gear B. Gear B spins on its own axis and is also attached to the ground. The torque between the gears is equal and each gear has equal but opposite angular momentum. When considering the Earth, the conservation of angular momentum becomes a factor, and it is up to the individual to decide which axis to use. However, the torque and angular momentum are the same regardless of the chosen axis. The no-slip condition dictates that the total angular momentum of the system is always zero. The torque of the motor is equal and opposite to the torque of the couple formed by the contact forces at the axles.
  • #1
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Say we have a motor attached to the Earth with gear A, that drives identically sized gear B. Gear B spins on its own axis and but is also attached to ground. Torque between gears is equal.

Technically each gear has equal but opposite AM right, but If I take Earth into account, how is Ang. Momentum conserved?

Which axis do I use? Axis of A or B?

I understand AM for simple problems, but this got me thinking. Maybe this is a bad example and requires clarity so please let me know.

Thank you
 
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  • #2
You can use whatever coordinate system you want, it's axes don't need to coincide with the rotation axes of the gears.

Anyway, you'll need to give more details about your setup. What do you mean by 'attached to the ground', do you mean the axle is fixed? Is there friction at either of the axles? If one gear is being driven by an external motor, then that's one pathway by which angular momentum is being transferred to the system.

When writing ##\sum \vec{\tau} = \dot{\vec{L}}## for the system of two gears w.r.t. some coordinate system, don't forget the torques of the contact forces between the gears and the axles in your equation!
 
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  • #3
etotheipi said:
You can use whatever coordinate system you want, it's axes don't need to coincide with the rotation axes of the gears.

Anyway, you'll need to give more details about your setup. What do you mean by 'attached to the ground', do you mean the axle is fixed? Is there friction at either of the axles? If one gear is being driven by an external motor, then that's one pathway by which angular momentum is being transferred to the system.

When writing ##\sum \vec{\tau} = \dot{\vec{L}}## for the system of two gears w.r.t. some coordinate system, don't forget the torques of the contact forces between the gears and the axles in your equation!

I was assuming no friction. Gears and motor are attached to ground as shown, fixed rigidly. I don't exactly how you would shown AM is conserved here as all my examples in my old physics books are about one simple axis
IMG_20201025_114514776.jpg
 
  • #4
Have a look at the forces I've labelled in, assuming (to keep things simple) that the gears are massless, the axle frictionless, and the forces between the gears are vertical.

The forces exerted by each gear on each other are shown in red, and the contact forces from the hinge on the gears are shown in green [remember that the net force on any gear must be zero, since its centre of mass is not accelerating!]. The motor also exerts a driving couple moment on gear A.

1603644640227.png


Notice that the green contact forces form a couple, whose torque on the 2-gear system is the same in any coordinate system. The couple moment exerted by the motor on the gear A, and by extension the 2-gear system, is also the same in any coordinate system. The total torque of the two red forces is zero in any coordinate system.

The no-slip condition tells you that, if the gears are identical, the total angular momentum of the system is always zero. This means that the torque of the motor is exactly opposite to the torque of the couple formed by the two contact forces at the axles!
 
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  • #5
etotheipi said:
Have a look at the forces I've labelled in, assuming (to keep things simple) that the gears are massless, the axle frictionless, and the forces between the gears are vertical.

The forces exerted by each gear on each other are shown in red, and the contact forces from the hinge on the gears are shown in green [remember that the net force on any gear must be zero, since its centre of mass is not accelerating!]. The motor also exerts a driving couple moment on gear A.

View attachment 271545

Notice that the green contact forces form a couple, whose torque on the 2-gear system is the same in any coordinate system. The couple moment exerted by the motor on the gear A, and by extension the 2-gear system, is also the same in any coordinate system. The total torque of the two red forces is zero in any coordinate system.

The no-slip condition tells you that, if the gears are identical, the total angular momentum of the system is always zero. This means that the torque of the motor is exactly opposite to the torque of the couple formed by the two contact forces at the axles!

So are you taking the motor torque about the axis of A and saying it is equal and opposite to the couple in green about the same axis? If so then I see my mistake and how it balances.
 
  • #6
alkaspeltzar said:
So are you taking the motor torque about the axis of A and saying it is equal and opposite to the couple in green about the same axis? If so then I see my mistake and how it balances.

That works, yes. But the 'motor' can be replaced equivalently by two oppositely directed, tangential forces on opposite sides of the gear A, i.e. a force couple. That means that the torque of the motor is the same when calculated w.r.t. any coordinate system, and consequently there's no benefit in this case in picking a coordinate system whose origin is at the centre of A, as opposed to anywhere else.

This is a useful property of couples. If in one coordinate system the torque of a couple is$$\vec{\tau} = \vec{r}_1 \times \vec{F} + \vec{r}_2 \times (-\vec{F})$$then the torque w.r.t. another coordinate system whose origin is displaced by a vector ##\vec{R}## will be$$\vec{\tau}' = (\vec{r}_1 - \vec{R}) \times \vec{F} + (\vec{r}_2 - \vec{R}) \times (-\vec{F}) = \vec{r}_1 \times \vec{F} - \vec{r}_2 \times \vec{F} = \vec{\tau}$$i.e. ##\vec{\tau} = \vec{\tau}'##.
 
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  • #7
I am not sure, but I would say that each gear has its own angular momentum vector, which is aligned with each axis and pointing each in opposite directions in this case.
I believe that, althought the rotational velocity is the same for both gears in this case, their moments of inertia could be very different if they were constructed from different materials.

If we could disconnect the gears from each other, the lighter one would have less angular momentum.
If we increase the rotational speed of that lighter gear with lower moment of inertia in order to make the magnitude of the angular momentum of both gears equal, then try connecting both rotating gears, the angular momentum of the system would increase (one gear would slow down and the other would speed up).

Copied from:
http://hyperphysics.phy-astr.gsu.edu/hbase/conser.html#conamo

"The angular momentum of an isolated system remains constant in both magnitude and direction. The angular momentum is defined as the product of the moment of inertia I and the angular velocity. The angular momentum is a vector quantity and the vector sum of the angular momenta of the parts of an isolated system is constant. This puts a strong constraint on the types of rotational motions which can occur in an isolated system. If one part of the system is given an angular momentum in a given direction, then some other part or parts of the system must simultaneously be given exactly the same angular momentum in the opposite direction."
 
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  • #8
Lnewqban said:
If we could disconnect the gears from each other, the lighter one would have less angular momentum.
If we increase the rotational speed of that lighter gear with lower moment of inertia in order to make the magnitude of the angular momentum of both gears equal, then try connecting both rotating gears, the angular momentum of the system would increase (one gear would slow down and the other would speed up).
Yes, this is true.

But are you implicitly assuming that the gears will continue to spin in place about their fixed axes? Have you accounted for the forces required to make this so?
 
  • #9
jbriggs444 said:
Yes, this is true.

But are you implicitly assuming that the gears will continue to spin in place about their fixed axes? Have you accounted for the forces required to make this so?
Yes.
No.
Please, explain.
 
  • #10
Lnewqban said:
Yes.
No.
Please, explain.
So yes, you are assuming that the gears will continue to spin in place about their fixed axes. And no, you have not accounted for the forces required to make this so. You want an explanation of what forces we are talking about.

Look back at post #4. And remember the circumstance here. We have two gears which have been separated. They are spinning so as to have equal and opposite angular momenta about their respective centers. But they are not spinning at the corresponding rates to be able to mesh without slipping.

So when we bring the gears together so that they mesh, the one will be exerting a force on the other. Of course, by Newton's third law, the other will be exerting a force on the one.

But by your assumption, the gears remain spinning in place. That means that there is an external force from the axles on each of the gears. If you want to consider conservation of angular momentum on the two-gear system, you have to account for that external net torque.

If you do the accounting properly, the fact that the both gears gain counter-clockwise angular momentum abound their respective centers and that the system as a whole gains counter-clockwise angular momentum about any specified axis is consistent with the existence of an external counter-clockwise torque.
 
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  • #11
jbriggs444 said:
... Look back at post #4. And remember the circumstance here. We have two gears which have been separated. They are spinning so as to have equal and opposite angular momenta about their respective centers. But they are not spinning at the corresponding rates to be able to mesh without slipping.
...
I believe that I understand post #4:
Two massless gears of identical shape rotating in opposite directions, moved by a motor of constant angular speed.
I agree with the statement from @etotheipi about the total angular momentum of that system to be zero.

Sorry, I don't understand your quote above regarding both angular velocities being equal at the moment of separation but not at the moment of reunion of the spining gears.
Are you referring to the ideal case shown in post #4 or to the deviation from it that I imagined and tried to explain in post #7?
 
  • #12
Lnewqban said:
Sorry, I don't understand your quote above regarding both angular velocities being equal at the moment of separation but not at the moment of reunion of the spining gears.
My understanding was that the concern is about the delta between just prior to the moment of reunion and just after.

Ahhh, but you are interested in comparing the angular momenta of the two gears, one to the other. Clearly those will be unequal after reunion. Both gears are subject to torques pointing in the same direction. Either both clockwise or both counterclockwise. One gear will slow down. The other gear will speed up. If their angular momenta were equal (but opposite) before reunion (which they are by design) then they must be unequal after reunion.

The moment of reunion is an ugly one to ask about since that is an instant at which an impulsive change of angular momentum takes place.
 
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  • #13
I see it clearly now.
Thank you, jbriggs444
 
  • #14
etotheipi said:
That works, yes. But the 'motor' can be replaced equivalently by two oppositely directed, tangential forces on opposite sides of the gear A, i.e. a force couple. That means that the torque of the motor is the same when calculated w.r.t. any coordinate system, and consequently there's no benefit in this case in picking a coordinate system whose origin is at the centre of A, as opposed to anywhere else.

This is a useful property of couples. If in one coordinate system the torque of a couple is$$\vec{\tau} = \vec{r}_1 \times \vec{F} + \vec{r}_2 \times (-\vec{F})$$then the torque w.r.t. another coordinate system whose origin is displaced by a vector ##\vec{R}## will be$$\vec{\tau}' = (\vec{r}_1 - \vec{R}) \times \vec{F} + (\vec{r}_2 - \vec{R}) \times (-\vec{F}) = \vec{r}_1 \times \vec{F} - \vec{r}_2 \times \vec{F} = \vec{\tau}$$i.e. ##\vec{\tau} = \vec{\tau}'##.
Isnt the math wrong here... if i have a couple shouldn't Fxr1 be added to FxR2? Seems to be an extra minus sign.
 
  • #15
I definitely did it right, promise!
 
  • #16
etotheipi said:
I definitely did it right, promise!
what am i doing wrong. If i have a couple, two equal forces F, I have to multiply them via their radius and add together. See diagram below

If there is another axis, i agree i can find the same torque, but my formula would be the distance from the new origin to existing axis Rnew as such that;

(Rnew + r1)F - (Rnew -r2)F.
 

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  • #17
alkaspeltzar said:
what am i doing wrong. If i have a couple, two equal forces F, I have to multiply them via their radius and add together. See diagram below

If there is another axis, i agree i can find the same torque, but my formula would be the distance from the new origin to existing axis Rnew as such that;

(Rnew + r1)F - (Rnew -r2)F.
we might be saying same thing. I see you wrote in vector format where i am not. IF you have force Vector F, but it is opposite of the other force vector F, then the negative in front makes it makes sense to what i wrote above
 
  • #18
Yes, I think that it is the discrepancy in sign convention. Choosing to put the signs into the values and leaving the formula pristine. Or putting the signs into the formula so that all of the values can be positive.
 
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  • #19
jbriggs444 said:
Choosing to put the signs into the values and leaving the formula pristine. Or putting the signs into the formula so that all of the values can be positive.

Only the first option is correct reasoning, second option is just fudging 🤭
 

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