Does Newton's Third law apply to torque/rotation?

[alkaspeltzar]

Here:
torque_by_GROUND_on_B = -torque_by_A_on_B = 50

So the ground applies a horizontal force of 10lbs at the 5 inches about B. This is opposite and equal to the torque by A on B, about B. ANd if you consider gear B, these are the opposite and equal torques for it. Its like Gear A is now acting similar to the ground, and therefore the N3L balance.

Right?

 

[russ_watters]

So the ground applies a horizontal force of 10lbs at the 5 inches about B. This is opposite and equal to the torque by A on B, about B. ANd if you consider gear B, these are the opposite and equal torques for it. Its like Gear A is now acting similar to the ground, and therefore the N3L balance.

Right?

Sounds about right, yes.

 

[alkaspeltzar]

Sounds about right, yes.

THank you

I think i see what is going on here. In the end, torque is dependent on the point of rotation and everything about it, like summing the moments has to be done around the same axis. However, because of conservation of ang momentum, and N3L, it all has to balance one way or another.

This example probably was bad as i didn't start with good unknowns, but in the end i see it all. Makes me understand why in many cases we don't have to care about all the reactions because we identify our system to make life easier.

Thanks again

 

[A.T.]

So the ground applies a horizontal force of 10lbs at the 5 inches about B. This is opposite and equal to the torque by A on B, about B. ANd if you consider gear B, these are the opposite and equal torques for it. Its like Gear A is now acting similar to the ground, and therefore the N3L balance.

Right?

Wrong. The torque_by_A_on_B and the torque_by_GROUND_on_B are not a N3L pair. This is not N3L, but a N2L balance. You are still confusing the laws you are applying.

 

[alkaspeltzar]

Wrong. The torque_by_A_on_B and the torque_by_GROUND_on_B are not a N3L pair. This is not N3L, but a N2L balance. You are still confusing the laws you are applying.

Probably be clearer if you drew it out as I asked nicely but since I have tried and still getting it wrong I guess I just won't care. Your replys are short but leave a lot to be explained to someone trying to learn

 

[A.T.]

Probably be clearer if you drew it out as I asked nicely ...

I have given you all the N3L pairs, for all your diagrams. Nowhere in there are torque_by_A_on_B and torque_by_GROUND_on_B a N3L pair.

 

[alkaspeltzar]

Yes, it does.

The point of N3 is the conservation of momentum. The point of the torque version is the conservation of angular momentum. Angular momentum requires the specification of one single axis about which all angular momenta are measured. That is just how angular momentum works.

Dale, can we talk or expalin where the n3l pair is for the torque B has on it? I get the point about being around a single axis, but do I just ignore what is happening to hear B

 

[alkaspeltzar]

I have given you all the N3L pairs, for all your diagrams. Nowhere in there are torque_by_A_on_B and torque_by_GROUND_on_B a N3L pair.

I asked though what about gear B, do we just ignore the torque about B as it is around an axis not related to the problem?

 

[Dale]

Dale, can we talk or expalin where the n3l pair is for the torque B has on it? I get the point about being around a single axis, but do I just ignore what is happening to hear B

No, you don't ignore what is happening to gear B. You must include all torques on all objects in the system. But you must fix the axis of interest. Which axis do you want to know if there is torque about? Then you calculate all torques about that point.

 

[A.T.]

I asked though what about gear B, do we just ignore the torque about B as it is around an axis not related to the problem?

"About B" makes no sense. Do you mean about the center of B? You can use any point, as long it's the same for both torques of an N3L pair.

 

[PeroK]

"About B" makes no sense. Do you mean about the center of B? You can use any point, as long it's the same for both torques of an N3L pair.

This misconception, and the OP's refusal to address it, has persisted for 60 posts now.

 

[alkaspeltzar]

No, you don't ignore what is happening to gear B. You must include all torques on all objects in the system. But you must fix the axis of interest. Which axis do you want to know if there is torque about? Then you calculate all torques about that point.

i want to know the torques about the center of A

 

[russ_watters]

"About B" makes no sense. Do you mean about the center of B? You can use any point, as long it's the same for both torques of an N3L pair.

Sorry, I think I added that: I've been calling the mounting point at the center of gear B, "Point B".

Nowhere in there are torque_by_A_on_B and torque_by_GROUND_on_B a N3L pair.

This might be another interpretation by me that isn't stated, but I interpreted the new drawing to show "Object B" is really a wheel and a gear, connected by a shaft -- but with the scenario morphing I don't know for sure. If that's the case, then a point on the shaft (Point B) has two equal and opposite torques applied to it. If "Object B" is truly one object (a wheel and a gear at the same time?), then it doesn't even need to be mounted on a shaft at all.

 

[A.T.]

i want to know the torques about the center of A

Then why do you keep asking about the torques "about B". Are you confusing "torque about ..." and "torque on ..."?

 

[Dale]

i want to know the torques about the center of A

View attachment 276890

So the torque of the ground force at A is 0, the torques at the gear are 10 in lb and -10 in lb and the torque at the center of B is 60 in lb. So the net torque on the system is 60 in lb about the center of A.

 

[alkaspeltzar]

Then why do you keep asking about the torques "about B". Are you confusing "torque about ..." and "torque on ..."?

Gear A DRIVES Gear B right?

There are contact forces between the two yes? Equal and opposite.

I see a torque arm of 5 inches, with a force of 10 lbs acting on gear B about its center. If there is a N3L for torque, i am trying to understand where does one account for this?

If i take all the torques about the center of gear A, i can see how angular momentum balances, but still Gear B has a torque applied to it, and i don't see where its equal and opposite is?

Perhaps someone ccould sketch on paper and show me what i am missing to help with all the 'about this or that"

 

[alkaspeltzar]

I think I see it now.. sorry I only had to sketch it like 30 times to see it. Sorry everyone. Let me sleep on it

 

[A.T.]

Gear A DRIVES Gear B right?

This is completely irrelevant to the application of N3L.

i want to know the torques about the center of A

I see a torque arm of 5 inches, with a force of 10 lbs acting on gear B about its center.

Make up your mind about which point you take the torques.

 

[alkaspeltzar]

This is completely irrelevant to the application of N3L.
Make up your mind about which point you take the torques.

there's a difference between being rude and trying to understand something you're confused about, I'm sorry I just don't get it and it took me a while to sketch it out and I think now I see how I have to apply Newton's third law to the single point and look at the torques there. Again I'm sorry it's not like I have a classroom to sketch us out in

 

[alkaspeltzar]

It's been almost 20 years since I've been in a physics classroom sometimes the easiest things and the fundamental things are the toughest to remember even though I can do the work on a daily basis so please bear that in mind I'm not trying to be stubborn

 

[alkaspeltzar]

Okay, I took all the torques about center of gear A, and I agree and got the following as @AT instructed. Note*** I assume that gear B is connected to a shaft mounted to the ground which wasn't shown in my sketch.

When taken around the same point A: torque_by_MOTOR_on_A = -torque_by_A_on_MOTOR
torque_by_B_on_A = -torque_by_A_on_B
torque_by_GROUND_on_B = -torque_by_B_on_GROUND

Then all agree with N3L. But then if I break down the system more, and look specifically at gear B, I see it has a torque about its axis which is why it rotates. Plus this is the reason why angular momentum balances between the earth/gears as a system. When you calc Ang momentum of the system around A, you see equal but opposite values. Gear B about point A has Ang. Momentum opposite of the Earth portion about A.. And it has to balance as @Dale said.

Sorry took me so long to see it. I realized I was confusing systems and torques by analyzing different points at once

 

[vanhees71]

I haven't read the entire thread. So maybe I just repeat what has been already said.

It's always good to envoke some symmetry arguments when discussion such fundamental questions, i.e., the profound relation between symmetries and conservation laws (Noether 1918).

From this point of view Newton's Laws are just substituted by the symmetries of the Galilei-Newtonian spacetime manifold. Newton's Lex 3 is nothing else than spatial translation symmetry and symmetry under Galilei boosts. You end up that for a closed system the total momentum must be conserved and the center of mass must move uniformly (i.e., with constant velocity against any inertial frame of reference). The possible forces are conservative interaction forces (with time-independent potentials due to temporal translation invariance). If you restrict yourself to the usual "two-body forces" you end up with Newton's Lex 3 for particle pairs. For more general forces it's not as simple, but these usually don't occur in practical applications of classical mechanics.

Now space is assumed to be a Euclidean affine manifold and thus also being isotropic, i.e., a closed system must be invariant under rotations around any arbitrary point of space. This implies the conservation of total angular momentum (wrt. to any arbitrary reference point!) and thus the sum over all torques around any arbitrary reference point must vanish.

 

[alkaspeltzar]

Okay, so i have tried to let this sink in and now i know better what I am asking.

Imagine a plate, that has a motor connected to gear A and gear A is driving gear B. Gear B is also connected to plate. Disregard friction. Rotations are as show.

Now when the motor starts up, it put in 10inch-pounds. Let's say all the power goes to Gear B to make it spin( i know there would be alittle to gear A but i am ignoring it)

Wouldn't for ang mometum to be conserved, the entire system/plate have to rotate CW, which is opposite of CCW gear B?

Therefore if I take the sum of the torques about center of gear A, shouldn't i get the torque that makes system about A spin? I am not getting it. For some reason they cancel out.

I know the contact forces at the gears provide no net torque on the system about A. But if the gear B applies a force to the axle of 10lbs, and the axle provides 10 lbs back, those torques cancel. So what am i doing wrong

 

[russ_watters]

Your description of this new system is kind of a mess, and I don't think it's physically possible as described. A couple of things:

1. 10 lb is a force, not a torque.
2. If an object spins at constant speed, the net torque on it is zero. For what you described, the torques don't appear to sum to zero. I don't want to tell you how to design your own system, but it looks to me like object B needs a brake.

 

[Dale]

So what am i doing wrong

I am not sure, but I think you are neglecting the torque provided by the motor. The motor does not exert a force, it exerts a torque.

 

[alkaspeltzar]

Your description of this new system is kind of a mess, and I don't think it's physically possible as described. A couple of things:

1. 10 lb is a force, not a torque.
2. If an object spins at constant speed, the net torque on it is zero. For what you described, the torques don't appear to sum to zero. I don't want to tell you how to design your own system, but it looks to me like object B needs a brake.

I corrected it above, the motor puts in 10in-lbs torque

And i wasnt assuming constant speed today. Assuming everything at rest then the motor turns on, with all the torque and energy going to speed up Gear B.

So i should see a torque about the center of gear A to balance ang. momentum right? IF you are curious as to the extra force on each gear, i am looking at contact forces, force from axle, force on axle

 

[alkaspeltzar]

I am not sure, but I think you are neglecting the torque provided by the motor. The motor does not exert a force, it exerts a torque.

see my reply to russ

 

[Dale]

IF you are curious as to the extra force on each gear, i am looking at contact forces, force from axle, force on axle

At the axle of A there is a force from the bearings, but there is also a torque from the motor. The force from the bearings has no torque about the axle, but the motor torque has torque about the axle.

 

[alkaspeltzar]

At the axle of A there is a force from the bearings, but there is also a torque from the motor. The force from the bearings has no torque about the axle, but the motor torque has torque about the axle.

I accounted for that i think.

I have a force of 10 lbs at Axle of B, creating a CCW torque of 60 about A. Then i have the force from the Gear B on the axle of 10 lbs, creating a CW torque of 60 about A. The motor adds 10 on gear A about A, the reaction is -10 on system about A.

When i add it all, i get zero, but i know that is not right. Since we are assuming it starting up, all the energy going to gear B, there has to be a net torque on the system about A.

 

[russ_watters]

And i wasnt assuming constant speed today. Assuming everything at rest then the motor turns on, with all the torque and energy going to speed up Gear B.

Oy, ok, well that complicates things a lot, and it's still wrong. The objects presumably have mass and moment of inertia, so the power/torque goes into speeding-up both objects. And since only the motor is hard-connected to the "plate", and it applies a torque, the plate must rotate as well.
[edit]
And the actual torque/force balance can't be calculated from the information provided.

 

[alkaspeltzar]

Oy, ok, well that complicates things a lot, and it's still wrong. The objects presumably have mass and moment of inertia, so the power/torque goes into speeding-up both objects. And since only the motor is hard-connected to the "plate", and it applies a torque, the plate must rotate as well.
[edit]
And the actual torque/force balance can't be calculated from the information provided.

Couldnt i assume the plates are massless...oh wait, then i suppose how would you know if the plate rotated CW while the gears spun OR in the other case what if gear A drove the whole plate CCW and gear B barely turned or generally what is gear A just turned everything because gear B had so much mass it and the plate rotated CW.

So really, i am trying to do a simple force/torque balance with a crappy problem :) Makes sense why it works when i have constant speed/zero acceleration. From a statics point, that makes a heck of lot more sense and explains a lot of my confusion yesterday.

I am assuming alot, best better top or you know what!

 

[russ_watters]

Couldnt i assume the plates are massless...oh wait, then i suppose how would you know if the plate rotated CW while the gears spun OR in the other case what if gear A drove the whole plate CCW and gear B barely turned or generally what is gear A just turned everything because gear B had so much mass it and the plate rotated CW.

If the plates are massless but accelerating they violate Newton's 2nd Law; F=ma.

So really, i am trying to do a simple force/torque balance with a crappy problem :)

You're making an awfully complicated simple problem.

 

[alkaspeltzar]

If the plates are massless but accelerating they violate Newton's 2nd Law; F=ma.

You're making an awfully complicated simple problem.

i was thinking of a situation and it lead to me wondering. Best i guess if i follow the books examples.

Thank you for the insight, explains a lot of my problem. Also explains why things don't look like i expect. I know how it works in static situation. And know my gear ratios, energy balance etc with transmissions. I will let this go.

Have a good night!

 

[A.T.]

Wouldn't for ang mometum to be conserved, the entire system/plate have to rotate CW, which is opposite of CCW gear B?

If gear B gains angular momentum, the rest of the isolated system will gain the opposite angular momentum.

 

[Dale]

I have a force of 10 lbs at Axle of B, creating a CCW torque of 60 about A. Then i have the force from the Gear B on the axle of 10 lbs, creating a CW torque of 60 about A. The motor adds 10 on gear A about A, the reaction is -10 on system about A.

If you specify the motor torque then the other forces are an unknown that you have to solve for and vice versa. You cannot set those arbitrarily or independently. Also, you need to know their moments of inertia.

 

[alkaspeltzar]

If gear B gains angular momentum, the rest of the isolated system will gain the opposite angular momentum.

So that's what I am trying to figure out is where does the torque around the rest of the system come from to gain the opposite angular momentum.

If I take the torques about A, shouldn't that work? When I work it out, with the forces above, something must be wrong as the torques sum to zero

 

[jbriggs444]

So that's what I am trying to figure out is where does the torque around the rest of the system come from to gain the opposite angular momentum.

If I take the torques about A, shouldn't that work? When I work it out, with the forces above, something must be wrong as the torques sum to zero

What scenario are we talking about? What forces are we talking about? What are you counting as one part of the system? What are you counting as the "rest of the system"? Where is your list of torques and why did you expect the total to be non-zero?

 

[alkaspeltzar]

What scenario are we talking about? What forces are we talking about? What are you counting as one part of the system? What are you counting as the "rest of the system"? Where is your list of torques and why did you expect the total to be non-zero?

See above with pictures. It was clear

I drew a plate with gears attached and motor. I specified motor torque and interaction forces. Curious as gear B rotates, what torque is applied to the entire system around A to balance Ang. Momentum

 

[jbriggs444]

So we are talking about the drawing in #73. And you are summing torques around point A at the center of the driven gear A.

But you are responding to a post by @A.T. where he speaks of an "isolated system". There is no isolated system in post #73.

We can turn this into an isolated system composed of three entities. There is the ground, there is gear A. There is gear B.

We wish to identify the torques on gear B about the reference point at the center of gear A.

We begin by listing the forces on gear B. There are two. The force from the teeth on gear A where they mesh with gear B. And the force of the axle where gear B is fixed to the ground. The two forces are equal and opposite. The moment arms are not equal. So the torques are not equal and opposite. So the angular momentum of gear B changes over time.

We can apply Newton's third law for forces and identify the third law partner forces: The force from the teeth on gear B on gear A and the force of gear B on the ground at its axle.

Both force pairs are contact forces. The points of application of the two members of the one force pair are co-located. The points of application of the two members of the other force pair are also co-located. Newton's third law for torque follows trivially in this case. The moment arm for the two members of the force pair are equal. The forces are equal and opposite. So the torques will be equal and opposite.

One can see that the total torque on the three pieces of the isolated system will necessarily sum to zero as long as only contact forces are involved. For every torque in the sum, there is an equal and opposite torque somewhere else in the sum.

 

[alkaspeltzar]

So we are talking about the drawing in #73. And you are summing torques around point A at the center of the driven gear A.

But you are responding to a post by @A.T. where he speaks of an "isolated system". There is no isolated system in post #73.

We can turn this into an isolated system composed of three entities. There is the ground, there is gear A. There is gear B.

We wish to identify the torques on gear B about the reference point at the center of gear A.

We begin by listing the forces on gear B. There are two. The force from the teeth on gear A where they mesh with gear B. And the force of the axle where gear B is fixed to the ground. The two forces are equal and opposite. The moment arms are not equal. So the torques are not equal and opposite. So the angular momentum of gear B changes over time.

We can apply Newton's third law for forces and immediately identify the third law partner forces: The force from the teeth on gear B on gear A and the force of gear B on the ground at its axle.

Both force pairs are contact forces. The points of application of the two members of the one force pair are co-located. The points of application of the two members of the other force pair are also co-located. Newton's third law for torque follows trivially in this case. And one can see that the total torque on the three pieces of the isolated system will necessarily sum to zero as long as only contact forces are involved.

Yes in my picture I was showing it isolated with the ground...now you are following correctly.

Okay, so we shown gear B has torque applied to it. I agree.

Now where I am confused is the ground about A would have to have a torque applied to it such that it gains Ang. Momentum opposite of that of gear B. When I do a force balance as shown, I am not seeing where the ground about A has the unbalanced torque. When I sum it all together, based on my picture in #73, I get zero

 

[russ_watters]

So that's what I am trying to figure out is where does the torque around the rest of the system come from to gain the opposite angular momentum.

If I take the torques about A, shouldn't that work? When I work it out, with the forces above, something must be wrong as the torques sum to zero

You haven't fixed the problems with the scenario pointed out last night. But we might be back to N3L again: if a motor applies a torque to a shaft, the shaft applies an equal and opposite torque to the motor.

When people talk about forces/torque "summing to zero", they are generally not talking about N3L, they are talking about summing the forces/torque applied to one object.

 

[jbriggs444]

Yes in my picture I was showing it isolated with the ground...now you are following correctly.

Okay, so we shown gear B has torque applied to it. I agree.

Now where I am confused is the ground about A would have to have a torque applied to it such that it gains Ang. Momentum opposite of that of gear B. When I do a force balance as shown, I am not seeing where the ground about A has the unbalanced torque. When I sum it all together, based on my picture in #73, I get zero

If the gears have a zero moment of inertia then the motor torque must be zero and there is no problem to solve.

So you must be assuming that the gears have a non-zero moment of inertia.

If the moments of inertia of the two gears are identical then the gears gain no net angular momentum as they accelerate and your result of zero net torque on the ground must be correct.

So let us assume that the gears have unequal moments of inertia.

Now then, show us your calculations under these assumptions.

 

[alkaspeltzar]

If the gears have a zero moment of inertia then the motor torque must be zero and there is no problem to solve.

So you must be assuming that the gears have a non-zero moment of inertia.

If the moments of inertia of the two gears are identical then the gears gain no net angular momentum as they accelerate and your result of zero net torque on the ground must be correct.

So let us assume that the gears have unequal moments of inertia.

Now then, show us your calculations under these assumptions.

I'm assuming the big gear has greater than small gear moment of inertia. Then from there I am lost as to how to solve the problem to see the torque about the entire system. I would have assumed that there would have to be an equal and opposite torque as others said earlier

 

[jbriggs444]

I'm assuming the big gear has greater than small gear moment of inertia. Then from there I am lost as to how to solve the problem to see the torque about the entire system. I would have assumed that there would have to be an equal and opposite torque as others said earlier

Which torques do you expect to be equal and opposite to which other torques? Newton's third law for torques means that members of third law torque pairs are equal and opposite. It does NOT assure us, for instance, that the torque of motor on gear A is equal and opposite to the torque from the teeth of gear B on gear A.

 

[alkaspeltzar]

Which torques do you expect to be equal and opposite to which other torques? Newton's third law for torques means that members of third law torque pairs are equal and opposite. It does NOT assure us, for instance, that the torque of motor on gear A is equal and opposite to the torque from the teeth of gear B on gear A.

As AT expressed, since gear B would rotate and gain Ang momentum, the system as a whole would have to gain opposite amount

So I keep trying to figure out the torque about A the system would have such that it's angular momentum nwoukd be opposite and equal to the Ang momentum of gear B. There has to be a torque on the ground about A for this to happen correct?

 

[A.T.]

If I take the torques about A, shouldn't that work?

Not if A is moving in circles. There is no angular momentum conservation about non-inertial points.

 

[alkaspeltzar]

As AT expressed, since gear B would rotate and gain Ang momentum, the system as a whole would have to gain opposite amount

So I keep trying to figure out the torque about A the system would have such that it's angular momentum nwoukd be opposite and equal to the Ang momentum of gear B. There has to be a torque on the ground about A for this to happen correct?

Imagine the gear B is being turned Ccw. As a resultz the whole ground with gears about A should turn CW. Where is the torque about A?

 

[alkaspeltzar]

Not if A is moving in circles. There is no angular momentum conservation about non-inertial points.

A is fixed the ground so it isn't rotating in circles, it is the axis

 

[jbriggs444]

As AT expressed, since gear B would rotate and gain Ang momentum, the system as a whole would have to gain opposite amount

So I keep trying to figure out the torque about A the system would have such that it's angular momentum nwoukd be opposite and equal to the Ang momentum of gear B. There has to be a torque on the ground about A for this to happen correct?

When you say "system as a whole", you appear to mean "rest of the system" -- i.e. gear A plus the ground.

Yes, if the ratios of the moments of inertia of the two gears do not match the gear ratio (I've belatedly realized that having the moments of inertia match is not the right criterion) and if the gears are accelerating then there must be a net torque on the ground.

Whether this means that the ground is rotating depends on whether we decide to treat the moment of inertia of the ground as being infinite. Regardless, it will have a non-zero rate of change of angular momentum.

 

[alkaspeltzar]

When you say "system as a whole", you appear to mean "rest of the system" -- i.e. gear A plus the ground.

Yes, if the ratios of the moments of inertia of the two gears do not match the gear ratio (I've belatedly realized that having the moments of inertia match is not the right criterion) and if the gears are accelerating then there must be a net torque on the ground.

Whether this means that the ground is rotating depends on whether we decide to treat the moment of inertia of the ground as being infinite. Regardless, it will have a non-zero rate of change of angular momentum.

Okay, if the gears like B are accelerating then how do you determine the torque about A( with the ground connected) such that you can determine how it rotates oppositely?

That's my question. If gear B is accelerating rotationally, what torque balances out the entries assembly( plus ground).