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

[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).

 

[russ_watters]

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?

I think you are still not understanding when to apply Newton's 3rd Law and when to apply Newton's 2nd law. Please see the below scenario, with five forces labeled (magnitudes not to scale), and answer these questions:
1. Identify any Newton's 3rd Law force pairs.
1a. Do any of the identified N3L force pairs "sum to zero"?
2. What is the net force on the block?
3. Excluding your answer to 1a, do any forces "sum to zero" in this scenario?

 

[russ_watters]

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?

You've specified the motor torque in the problem statement.

 

[alkaspeltzar]

I think you are still not understanding when to apply Newton's 3rd Law and when to apply Newton's 2nd law. Please see the below scenario, with five forces labeled (magnitudes not to scale), and answer these questions:
1. Identify any Newton's 3rd Law force pairs.
1a. Do any of the identified N3L force pairs "sum to zero"?
2. What is the net force on the block?
3. Excluding your answer to 1a, do any forces "sum to zero" in this scenario?
View attachment 276974

I get Newton's 3 law. Opposite and equal and not applied to same object.

Jbriggs444 is getting what I'm asking. Please follow his comments

 

[jbriggs444]

I get Newton's 3 law. Opposite and equal and not applied to same object.

That's not Newton's 3rd.

It's equal and opposite and applied symmetrically between the same pair of objects.

 

[A.T.]

A is fixed the ground ...

Then it's not an isolated system, unless you include the entire Earth.

 

[alkaspeltzar]

Then it's not an isolated system, unless you include the entire Earth.

Assume its isolated. Whether it be the Earth or plate

 

[alkaspeltzar]

That's not Newton's 3rd.

It's equal and opposite and applied symmetrically between the same pair of objects.

That what I meant. Please respond to my earler question. Post 95

 

[A.T.]

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

Assume its isolated. Whether it be the Earth or plate

If the entire thing with plate is isolated, then you cannot guarantee that A is inertial, unless you specify that A is the center of mass of the entire isolated system.

 

[jbriggs444]

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).

If you are using reference point A then there are two non-zero torques on the ground. One from the motor and one from the axle from gear B.

If you are using reference point B then there are two non-zero torques on the ground. One from the motor's torque and one from the axle from gear A.

The standard approach to problems of this sort is to start writing down equations. For instance, for the angular acceleration of gear A based on the sum of torques on it divided by its moment of inertia.

Can you write down some equations for us? Feel free to make the ground infinitely massive so that the positions of the axles at the center of A and B do not move. [As pointed out previously, an infinitely massive ground can still act as a source or sink for angular momentum]. With enough equations in hand, we can solve for torques and angular accelerations.

 

[alkaspeltzar]

If you are using reference point A then there are two non-zero torques on the ground. One from the motor and one from the axle from gear B.

If you are using reference point B then there are two non-zero torques on the ground. One from the motor's torque and one from the axle from gear A.

The standard approach to problems of this sort is to start writing down equations. For instance, for the angular acceleration of gear A based on the sum of torques on it divided by its moment of inertia.

Can you write down some equations for us? Feel free to make the ground infinitely massive so that the positions of the axles at the center of A and B do not move. [As pointed out previously, an infinitely massive ground can still act as a source or sink for angular momentum]. With enough equations in hand, we can solve for forces and accelerations.

jbriggs444, hold on, i think you might have found my error
"If you are using reference point A then there are two non-zero torques on the ground. One from the motor and one from the axle from gear B."

I am using reference point A, but isn't there the reaction force from the axle on the Gear B. If you see my drawing in post #73 that is what i was trying to show. Then everytime i figure the torques about A, it would sum to zero.

Can i not use the reaction from the axle on B about A as part of the calculation? If not, can you tell me why?

I think this is what is creating the error

 

[jbriggs444]

jbriggs444, hold on, i think you might have found my error
"If you are using reference point A then there are two non-zero torques on the ground. One from the motor and one from the axle from gear B."

I am using reference point A, but isn't there the reaction force from the axle on the Gear B. If you see my drawing in post #73 that is what i was trying to show. Then everytime i figure the torques about A, it would sum to zero.

Can i not use the reaction from the axle on B about A as part of the calculation? If not, can you tell me why?

I think this is what is creating the error

You are still not showing any equations, any calculations and not even any list of torques.

If you are adding up torques on gear B about reference point A then there should most definitely be a non-zero torque from the ground through B's axle. Since gear A is driven clockwise, it exerts a downward (toward us, out of the page) linear force on gear B. Since the center of gear B remains motionless, pinned to the ground by its axle, it follows that the axle is exerting an upward (away from us, into the page) linear force on gear B. Since this force has a non-zero moment arm about point A, it embodies a non-zero torque about the axis at point A. So the axle puts a non-zero torque on gear B about point A.

Edit: There is an axle from B driven into the ground, right?

The sum of torques on all three objects will sum to zero regardless of what reference point is used. That would not be an error.

Unless you start showing some work, there is not much we can do to identify errors in that work.

 

[russ_watters]

jbriggs444, hold on, i think you might have found my error
"If you are using reference point A then there are two non-zero torques on the ground. One from the motor and one from the axle from gear B."

I am using reference point A, but isn't there the reaction force from the axle on the Gear B. If you see my drawing in post #73 that is what i was trying to show. Then everytime i figure the torques about A, it would sum to zero.

You evidently have the force on the plate at B facing the wrong direction and you also don't know the magnitude. The torque on the plate about A from that force and the motor torque do not sum to zero.

 

[alkaspeltzar]

So which force is wrong? Here is my picture again edited with more details.

Torques about A if have are as follows:
Motor torque of 10 inlbs
Torque on plate of -10 inlbs.
Contact forces at gear interface create zero torque about A, so we can ignore those
Force on axle at B, which is 10, creates a torque about A that is 60inlbs
Force on B from axle, which is 10, creates a torque about A that is -60 inlbsWhen i sum all this up, it is zero. That is not right, as the plate/entire assembly about A should rotate opposite of the rotation of gear B

 

[russ_watters]

Contact forces at gear interface go to zero, sowe can ignore those

Huh?

Force on axle at B about A is 60inlbs
Force on B from axle about A is -60 inlbs

Those are wrong and you haven't really attempted to properly calculate them yet.

 

[alkaspeltzar]

Huh?

Those are wrong and you haven't really attempted to properly calculate them yet.

i cleaned up, see edits

 

[A.T.]

When i sum all this up, it is zero. That is not right, as the plate/entire assembly about A should rotate opposite of the rotation of gear B

To get the angular acceleration of the plate you sum only the torques acting on the plate, not all torques that exist. You seem to have problems with applying N2L.

 

[alkaspeltzar]

To get the angular acceleration of the plate you sum only the torques acting on the plate, not all torques that exist. You seem to have problems with applying N2L.

SOB, I was treating the force on gear B and the plate the same. There is force on B due the axle, but it doesn't create a torque about A for the entire plate.

I just applied it wrong in my freebody...fml!

ARGH! Thanks

 

[russ_watters]

Force on axle at B, which is 10, creates a torque about A that is 60inlbs
Force on B from axle, which is 10, creates a torque about A that is -60 inlbs

Again: Those are wrong and you haven't really attempted to properly calculate them yet.
[Edit]
The gears still have mass and are still accelerating, right? You haven't included that in your calculations.

 

[A.T.]

I just applied it wrong in my freebody...fml!

You should use the exploded view in your FBDs to avoid confusing yourself:
https://peer.asee.org/the-exploded-...pproach-to-teaching-the-free-body-diagram.pdf

 

[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 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.

Thank everyone. I know my problem has missing details but I was trying to solve a general idea I had, which was why does the plate have to rotate one way if gear B is accelerated there other.

If I follow @jbriggs44 above setup and what @At pointed out it makes sense now.

If I calc the torque about A for gear B, there is a positive torque. But then I have to look at the plate/system, taking the equal and opposite forces, both at the gear interface, the axle, and opposite motor torque. I see a torque about A for the plate. This is why as the gear B goes one way, the plate goes the other.

And if my system is isolated, that balances out the entire system, which is what I get. As an isolated system, everything should balance out! Duh

Sorry for 90-100 drawn out comments. You been very helpful all.

I was breaking it apart but trying to do it as a whole system. It was causing me to see it two different ways and I cannot do that.

Thanks again.