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

[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.

 

[alkaspeltzar]

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

Hey we got there guess that's all that matters now

 

[Shane Kennedy]

Summary:: There are similar 1st and 2nd laws for torques as there are for forces, but it doesn't seem like this applies to N3L?

Propellor driven (drawn?) aircraft with big engines have a tendency to flip over as a result of the "equal and opposite reaction. Pilots have to be careful not to apply too much power until they are traveling fast enough for the ailerons to counteract it.

 

[alkaspeltzar]

Propellor driven (drawn?) aircraft with big engines have a tendency to flip over as a result of the "equal and opposite reaction. Pilots have to be careful not to apply too much power until they are traveling fast enough for the ailerons to counteract it.

True, but its harder to see when there are multiple axis which is why I was trying to think/form the gear example.

But in the end, torque is about a constant axis or point of rotation and I mixing them up. Also if the gears are attached to Earth then it too isn't a closed system so we typically ignore Ang conservation and equal/opposite forces etc as they don't apply

Thanks for good example

 

[Shane Kennedy]

True, but its harder to see when there are multiple axis which is why I was trying to think/form the gear example.

But in the end, torque is about a constant axis or point of rotation and I mixing them up. Also if the gears are attached to Earth then it too isn't a closed system so we typically ignore Ang conservation and equal/opposite forces etc as they don't apply

Thanks for good example

Another example is a planetary gearbox. If just one part is fixed, then the other parts will share the power. In the case of the planets being fixed, the inner and outer will rotate in opposite directions with the opposite (to input direction) torque being experienced by the gearbox mounting

 

[atyy]

Newton's third law is generalized to the conservation of momentum. When Noether's theorem is valid momentum conservation arises from spatial translational invariance, while angular momentum conservation arises from rotational invariance.

The following two excerpts are from Kleppner & Kolenkow (K&K), Introduction to Mechanics, 2nd Ed.

[K&K, Eq 7.7 says torque = rate of change of angular momentum]
"In fact, Eq. (7.7) follows directly from Newton’s second law. Only when we talk about extended systems does angular momentum assume its proper role as a new physical concept."

[K&K, From Section 7.6, p260]
"In Chapter 4 we showed that the translational motion of a system of particles is simple to describe if we distinguish between external forces and internal forces acting on the particles. The internal forces cancel by Newton’s third law, and the momentum changes only because of external forces. This leads to the law of conservation of momentum: the momentum of an isolated system is constant.

In describing rotational motion it is tempting to follow the same procedure by distinguishing between external and internal torques. Unfortunately, there is no way to prove from Newton’s laws that the internal torques add to zero. Nevertheless, it is an experimental fact that internal torques always cancel because the angular momentum of an isolated system has never been observed to change. We shall discuss this more fully in Chapter 8, and for the remainder of this chapter we shall simply assume that only external torques change the angular momentum of a rigid body."

[K&K, Section 8.5, p311]
"The situation shown in figure (a) corresponds to the case of central forces.We conclude that in the particular case of central force motion the conservation of angular momentum follows from Newton’s laws. However, Newton’s laws do not explicitly require forces to be central. We must conclude that Newton’s laws have no direct bearing on whether or not the angular momentum of an isolated system is conserved because these laws do not exclude the situation shown in figure (b). It is possible to take exception to the argument above on the following grounds: although Newton’s laws do not explicitly require forces to be central, they implicitly make this requirement because in their simplest form Newton’s laws deal with particles. Particles are idealized masses that have no size and no structure. In this case, the force between isolated particles must be central, since the only vector defined in a two-particle system is the vector rjk from one particle to the other. ...

Because the spin of an electron defines an additional direction in space, the force between two electrons need not be central. ...

There are other possibilities for non-central forces. Experimentally, the force between two charged particles moving with respect to each other is not central; the velocities define additional axes on which the force depends.

The situation, in brief, is that Newtonian physics is incapable of predicting conservation of angular momentum but no isolated system has yet been encountered experimentally for which the total angular momentum is not conserved. We conclude that conservation of angular momentum is an independent physical law, and until a contradiction is discovered, our physical understanding must be guided by it."

 

[vanhees71]

The conservation of angular momentum, seen from the perspective of Noether's theorems, is not an additional physical law but follows from the isotropy of space. That's why it holds in Newtonian as well as special relativistic physics, because in both spacetime models rotations build a subgroup of the space-time symmetry group (Galilei group and special orthochronous Poincare group, respectively).

 

[atyy]

This is what my physics books says. It says that thru N3L, internal torques on a system should sum to zero. But doesn't say there is a N3L strictly for torque

N3L alone does not mean that internal torques sum to zero.

One needs the additional assumption that all forces are central. Newton's laws (including N3L) plus this additional assumption allows angular momentum conservation to be derived.

See also the quotes from Kleppner & Kolenkow's textbook in post #125.

In addition to K&K, here are additional references that derive angular momentum conservation with the explicit statement of the central force assumption.
http://www.physics.usu.edu/Torre/3550_Fall_2012/Lectures/04.pdf
https://sites.astro.caltech.edu/~golwala/ph106ab/ph106ab_notes.pdf (p37-38, Eq 1.22, 1.23)

 

[vanhees71]

Yes, and the Galilei group thus enforces that you must have all interaction forces between two particles must be central. By definition a closed many-body system obeys all the symmetries and thus all 10 conservation laws resulting from the conserved "Noether quantities" resulting from the 10 independent one-particle subgroups, i.e., energy (homogeneity of time), momentum (homogeneity of space), angular momentum (isotropy of space), and center-of-mass velocity (invariance under Galilei boosts).