Free body diagram of a wheel driven by a motor

[EM_Guy]

So now are the two forces from rim tension or compression oriented in opposite directions?

I think so. They are both tangential to the point touching the ground - in the limit. If we are not quite in the limit, then the two forces are not quite in opposite directions.

 

[jbriggs444]

I think so. They are both tangential to the point touching the ground - in the limit. If we are not quite in the limit, then the two forces are not quite in opposite directions.

And if we are not quite in the limit the force from the spoke at that point is not quite zero.

If we take the limit, the force of the spoke at that point is zero as is the force from compression or tension. The force from the ground is non-zero. The mass at the point is zero. We can conclude from this that acceleration is infinite and the wheel deforms. Accordingly, it is unrealistic to take the limit.

 

[EM_Guy]

And if we are not quite in the limit the force from the spoke at that point is not quite zero.

The spoke? From the earlier discussion, isn't it clear that the spoke might be in tension or it might be in compression or (at just the right angular speed) it might be stress free. But I was speaking of the tension or compression forces from the rim.

The mass at the point is not zero. I've said that it is m, and we can call it a point mass. Or if you prefer, we can call it dm - a differential mass.

Of course, in reality, the wheel (or tire) touches the ground at more than one point, and there is some measure of deformation of the tire or wheel where it is touching the ground. But I'm conducting a thought experiment in which the wheel is a rigid object, and in which the wheel touches the ground at a single point.

I'm still not sure what point you are making - other than to say that the ideal assumptions of my thought experiment are not realistic. If that is your point, then I acknowledge it.

 

[jbriggs444]

The spoke? From the earlier discussion, isn't it clear that the spoke might be in tension or it might be in compression or (at just the right angular speed) it might be stress free. But I was speaking of the tension or compression forces from the rim.

Either you have a spoke at the point or you don't. If you don't then don't talk about one. If you do then you have to justify how you happen to have a spoke right at the infinitesimal mass element you are considering.

The mass at the point is not zero. I've said that it is m, and we can call it a point mass. Or if you prefer, we can call it dm - a differential mass.

Point mass already has a meaning in physics which is distinct from this. We must not use that term.
Differential mass is not the choice that I would make. I would accept "infinitesimal mass".

Of course, in reality, the wheel (or tire) touches the ground at more than one point, and there is some measure of deformation of the tire or wheel where it is touching the ground. But I'm conducting a thought experiment in which the wheel is a rigid object, and in which the wheel touches the ground at a single point.

Then you either need to have the mathematical sophistication to be able to deal with calculus and limits or use a different model.

I'm still not sure what point you are making - other than to say that the ideal assumptions of my thought experiment are not realistic. If that is your point, then I acknowledge it.

No that's not it. You are free to make idealizations. But you have to do so fairly. It is cheating to say that the mass in the infinitesimal interval under consideration is non-zero but that the angle made by that interval is zero.

It is also unfair to claim that the rim has no effect and then object that a model consisting of tennis shoes on spokes is inapt.

 

[EM_Guy]

Either you have a spoke at the point or you don't. If you don't then don't talk about one. If you do then you have to justify how you happen to have a spoke right at the infinitesimal mass element you are considering.

It seems that we are talking past one another here. Yes, there is a spoke at this point. We have a wheel with several spokes, by which the rim and the center are connected. I am considering a situation in which one of those spokes is vertical, and that the section of wheel touching the ground is the same section of wheel that is connected to that spoke.

Point mass already has a meaning in physics which is distinct from this. We must not use that term.
Differential mass is not the choice that I would make. I would accept "infinitesimal mass".

Fine. Then we will call it an infinitesimal mass, though I am not sure what the difference is between a differential mass and an infinitesimal mass. Is this a major concept that needs to be considered in order to have this discussion? It seems to me like this is just an issue of semantics. I'm an engineer. I'm not a mathematician or a physicist. I'm just trying to understand dynamics and kinematics of the rolling motion of the wheel.

No that's not it. You are free to make idealizations. But you have to do so fairly. It is cheating to say that the mass in the infinitesimal interval under consideration is non-zero but that the angle made by that interval is zero.

Okay, so we can also have an infinitesimal angle (though I would call it a differential angle). In this case, the tension or compression forces by the rim each have a differential vertical component. The horizontal components are in opposite directions. The question is: Are those horizontal components equal in magnitude? We can also ask the question whether the differential vertical components are equal in magnitude.

 

[jbriggs444]

It seems that we are talking past one another here. Yes, there is a spoke at this point. We have a wheel with several spokes, by which the rim and the center are connected. I am considering a situation in which one of those spokes is vertical, and that the section of wheel touching the ground is the same section of wheel that is connected to that spoke.

Unless the spokes are infinitely dense, that section of the wheel is not infinitesimal.

Fine. Then we will call it an infinitesimal mass, though I am not sure what the difference is between a differential mass and an infinitesimal mass.

You offered the opportunity to choose a term. I took the opportunity. As you say, a rose by any other name...

Okay, so we can also have an infinitesimal angle (though I would call it a differential angle). In this case, the tension or compression forces by the rim each have a differential vertical component. The horizontal components are in opposite directions.

OK. So we are back on the same page then? You agree that the net contribution of rim tension or compression across an infinitesimal section of rim has a non-zero (albeit infinitesimal) radial component?

The question is: Are those horizontal components equal in magnitude?

If the wheel is driven and if the spoke does not support shear stresses then it is clear that the two magnitudes must not be equal.

 

[A.T.]

Therefore, T = ma.

And that means T causes the acceleration?

Lets have:

F₁ = 1 N
F₂ = 1 N
F₃ = -2 N
F₄ = 1 N

F₁ = ma
F₂ = ma
F₄ = ma

So which of the 3 causes the acceleration?

And it is not arbitrary; it is deduced from the FBD of the entire unicycle / person system.

Maybe not arbitrary, but it seems quite handy-wavy and philosophical, rather than of any physical relevance.

 

[EM_Guy]

Maybe not arbitrary, but it seems quite handy-wavy and philosophical, rather than of any physical relevance.

Sure it has physical relevance. Any given vector (representing a physical force) can be decomposed into any number of components. The resultant vector is the linear combination of its components. I have broken down the force in the spoke into two components: a -Mg component and a T component. T is a function of the angular speed of the wheel. T = ma because -Mg - mg + N = 0. I acknowledge that when T - Mg = 0, then N - mg = ma. It still remains true that T = ma. This isn't a net tension. It is one component of the force in the spoke.

 

[EM_Guy]

If the wheel is driven and if the spoke does not support shear stresses then it is clear that the two magnitudes must not be equal.

This is not clear to me. It is not even clear to me that the spoke does not support shear stresses. When you have rolling motion, I would think that the torque about the axle and the moment due to static friction would cause sheer stress in the spokes. But for simplicity, let's say that the spokes do not support sheer stress. How would we describe these forces from the rim on the section of wheel touching the ground?

 

[jbriggs444]

This is not clear to me. It is not even clear to me that the spoke does not support shear stresses. When you have rolling motion, I would think that the torque about the axle and the moment due to static friction would cause sheer stress in the spokes. But for simplicity, let's say that the spokes do not support sheer stress. How would we describe these forces from the rim on the section of wheel touching the ground?

Again, you come back to the model with tennis shoes on spokes. If that is the model you want, just say so.

A bicycle wheel is an interesting example of a way to take spokes that do not support shear and arrange them so that torque can nonetheless be transmitted from hub to rim. Take a look at one. The spokes are not radial but are slanted and woven. The angle allows spoke tension to have a tangential component that is affected by torque.

 

[EM_Guy]

Again, you come back to the model with tennis shoes on spokes. If that is the model you want, just say so.

That is not the model I want. How did I "come back" to this model?

 

[jbriggs444]

That is not the model I want. How did I "come back" to this model?

You have mass units on the end of spokes supporting shear stresses and unaffected by the rim. That is precisely the situation for tennis shoes on spokes.

 

[EM_Guy]

You have mass units on the end of spokes supporting shear stresses and unaffected by the rim.

I didn't say that they are unaffected by the rim. Can you not have mass units on the end of spokes supporting shear stresses that are also affected by the rim?

 

[jbriggs444]

I didn't say that they are unaffected by the rim. Can you not have mass units on the end of spokes supporting shear stresses that are also affected by the rim?

But you do not seem to want to accept that the rim does anything. If it does nothing, it might as well not be there.

If we take a model where each spoke transmits part of the torque from hub to rim and where the contact force from the ground exerts a force at a single point, what can we say about the change in tension or compression of the rim across the point of contact with the ground?

 

[EM_Guy]

But you do not seem to want to accept that the rim does anything.

I didn't say that. I acknowledged that we have two tangential forces of the rim on the infinitesimal section of wheel touching the ground as well as two normal forces of the rim on the section of wheel touching the ground.

If we take a model where each spoke transmits part of the torque from hub to rim and where the contact force from the ground exerts a force at a single point, what can we say about the change in tension or compression of the rim across the point of contact with the ground?

I don't know. First, it is not clear to me - even in free space (no ground) whether the torque on the axle causes the rim to be in greater compression or tension? On one hand the torque transmitted through each spoke would seem to compress (push) the section of rim ahead of it, while also pulling the section of rim behind it. Therefore, without considering the ground, I don't think the torque on the axle changes the compression / tension of the rim.

But for the section of wheel touching the ground, right as that section starts to touch the ground, I think that there is increased compression (torque through spoke pushing the rim, while static friction is pushing the rim in the opposite direction). But then as that section of rim leaves the ground, the spoke seems to be pulling that section of wheel, and - in a sense - the static friction force is also pulling that section of wheel - causing increased tension (or decreased compression) in that part of the rim. Meanwhile the normal component of the force of the rim (not yet touching the ground) on the point touching the ground would be downwards, while the normal component of the force of the rim (just previously touching the ground) on the point touching the ground would be upwards.

 

[A.T.]

It still remains true that T = ma.

And that means T causes the acceleration?

Lets have:

F1 = 1 N
F2 = 1 N
F3 = -2 N
F4 = 1 N

F1 = ma
F2 = ma
F4 = ma

So which of the 3 forces causes the acceleration?

 

[EM_Guy]

The summation of forces causes acceleration.

$∑F = ma$

It turns out in this case that $∑F = T$

 

[A.T.]

The summation of forces causes acceleration.

That is also my interpretation. Although I would rather say "determines", instead of "causes".

It turns out in this case that ∑F=T

It doesn't "turn out". You simply choose to decompose a physical force such that one component is equal to the net force. There is nothing significant about the fact that you can find such a decomposition, because as you correctly noted yourself: Any given vector (representing a physical force) can be decomposed into any number of components..

 

[EM_Guy]

That is also my interpretation

Well, that's good! I'm glad we agree on Newton's 2nd Law! :)

It doesn't "turn out". You simply choose to decompose a physical force such that one component is equal to the net force. There is nothing significant about the fact that you can find such a decomposition, because as you correctly noted yourself: Any given vector (representing a physical force) can be decomposed into any number of components..

But I did so - taking two FBDs into account. I could have decomposed the Fspoke arbitrarily into any number of components. But my choice was not arbitrary.

 

[jbriggs444]

First, it is not clear to me - even in free space (no ground) whether the torque on the axle causes the rim to be in greater compression or tension? On one hand the torque transmitted through each spoke would seem to compress (push) the section of rim ahead of it, while also pulling the section of rim behind it. Therefore, without considering the ground, I don't think the torque on the axle changes the compression / tension of the rim.

Yes, we agree.

But for the section of wheel touching the ground, right as that section starts to touch the ground, I think that there is increased compression (torque through spoke pushing the rim, while static friction is pushing the rim in the opposite direction).

If there are (for instance) 24 spokes attached to the rim, what fraction of the total torque from the hub would you expect the spoke nearest the bottom to be applying to the rim?

If there is one contact patch of the tire on the road, what fraction of the total torque from the road would you expect from the contact patch?

 

[EM_Guy]

torque from hub applied to the rim: $\frac{1}{24}$^th

torque (or force) from friction applied to contact patch: the entire frictional force

 

[jbriggs444]

So if there happens to be a spoke near the contact patch, one would expect a difference between compression/tension before and after the contact patch/spoke area roughly equal to 23/24 of the frictional force. And one would expect a delta of 1/24 of the frictional force around each of the other 23 spokes.