Free body diagram of a wheel driven by a motor

[EM_Guy]

Neglecting drag, I'm trying to understand how a wheel driven by a motor on a level surface rolls without sliding and experiences constant velocity.

I'm trying to construct a free-body diagram.

The wheel has a weight and a normal force acting on it in the vertical directions. The sum of these forces must equal zero.

The wheel is rolling. So, the point at which the wheel is touching the ground has zero velocity. This is a kinematic constraint. If the wheel is rolling to the right, then the ground is exerting a static frictional force on the wheel that is directed to the left (I think). Since the wheel is rolling to the right, the wheel exerts a frictional force on the ground that is directed to the right, and by Newton's third law, the ground exerts a force on the wheel that acts to the left. Is this correct?

The motor is exerting a moment on the wheel. Since the wheel is rolling to the right, the torque would be clockwise. But I'm not sure if the moment exerted on the wheel about the center of the wheel is the same as the moment exerted on the wheel at the point at which the wheel is touching the ground.

This accounts for all the forces and moments acting on the wheel.

This is equivalent to a wheel of mass m and moment of inertia I whose center of mass has a linear acceleration to the right and the wheel has a clockwise angular acceleration. However, I'm saying that the wheel is moving at constant velocity. So both the linear and angular acceleration must be zero.

I think I'm getting screwed up on the moment exerted by the motor on the wheel. The moment about any point on the wheel is the sum of the cross products of displacement vectors and applied forces. So, the moment exerted by the motor about the center of the wheel is not the moment of the motor about the point at which the wheel is touching the ground.

I can't seem to set up sum(F) = ma and sum(M) = I*alpha equations here.

 

[A.T.]

However, I'm saying that the wheel is moving at constant velocity. So both the linear and angular acceleration must be zero.

Add rolling resistance.

 

[Doc Al]

If the wheel is rolling to the right, then the ground is exerting a static frictional force on the wheel that is directed to the left (I think). Since the wheel is rolling to the right, the wheel exerts a frictional force on the ground that is directed to the right, and by Newton's third law, the ground exerts a force on the wheel that acts to the left. Is this correct?

No. Try this. If there were no friction, which way would the wheel (where it makes contact with the ground) move? Friction must oppose that motion.

 

[EM_Guy]

If there was no friction, then the wheel would be sliding to the right. So, the frictional force of the ground on the wheel is directed to the left.

 

[EM_Guy]

Add rolling resistance.

Is it necessary to add rolling resistance? Let's say for a second that the rolling resistance is zero. And let's say that the motor is off. So, we only have static friction acting on the wheel in the horizontal direction.

 

[A.T.]

No. Try this. If there were no friction, which way would the wheel (where it makes contact with the ground) move? Friction must oppose that motion.

Kinetic friction opposes the relative motion at the contact. The direction of static friction depends on the forces and torques applied otherwise, and can go either way or be zero.

 

[A.T.]

Let's say for a second that the rolling resistance is zero. And let's say that the motor is off. So, we only have static friction acting on the wheel in the horizontal direction.

Without resistance or propulsion the wheel has a constant speed and static friction is zero.

 

[EM_Guy]

So, if we have pure rolling to the right and a balanced wheel, the point of contact has an instantaneous velocity of zero. But that particle is about to have a nonzero velocity directed upwards of left.

The end goal here is to set up a set of equations to correspond with the free-body diagram.

 

[EM_Guy]

Without resistance or propulsion the wheel has a constant speed and static friction is zero.

Okay. So, then I add the moment of the motor. So we have a CW torque. Now I'm confusing myself. Is the static friction of the ground on the wheel acting to the right?

So, would it be right to say that the moment about the center of the wheel minus the r x fs = I*alpha (where fs is static friction and r is the radius)?

Furthermore, is it correct then to say that fs = m*a?

Finally, is it correct to say that as long as a motor is providing torque then you will not have constant linear velocity if you do not take rolling resistance and/drag into account?

 

[Doc Al]

Kinetic friction opposes the relative motion at the contact. The direction of static friction depends on the forces and torques applied otherwise, and can go either way or be zero.

I was assuming, as stated, that the wheel was being driven to the right by a motor at its axle. And that it was not slipping. So, ignoring complications of rolling resistance, etc., there will be a static friction force acting to the right, accelerating the wheel. (But you're right, in general.)

 

[EM_Guy]

Without resistance or propulsion the wheel has a constant speed and static friction is zero.

I think I finally get it. I envisioned the system of particles that comprises the bottom part of the wheel. Thinking about the free body diagram of that system of particles...

Even when the moment (propulsion) is zero (and thus the static friction acting to the right is zero), the bottom of the wheel (in the frame of reference of center of the wheel) is moving to the left with some speed. Therefore, there is an tension force acting upward on the bottom of the wheel causing it to accelerate upward. Before I considered this tension force, I was trying to figure out how that part of the wheel ever moved.

Thanks for the help.

 

[A.T.]

Therefore, there is an tension force acting upward on the bottom of the wheel causing it to accelerate upward.

I'm pretty sure the bottom of the wheel is under vertical compression, not tension. But the net force on any part of the wheel (at constant speed) is towards the center, hence upwards for the bottom part.

 

[EM_Guy]

I'm pretty sure the bottom of the wheel is under vertical compression, not tension. But the net force on any part of the wheel (at constant speed) is towards the center, hence upwards for the bottom part.

If you draw a FBD of the small section of the wheel that is touching the ground, these are the forces that I have on that FBD:
1. Weight of that section.
2. A downward normal force on that section by the part of the wheel that is connected to but above that section.
3. A normal force of the ground on that section which is equal and opposite to the sum of 1 & 2.
4. A leftward force on that section that is due to the moment about the center of the wheel.
5. A static friction force on that section that is equal and opposite to 4.
6. This is where I think the tension force comes in. The section of the wheel that is touching the ground is connected to the center of the wheel. This is somewhat like a ball on a string, no? So I think this is the upward tension force.

I think that the vector sum of 1, 2, and 3 equal zero. I think 2 and part of 3 account for the compression on the bottom of the wheel.

I think that the vector sum of 4 and 5 equal zero. Since we are talking about static friction and pure rolling, I think this must be the case.

So, this part of the wheel has an instantaneous velocity of zero. And the vector sum of the forces 1 through 5 I think is zero. But the center of the wheel has a nonzero velocity to the right. Or from the frame of reference of the center of the wheel, the section of the wheel that we are examining has a nonzero velocity to the left. And this section of the wheel is connected to the center of the wheel. So, I think these are in tension giving us force #6, directed upward. Thus, the section of the wheel in touch with the ground that has zero velocity has an acceleration directed upward (from the frame of reference of the ground). From the frame of reference of the center of the wheel, the section of the wheel in touch with the ground has a leftward velocity with an upward acceleration.

 

[EM_Guy]

If we think of a bicycle wheel, I think we would say that the section of tire touching the ground is in compression, but I think we would say that the spokes are in tension, no?

 

[A.T.]

If you draw a FBD of the small section of the wheel that is touching the ground, these are the forces that I have on that FBD:
1. Weight of that section.
2. A downward normal force on that section by the part of the wheel that is connected to but above that section.
3. A normal force of the ground on that section which is equal and opposite to the sum of 1 & 2.

If 2 is downwards then you have vertical compression, not tension. It the spoke was under tension, it would pull upwards.

I think that the vector sum of 1, 2, and 3 equal zero.

Based on what?

 

[EM_Guy]

All the matter connected to the section of interest and above it has its own weight. Let object A be on top of object B, and let object B have zero velocity and let object B be touching the ground.

Drawing a FBD of object B, we have object B's weight, a downward normal force on object B from object A (which is equal in magnitude to the weight of A), and an upward directed normal force on object B from the ground.

Now if object B is connect to object A - such that the two objects together form one rigid body, and if object B has a nonzero velocity, then there is an additional tension force, no?

 

[A.T.]

Now if object B is connect to object A - such that the two objects together form one rigid body, and if object B has a nonzero velocity, then there is an additional tension force, no?

Acceleration is relevant for forces, not velocity. And no idea what you mean by "additional tension force".

 

[EM_Guy]

Acceleration is relevant for forces, not velocity. And no idea what you mean by "additional tension force".

That object B has a nonzero velocity is a kinematic constraint. Object B is connect to Object A. Object A has an instantaneous velocity of zero. Object B has a nonzero instantaneous velocity at the same instant in a direction that is perpendicular to the displacement vector between Object B and Object A. This is analogous then to swinging a ball that is attached to a string. Or analogous to a wheel. The section of tire touching the ground (Object A) has zero velocity. The center of mass of the wheel (Object B) has a nonzero velocity. There is not tension in the spoke that connects the wheel center to the wheel edge?

Examining the kinematics in the frame of reference of the center of the wheel, the wheel edge that is touching the ground has a nonzero velocity to the left, and it is accelerating toward the center of the circle. This normal acceleration is proportional in magnitude to the square of the velocity. Some force must be causing this acceleration. I'm saying it is a tension force in the spoke (or if the wheel is solid - in the material of the wheel connecting the center of the wheel to the edge of the wheel).

 

[A.T.]

Examining the kinematics in the frame of reference of the center of the wheel, the wheel edge that is touching the ground has a nonzero velocity to the left, and it is accelerating toward the center of the circle.

Right

Some force must be causing this acceleration.

It's called "net force", which is the sum of all forces.

I'm saying it is a tension force in the spoke

Wrong. The spoke pushes down on the edge, but less than the ground pushes up. So the net force on the edge is up, towards the center of the wheel.

 

[EM_Guy]

Wrong. The spoke pushes down on the edge, but less than the ground pushes up. So the net force on the edge is up, towards the center of the wheel.

I know about net force. I am breaking down the individual forces. I'm not seeing this. I get it and agree that the net force acting on the edge is up, for the edge is accelerating upward. But how to tell if the spoke is in compression or tension.

I suppose we can do a FBD of a differential element of the spoke.

1. The differential element has its own weight.
2. The differential element has a normal downward force acting on it (by everything above it that is attached to it). The moments due to the motor and static friction do not impact this particular normal downward force, right?
3. The differential element has a normal upward force acting on it (by the section of spoke beneath it that is attached to it). Again, the moments due to the motor and static friction do not impact this particular normal upward force, right?
4. A leftward torsional force on it due to the moment of the motor about the center?
5. A rightward torsional force on it due to the moment of the static friction force on the edge of the wheel?

Kinematic constraint: In the frame of reference of the wheel, this element is moving to the left with some velocity, implying that it has acceleration towards the center of the circle. (Note also that if the moment is causing the wheel to experience angular acceleration - and why wouldn't it, then this differential element also has tangential or horizontal acceleration).

It appears that the spoke is under compression, not tension, and that the compression forces are force #2 and part of force #3. Force #3 is greater than the sum of forces #2 and #1, thus causing the differential element to accelerate in toward the center.

Furthermore, it looks like all the spokes (above and below the center of the wheel) are under compression. In order for the spokes to be under tension, some forces would need to be applied in the outward radial direction (pulling the wheel apart).

I think I see now, but if I'm still off course, please let me know. Thanks for patiently explaining this to me.

 

[A.T.]

But how to tell if the spoke is in compression or tension.

If the wheel is stress free in zero g, then it will be radially compressed between the loaded axis and ground. That's the normal case that I was considering so far. But you could build a pre-tensioned wheel, that would have radial tension in all directions, up to a certain axis load.

 

[EM_Guy]

If the wheel is stress free in zero g, then it will be radially compressed between the loaded axis and ground. That's the normal case that I was considering so far. But you could build a pre-tensioned wheel, that would have radial tension in all directions, up to a certain axis load.

But in that case (of a pre-tensioned wheel), if you increase the load on the loaded axis, then will the tension in wheel decrease with increased load, up until a certain point, above which the pre-tensioned wheel is then compressed more and more with increased load?

 

[A.T.]

But in that case (of a pre-tensioned wheel), if you increase the load on the loaded axis, then will the tension in wheel decrease with increased load, up until a certain point, above which the pre-tensioned wheel is then compressed more and more with increased load?

Yes, that's what will happen between the axis and ground of a rigid pre-tensioned wheel. If the spokes are made of tensioned elastic strings, the string below the axis will have reduced tension or even go slack.

 

[EM_Guy]

Wrong. The spoke pushes down on the edge, but less than the ground pushes up. So the net force on the edge is up, towards the center of the wheel.

I just talked this problem over with a colleague of mine (who is a mechanical engineer). This is what we came up with:

Consider the spoke directly under the center of the wheel and the section of the wheel that is touching the ground. Let M = the weight of the vehicle being loaded on the axle, and let m = the weight of the section of the wheel that is touching the ground. Let the spoke be massless.

Can we not consider the force in the spoke as the superposition of compressive and tension forces?

So, if the car is just sitting there (omega = 0), then we have Mg pushing down on the spoke and N pushing up on the spoke. N = Mg.

If we forget the vehicle and forget gravity and are in total free space outside of any gravitational field and if we spin this wheel, then the spokes will be under tension, and the mass on the edge of the wheel will be accelerating towards the center of the circle (assuming omega = constant).

Can we not superimpose these two together?

In a free body diagram of the little section of wheel touching the ground, we have Mg pressing down on it, N pushing up on it, and since we are spinning, we have a tension force pulling up on it (the spoke is pulling the mass up). So, -Mg + N + T = ma = mv^2/r. -Mg + N = 0. So, T = ma.

Now, if the spoke has a cross-sectional area of A, then the stress in the spoke would be compressive stress = Mg/A - mv^2/r/A. If mv^2/r > Mg, then the spoke is in tension. If mv^2/r = Mg, then the spoke is not under any stress. If mv^2/r < Mg, then the spoke is in compression. But whether the spoke is in compression or tension, it is the tension component that is causing the piece of wheel touching the ground to accelerate upwards.

If -Mg + N > 0, then the whole vehicle would be lifting up off the ground. The vehicle is staying on the ground, so - Mg + N = 0.

If you watch a drag race, notice how the tires are under compression when the cars are still or slowly moving. Notice how that changes when they start moving fast.

 

[EM_Guy]

Check out 1:48 - 1:54. Tires in compression to tires in tension.

 

[A.T.]

Now, if the spoke has a cross-sectional area of A, then the stress in the spoke would be compressive stress = Mg/A - mv^2/r/A. If mv^2/r > Mg, then the spoke is in tension. If mv^2/r = Mg, then the spoke is not under any stress. If mv^2/r < Mg, then the spoke is in compression.

Yes indeed, if the wheel spins fast enough, even the lowest spoke might come under tension. In this case the axle is supported by the tension of the spokes above it.

But whether the spoke is in compression or tension, it is the tension component that is causing the piece of wheel touching the ground to accelerate upwards.

The acceleration upwards is a function of the net force, and it makes little sense to attribute it to individual forces. Especially when the spoke is compressed and thus pushes down on the piece of wheel touching the ground.

 

[EM_Guy]

The acceleration upwards is a function of the net force, and it makes little sense to attribute it to individual forces. Especially when the spoke is compressed and thus pushes down on the piece of wheel touching the ground.

If you draw a FBD of the piece of wheel touching the ground, you have this equation:

-Mg + N + T = ma = mv^2/r

So, you are correct to say that the acceleration is a function of the net force. However, I'm contending that -Mg + N = 0, assuming that the car itself isn't lifting off the ground. Therefore, are we not left with T = mv^2/r?

 

[A.T.]

If you draw a FBD of the piece of wheel touching the ground, you have this equation:

-Mg + N + T = ma = mv^2/r

I get:

-Mg - mg + N = ma

-Mg is the total force from all other wheel parts (spoke and circumference) acting on the piece of wheel touching the ground.

-mg is gravity acting on the piece of wheel touching the ground.

N is the vertical ground reaction acting on the piece of wheel touching the ground (ignoring horizontal components here)

 

[EM_Guy]

I get:

-Mg - mg + N = ma

-Mg is the total force from all other wheel parts (spoke and circumference) acting on the piece of wheel touching the ground.

-mg is gravity acting on the piece of wheel touching the ground.

N is the vertical ground reaction acting on the piece of wheel touching the ground (ignoring horizontal components here)

I should have put -Mg - mg + N + T = ma = mv^2/r.

You are saying that there is no tension component.

Take the same wheel spinning at the same angular speed in free space (no ground, no friction, no gravitational field).

Then, the equation becomes:

T = ma = mv^2 / r.

According to you, the normal force exceeds the magnitude of (M+m)g, which is what causes the acceleration. But this would also cause the car itself to accelerate upward. Consider the FBD of the car itself (or unicycle in our case - for simplicity).

Vertical direction:

-(M+m)g + N = (M+m)a.

If N > (M+m)g, then the car accelerates off of the ground.

I maintain that N = (M+m)g, since the car is staying on the ground.

All this to say that you have to include T in the FBD of the piece of wheel touching the ground, and that this causes the piece of wheel to accelerate. Again, this does not imply that the spokes are necessarily under tension. Rather, I am saying that the squeezing of the spokes by the weight loaded on the axle and the normal force of the ground causes some measure of compression, but that at the same time the wheel is spinning with some angular speed. And thus, there is some measure of tension in the spokes too.

 

[A.T.]

All this to say that you have to include T in the FBD of the piece of wheel touching the ground, and that this causes the piece of wheel to accelerate.

I don't understand your causation logic. Let's say you have a mass of 1kg with 3 forces:

F₁ = 1N
F₂ = 1N
F₃ = -1N

So F_net = 1N and a = 1 m/s². Now tell me: Which of the three forces is causing the acceleration according to you? It is F₁ because F₂ and F₃ cancel, or is it F₂ because F₁ and F₃ cancel?

 

[EM_Guy]

How about we have that discussion after we agree on the right equation for the FBD? You seem to be acknowledging that as the angular speed of the wheel increases beyond a certain point, even the spokes on the bottom of the wheel could be in tension. But when we draw a FBD of the part of the wheel touching the ground, you include no tension in your diagram or in your equation.

 

[A.T.]

How about we have that discussion after we agree on the right equation for the FBD?

Well, draw one. Define the bodies and the interactions between them.

 

[EM_Guy]

See post #27 and my correction in post #29. There is the equation describing the FBD of the part of the wheel touching the ground. In post #28, you did not include a tension force.

 

[A.T.]

There is the equation describing the FBD of the part of the wheel touching the ground.

A FBD defines interacting bodies and labels the forces between them. Which terms correspond to which body exerting a force on the part of the wheel touching the ground?

 

[EM_Guy]

From post #29:

-Mg = downward force of vehicle exerted through the spoke onto the section of wheel touching the ground
-mg = weight of section of wheel touching the ground
N = normal upward force of ground acting on the section of wheel touching the ground
T = tension in spoke attaching the section of wheel touching the ground to the center of the wheel

 

[A.T.]

-Mg = downward force of vehicle exerted through the spoke onto the section of wheel touching the ground
T = tension in spoke attaching the section of wheel touching the ground to the center of the wheel

If the spoke is one body, then there is one force F_{spoke on section} exerted by the spoke on the section of wheel touching the ground. And that force is either downward or upward or zero.

 

[EM_Guy]

If the spoke is one body, then there is one force F_{spoke on section} exerted by the spoke on the section of wheel touching the ground. And that force is either downward or upward or zero.

There is one net force exerted by the spoke on the section of wheel touching the ground. It will be downward, upward, or zero. But that net force can be described as the superposition of two forces (as I have described).

 

[A.T.]

But that net force can be described as the superposition of two forces

You can just as well decompose the ground reaction force into two forces, one of which equals m*a, and claim that this is the force that "causes acceleration". It's arbitrary and meaningless.

 

[EM_Guy]

You can just as well decompose the ground reaction force into two forces, one of which equals m*a, and claim that this is the force that "causes acceleration". It's arbitrary and meaningless.

I believe you are off base here. But let me suspend that belief for a second and assume that you are correct.

I have said: -Mg - mg + N + T = ma = mv^2/r

You have said: -Mg -mg + N = ma

I have also said: -Mg - mg + N = 0 (Consider a FBD of the entire vehicle. The vehicle isn't lifting up off of the earth.)

You have not replied to this. Let me ask you point blank: Do you or do you not agree that this equation describes the FBD of the vehicle?
-Mg - mg + N = (M+m)ay

ay = acceleration in the y-direction.
M is a constant (the mass of vehicle - less the section of wheel touching the ground).
m is a constant (the mass of the section of wheel touching the ground).
g is obviously a constant.

Let's put it this way, I'm saying that the net upward force that the spoke exerts on the section of wheel touching the ground is this:

Fspoke = T - Mg

You are saying that the net upward force that the spoke exerts on the section of wheel touching the ground is this:

Fspoke = -Mg

Now, neither M nor g change as the angular speed of the wheel changes, so you are telling me then that the angular speed of the vehicle does not impact one bit the force of the spoke on the section of wheel touching the ground.

Yet, you also say:

if the wheel spins fast enough, even the lowest spoke might come under tension

Do you not think that you are contradicting yourself?

You certainly make assertions with confidence. Unless you are right and confident that you are right, you should not express yourself with so much confidence.

 

[A.T.]

You have said: -Mg -mg + N = ma

Ok, I see what you mean. I reused "-Mg" for "the total force from all other wheel parts (spoke and circumference) acting on the piece of wheel touching the ground." which is wrong.

I have said: -Mg - mg + N + T = ma = mv^2/r

This is basically decomposing F_{spoke on section} into -Mg and ma. Where you loose me is calling ma "spoke tension" (T) and claiming it is "causing the acceleration".

 

[EM_Guy]

This is basically decomposing F_{spoke on section} into -Mg and ma. Where you loose me is calling ma "spoke tension" (T) and claiming it is "causing the acceleration".

Okay, let's start at the beginning. We just have a wheel. The axle is unloaded. No vehicle. No gravity even. Just a wheel in space outside of any gravitational field.

Let us examine the force acting on one little section of the rim that is connected to a spoke.

If the wheel is not spinning, then the net force acting on that section of the rim is zero. We have no gravity, no normal force, no tension or compression in the spoke.

Okay, now, let's say that the wheel is spinning at some constant angular speed.

Now, if we consider a FBD of a small section of the wheel, we have this:

Fspoke = ma.

And we know that a = v^2/r from kinematic constraints. Fspoke = tension = T.

Okay...

Now, scenario two: We have a wheel on a unicycle on the Earth and a person sitting on the unicycle. No rolling, no motion.

FBD of small section of the wheel touching the ground:

Fspoke - mg + N = 0.

Fspoke = -Mg. The spoke is in compression. The weight of the unicycle and person sitting on the unicycle is transmitted through the spoke pushing the section of wheel on the ground down. The normal force is pushing it up.

Okay, now let the wheel roll at some angular speed.

Fspoke - mg + N = ma. (This is the equation corresponding to the FBD of the section of the wheel that is touching the ground).

But now Fspoke is the superposition of the weight of the unicycle and person sitting on the unicycle (causing compression) and the tension that arises due to the fact that you have a piece of a wheel connected to the rest of the wheel (which we are taking as a rigid body). Specifically, the piece of the wheel touching the ground is connected to the center of the wheel through the spoke.

Thus, Fspoke = T - Mg. (I am using superposition here. Mg due to the weight of the unicycle / rider compressing the spoke, while the angular speed of a rigid body causes tension in the spoke. Is the spoke under net tension or net compression? It depends how fast the wheel is spinning and the mass of the system).

Therefore,

T - Mg - mg + N = ma. (This is the equation corresponding to the FBD of the section of the wheel that is touching the ground - keeping in mind that I have done a simple substitution here for Fspoke).

Again, a = v^2/r.

At the same time, the unicycle and the rider are not accelerating upwards off the ground. This is the crux of my argument. So if we draw a FBD of the whole system, we get:

-Mg - mg + N = 0. (This is the equation corresponding to the FBD of the whole unicycle / rider system.)

Note that while the section of wheel that is touching the ground is accelerating upward, the system taken as a whole is not accelerating upward.

Therefore,

T = ma. (This is the equation corresponding to the FBD of the section of wheel that is touching the ground - simplified - given the equation corresponding to the FBD of the whole unicycle / rider system).

Note: This does not imply that the spoke is under any net tension.

a = v^2 / r.

Now, let's consider the compressive and tension stress in the spoke. Let negative correspond with compression and positive correspond with tension:

(T - Mg)/A = net tension stress.

But I have said that T = ma = mv^2/r.

So,

(mv^2/r - Mg) / A = net tension stress.

Therefore, if mv^2/r > Mg, the spoke is under positive net tension. If mv^2/r < Mg, the spoke is under positive net compression (aka negative net tension). If mv^2/r = Mg, the spoke is under no stress at all.

 

[EM_Guy]

The crux of my argument is that we can (and we do) have a situation in which forces are simultaneously trying to place the spoke under compression and tension.

Imagine a rigid box. You can imagine simultaneously compression forces applied to the box - trying the squeeze the box, while tension forces are trying to stretch the box. Is the box under compression or tension?

 

[A.T.]

Okay, let's start at the beginning. We just have a wheel. The axle is unloaded. No vehicle. No gravity even. Just a wheel in space outside of any gravitational field.

When you have only a single force acting on the section (like when spinning in space), then I can see how you can say that this force "causes the acceleration". But as soon you have multiple forces, then I don't see the physical significance of such attribution. You can represent any of the forces as ma and the remainder.

The crux of my argument is that we can (and we do) have a situation in which forces are simultaneously trying to place the spoke under compression and tension.

It's not a situation. It just the way you have chosen to decompose two compressive forces (for slow rotation).

Imagine a rigid box. You can imagine simultaneously compression forces applied to the box - trying the squeeze the box, while tension forces are trying to stretch the box. Is the box under compression or tension?

Let's say you push the box with 2N from left and push 1N from right. It's obviously under compression and accelerating to the right. But then you choose to represent the 1N right push force as a superposition of 2N push and 1N pull, then you call the 1N pull is a "tensional component" which "causes the acceleration". That's what it all boils down to, as far I can see.

 

[EM_Guy]

You are getting distracted by my box analogy. Go back to post # 41. Which, if any, of my equations are wrong and why?

 

[A.T.]

You are getting distracted by my box analogy. Go back to post # 41. Which, if any, of my equations are wrong and why?

It's not the equations, but your interpretation of them. The simple box example demonstrates its arbitrariness very well.

 

[EM_Guy]

So my equations are right, but my interpretations are wrong? Which interpretations are wrong?

You missed the point of the box analogy. If the box is under compression then you have two forces acting on it - squeezing it. The net force is thus zero, and the box doesn't accelerate, but it is under compression. If the box is under tension, then you have two forces acting on it - trying to pull it apart. The box doesn't accelerate, but the box is under tension.

Bottom line, if you are not going to take the time to go step by step through post #41, then I do not see that it will benefit either of us to continue this discussion.

 

[A.T.]

Which interpretations are wrong?

Not wrong, but arbitrary: The attribution of the total acceleration to some component of one of multiple forces.

If the box is under compression then you have two forces acting on it - squeezing it. The net force is thus zero, and the box doesn't accelerate,

Compression and acceleration aren't mutually exclusive, and since our wheel section is accelerating so should the box in the analogy.

 

[A.T.]

f mv^2/r = Mg, the spoke is under no stress at all.

Note that in this case the spoke exerts no force on the bottom wheel section. So the only force acting on the bottom wheel section is the force from the ground. And yet you still choose to attribute the acceleration not to that single force acting from the ground, but to some "tension component" of a spoke that isn't under tension and exerts no force?

 

[EM_Guy]

Not wrong, but arbitrary: The attribution of the total acceleration to some component of one of multiple forces.

It is not arbitrary. I have two FBDs - one for the entire unicycle / person system and one for the piece of wheel that is touching the ground. The two equations share terms.

And yet you still choose to attribute the acceleration not to that single force acting from the ground, but to some "tension component" of a spoke that isn't under tension and exerts no force?

T - Mg - mg + N = ma

Fspoke = T - Mg

So, you can say

Fspoke - mg + N = ma

And if Fspoke = 0, then

-mg + N = ma

Granted. When there is no stress in the spoke, N - mg causes the upward acceleration of the segment of the wheel touching the ground. This is true.

But what I have been saying is true too.

-Mg - mg + N = 0

Therefore, T = ma.

This tension isn't a "net tension" but a "tension component" in the Fspoke = T - Mg vector. And it is not arbitrary; it is deduced from the FBD of the entire unicycle / person system.

 

[jbriggs444]

Okay, let's start at the beginning. We just have a wheel. The axle is unloaded. No vehicle. No gravity even. Just a wheel in space outside of any gravitational field.

Let us examine the force acting on one little section of the rim that is connected to a spoke.

If the wheel is not spinning, then the net force acting on that section of the rim is zero. We have no gravity, no normal force, no tension or compression in the spoke.

Stop right there. That claim is unsupported. The rim may be under tension or compression.