Free body diagram of a wheel driven by a motor

  • #1
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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.
 

Answers and Replies

  • #2
A.T.
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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.
 
  • #3
Doc Al
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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.
 
  • #4
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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.
 
  • #5
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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.
 
  • #6
A.T.
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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.
 
  • #7
A.T.
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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.
 
  • #8
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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.
 
  • #9
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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?
 
  • #10
Doc Al
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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.)
 
  • #11
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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.
 
  • #12
A.T.
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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.
 
  • #13
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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.
 
  • #14
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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?
 
  • #15
A.T.
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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?
 
  • #16
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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?
 
  • #17
A.T.
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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".
 
  • #18
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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).
 
  • #19
A.T.
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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.
 
  • #20
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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.
 
  • #21
A.T.
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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.
 
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  • #22
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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?
 
  • #23
A.T.
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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.
 
  • #24
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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.
 
  • #25
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Check out 1:48 - 1:54. Tires in compression to tires in tension.
 

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