# Forces on wheel

1. Jun 15, 2010

### darkfrog

I am in a disagreement with someone on another forum.
There is wheel with a freewheeling hub constrained to roll on a particular surface by applying force to the hub in the horizontal x direction. This person claims the tangential force applied to a wheel from the surface will also act on the center mass or fulcrum in the -x direction. We both agree that friction acts against forward motion in the -x direction but my position is that this is the only force acting against it, the rolling resistance which is a constant for the type of wheel and is a limited finite value and independent of the acceleration of the wheel. He is trying to tell me that acceleration will increase the amount of force in the -x vector. Who is correct?

2. Jun 15, 2010

### SpaceCowboy57

there is a force of rolling friction at the point instantaneously touching the ground. if it werent for the fact that this force is applied tangetial to the wheel, the wheel wouldn't roll at all, it would just slide. The forces can still be added in the x direction (the force at the hub minus the force tangential) in order to find the net force on the wheel.

For rolling resistance, the force is constant. This is because the force acts tangential to the wheel, at the point where the wheel touches the ground. This point is known as the instantaneous center of zero velocity, and thus the friction force applied here is static friction which would be constant.

3. Jun 15, 2010

### K^2

Friendly hint. If that force was constant, the wheel's rotational acceleration would be constant as well. Obviously, that's not true.

I'm the guy on the other side of the argument, by the way.

4. Jun 15, 2010

### Sithdarth

d(d(θ)/dt)/dt = α (constant angular acceleration)

d(θ)/dt = αt + ω_o

θ = (1/2)α(t^2) + ω_o(t) + θ_o

Now for a rolling wheel with θ measured in radians we know θ*r = x or distance traveled by its center of mass relative to the surface it is rolling on. Setting θ_o to zero because it is arbitrary anyway we get:

x = θ*r = (1/2)α(t^2)*r + ω_o(t)*r

Then differentiating with respect to time twice:

d(d(x)/dt)/dt = α*r = a (constant linear acceleration of the center of mass from nothing but a constant angular acceleration and the constraint of rolling)

That constant linear acceleration of the center of mass of the wheel must come from a force. So lets work with that a little:

Torque = I*α = I*a/r

but F = m*a

so a = F/m

Torque = I*α = I*F/(m*r)

Or

F = Torque*(m*r)/I

And there you have it a force on the center of mass of a rolling wheel from a torque. To take it a step further:

Torque = F_t*r

F = F_t*m*r^2/I

Assuming the wheel is a perfectly uniform disk we know I = 1/2*m*r^2. So:

F = F_t*m*r^2/(1/2*m*r^2)

F = 2*F_t

So the F acting on the center of mass of the wheel is actually twice the tangential force in this case. For a real wheel it would depend on the moment of inertia of the wheel. The fact remains that because of the constraint of rolling a force on the edge of the wheel was translated to become a force on the center of mass of the wheel.

5. Jun 15, 2010

### darkfrog

But the torque vector is in the z direction. How can it counter the force in the x direction?

6. Jun 15, 2010

### Sithdarth

Don't confuse the direction of the torque with the direction of a force. The direction of a force determines many things. The direction of a torque does nothing but keep track of the sign information. It tells you if the force causing the torque is creating a positive of negative angular acceleration and that is all it does.

7. Jun 15, 2010

### K^2

Do you actually see him adding torque to a force anywhere?

No. He computes the torque-generating force, and then applies it to CM motion, which is proper way of dealing with rigid bodies.

8. Jun 15, 2010

### darkfrog

I don't want to rehash our disagreement here, I would like others to step in and explain their thoughts on this.
Ksquared, You have already been told that the force was constant but then disagreed with spacecowboy immediately.
Your reasoning that the rolling force cannot be constant because the wheel's angular acceleration is not constant is flawed. The constant of rolling resistance operates against the force to make the wheel turn. If I push the wheel in the x direction, the rolling resistance is in the -x direction but limited to Fmax. Since I can create forces in excess of Fmax, then I can increase angular acceleration. It is the difference between my force and Fmax and that can increase.

9. Jun 15, 2010

### K^2

Force you apply at fulcrum does not generate torque. The force applied at contact point generates torque. If that force is constant, torque is constant. If torque is constant, angular acceleration is constant.

Simple, no?

If you disagree, you'll have to show me another force generating torque.

10. Jun 15, 2010

### Sithdarth

You will note that I started with a constant angular acceleration. This angular acceleration was completely general. That is any angular acceleration will behave in this manner. You will note that the bottom of a wheel is at zero velocity relative to what ever it is rolling on. Thus static friction determines the maximum tangential force that can be applied to the wheel before it slips. Rolling resistance accounts for SOME BUT NOT ALL of this maximum force. Which means if I place a torque on the wheel it will undergo further angular acceleration above and beyond the angular acceleration from rolling resistance. As long as the tangential force on the wheel is less than the maximum static frictional force between the wheel and the surface the wheels angular acceleration and the linear acceleration of its center of mass will stay linked as described.

It might be easier to think of a car. Cars accelerate forward by increasing the torque on their wheels. The more torque they supply to the wheels the faster the center of mass of those wheels accelerates. If rolling resistance supplied the maximum amount of torque the wheels could support than the car could never accelerate. The instant it increased the torque on its tires they would begin to slip. In the real world the force of static friction between the tires and the surface (also called traction) has a maximum that is much greater than the rolling resistance. This allows the car to apply lots of torque to its tires, and therefore a large tangential force on the ground, before they being to slip. Thus the car accelerates forwards. This is a basic property of rolling that all wheels must obey regardless of where the torque comes from and what supplies it.

11. Jun 15, 2010

### darkfrog

Is a wheel with a freewheeling hub a rigid body? Doesn't the hub have it's own rolling resistance, different from the rolling resistance of the wheel on the ground? Once I overcome the rolling resistance of the lubricated bearings in the hub, the rolling resistance of the tire provides the necessary force as I push the wheel forward. Once my force exceeds the friction of the tire, I will get slippage. Either way, accelerating the wheel will not increase the force against me since the rolling resistance of the hub has been overcome.

12. Jun 15, 2010

### Sithdarth

Once your force exceeds that static friction between the tire and the surface you will get slippage. This is not the same as overcoming the rolling resistance of the tire. There is quite a bit of extra force that can be exerted between overcoming rolling resistance and causing tire slippage.

http://en.wikipedia.org/wiki/Rolling_resistance" [Broken]

http://en.wikipedia.org/wiki/Friction" [Broken]

You will notice that the coefficients of rolling friction are roughly 100 times smaller than the coefficients of static friction. Which means you can apply roughly 100 times more force to the wheel than is needed to overcome rolling resistance before the wheel slips. This 100 times more force must cause the angular velocity of the tire to increase if applied at an edge. However, as I have already shown the center of mass velocity and the angular velocity are related by the act of rolling. So if some of that 100 times more force is applied tangentially to the wheel it has no choice but to increase its angular velocity which is also known as an angular acceleration. Then according to my proof since the wheel is still rolling that causes an acceleration of the center of mass of the wheel. Which indicates a force on the center of mass of the wheel in excess of rolling resistance.

Last edited by a moderator: May 4, 2017
13. Jun 15, 2010

### K^2

So you think the hub applies torque to the wheel in direction of rolling? I'm asking you, where does the torque that forces wheel to roll come from?

14. Jun 15, 2010

### darkfrog

You keep ignoring that we have a freewheeling hub. Tangential forces that create angular velocity of the wheel are also forcing it to spin around the hub. If I am holding the axle, I will not feel this force that you say is acting on the center mass because that force is also tangential to the hub which has a decidedly lower rolling resistance than the tire.

Please stop posting and don't answer. Let people from the forums here answer. That's the reason we started this thread in the first place. You both are merely repeating the same arguments you did previously.

Last edited by a moderator: May 4, 2017
15. Jun 15, 2010

### Sithdarth

Look again at https://www.physicsforums.com/showpost.php?p=2762548&postcount=4" For ANY wheel that is rolling without slipping ANY increase in angular velocity generated by ANY tangential force produces a linear acceleration on the center of mass of that wheel. If you wish the wheel to be stationary while applying a tangential force to the wheel you must apply an additional force to the center of mass of the wheel. Have you actually grabbed the axis of a rolling wheel while it undergoes a substantial tangential force? Something like 20 or 30g worth of angular acceleration on the wheel. I'd guess you haven't because if you had you would know that holding back a wheel by the axle as it undergoes an angular acceleration requires a force above and beyond rolling resistance.

1) Welcome to a public forum where as long as they aren't rude or break rules people can say whatever they like.

2) My posting is preventing no one from answering. I'd wager if anyone had anything substantial to add they'd do it regardless of anything else that is in the thread.

3) We're repeating the same arguments because they are the correct arguments.

Last edited by a moderator: Apr 25, 2017
16. Jun 15, 2010

### K^2

Enough of the silliness.

Here is a diagram. darkfrog. Main mass M has a wheel of mass m, radius R, and moment of inertia I attached to it. An external force F is applied at the center of mass of M and directly to the right.

The bearing is to be assumed frictionless. The wheels rolls without slipping and without rolling resistance.

Please, label all other force you think should be there, and find their magnitude. After that, I'll post my version, and we compare notes.

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