# Rolling Ball

1. Dec 3, 2008

### monsterdivot

Hi All,

I'm trying to figure out how to model a rolling ball in the x y plane.

I've started with the basic equation of motion with constant acceleration

d = v$$_{}0$$t + 1/2at$$^{}2$$

v$$_{}0$$ is the initial velocity
a is acceleration
t is time

Assume that v is a vector at an angle $$\alpha$$ with the x axis.
If the plane is flat, a is is all due to the force of rolling resistance F$$_{}r$$. I represent F$$_{}r$$ as a vector in the opposite direction to v. Is this the correct way to treat rolling resistance? If so, does this mean that if we decompose the equation into x and y components, does this mean that F$$_{}r$$ should be decomposed into x and y components?

F$$_{}rx$$ = F$$_{}r$$sin$$\alpha$$

and

F$$_{}ry$$ = F$$_{}r$$cos$$\alpha$$

Given that I'm treating F as a vector it makes sense to do this, but it doesn't feel right to me. This means that the rolling resistance varies in x and y with the angle of the vector v.

Is this right? Also, is F independent of the speed of the ball?

Sorry about the appearance. The subscripts aren't working for some reason. Only the t squared should show as a superscript. the rest should be subscripts.

Last edited: Dec 3, 2008
2. Dec 4, 2008

### tiny-tim

Welcome to PF!

Hi monsterdivot! Welcome to PF!

I'm not sure what you mean by "rolling resistance" â€¦

a ball doesn't have an axle, so (unless either the surface or the ball itself deforms, which in exam questions never happens , for which case see http://en.wikipedia.org/wiki/Rolling_resistance), the only "rolling resistance" is air-resistance.

The best way to treat a rolling ball problem without deformation is to use conservation of energy.

3. Dec 4, 2008

### monsterdivot

I'm afraid I am talking about deformation of the ball and or the surface. This is not an exam, but real life.

The rolling resistance I'm refering to is also known as rolling friction. It is caused by the deformation of the ball or the surface and is what causes a ball to eventually stop rolling. For my case, air resistance is assumed to be negligible.

The essence of my question is how to model the force caused by rolling resistence when decomposing the problem into x and y components. Why even decompose them into x and y you may ask since the problem could be treated as a 2D problem. Well, I hope to add slopes in the xz and yz planes later.

Thanks

4. Dec 4, 2008

### arildno

These are extremely local effects, in which you cannot ignore the particular materials the ball and the surface are made of, along with their changing shapes. The dissipation of energy due to not fully elastic deformations cannot, at present, be theoretically determined. As with many other real-life phenomena, look up in a table instead, to find measured values.

5. Dec 4, 2008

### monsterdivot

I don't want to know an actual value for the rolling friction which would depend on the ball and the surface characteristics. I just want to know how to treat the force when decomposing the equation of motion into x and y components. I can really just plug in any value for that and it will just make the ball deccelerate slower or faster.

Let me state my question again.

I have a ball with an initial velocity v, rolling on a flat surface, with some rolling friction F acting on the the ball in opposition to the direction of v. The x and y axis are in the plane of the flat surface with v and F at an angle alpha with the x axis.

Treating v as a vector, its x component is vcos(alpha) and y component is vsin(alpha). Should F be decomposed into components of x and y like v is? This would mean that the x and y components of F are not equal unless alpha is 45 deg.

Intuitively, to me this seems wrong. If we are looking at the independent components of the motion in x and y the rolling resistance should be equal in both x and y. The rolling friction is essentially constant. Of course, my intuition is not very reliable. That is why I ask the question.

Thanks again for the help.

6. Dec 5, 2008

### MikeLizzi

In mechanical engineering, rolling friction is modeled as static friction at the point of contact of the ball and the surface. Its direction is tangent to and pointing opposite to the rotation of the ball. Its magnitude is a function of the normal force of the ball on the surface.

So, if the surface is oriented so that it has both an x and y component (like a hill) the friction force will have both an x and y component.

Will you be conditions for slipping? That is when the static friction changes to sliding friction.

7. Dec 5, 2008

### monsterdivot

I think you're refering to sliding friction not rolling resistance. I may have confused things by switching to the term rolling friction. See http://en.wikipedia.org/wiki/Rolling_resistance

I'm not considering sliding friction here assuming that there is enough friction to keep the ball from sliding. In my case, the ball has perfect roll.

Thanks

8. Dec 5, 2008

### MikeLizzi

You already know this, but just to refresh;

The coefficient of friction is typically given for two conditions. One when the objects are slipping past each other. That's sliding friction. The other when they are not. That's static friction. A ball rolling without slipping is doing so because of static friction. If you don't have static friction, then the ball doesn't roll. It just sits there and spins.

What you call rolling resistance is a consequence of static friction.

That's the way engineers handle it. If you try to model the behavior of a ball you can set it up as rolling up a hill without slipping. Then you use static friction. But if your model includes changes in the slope of the hill, the static friction you initially set may not be sufficient to maintain the non-slip condition. So a comprehensive model must take into account sliding friction too.

But you can always start with a static friction model. That will take you pretty far.

Additional Comment: the wikipedia article is a more detailed examination/explanation of what static friction really is.

Last edited: Dec 5, 2008
9. Dec 7, 2008

### MikeLizzi

I didn't want to stifle the dialog. I can show you haw to set up the equations if you are still working on it.