# Circular Motion, Gravity, Muddy Wheel

1. Feb 12, 2013

### porschedude

1. The problem statement, all variables and given/known data
A car is moving with constant velocity v, and has wheels of radius R. The car drives over
a clump of mud and the mud with mass m, and sticks to the wheel with an adhesive force of
f perpendicular to the surface of wheel. At what angle (theta) does the piece of mud drop off
the wheel?

Note that theta is the measure of the central angle of the circle.

2. Relevant equations
Fc=mv^2/R

3. The attempt at a solution
I am legitimately stumped on this problem, aside from my qualms with the question (the mud wouldn't drop off, it would fly off), this is what I have so far. I drew a free body diagram of the piece of mud. One force vector is pointing directly down. The other force vector (f) pointing toward the center of the wheel. Resolving into components:
ƩFx = fsin∅
ƩFy = fcos∅-mg

Thus, when recombining these components to determine the net force vector, I get
F = √ƩFx2+ƩFy2 in the direction of the center of the wheel, thus F=Fc=mv2/R

thus, after some algebra you can solve for theta. However, this yields only periodic values of theta that allow the force vector F to be equal to centripetal force, meaning, only at these periodic values of theta is the mud adhering to the wheel? I'm really lost, any help is much appreciated.

Last edited: Feb 12, 2013
2. Feb 12, 2013

### tms

How can the resultant be towards the center of the wheel? The two components of $f$ by themselves point to the center. Add gravity in another direction, and the sum can't point to the center.

3. Feb 12, 2013

### porschedude

Ah, nice catch. Ok, so how would you reccommend I determine the radial component of the net force

4. Feb 12, 2013

### porschedude

Nevermind, I resolved that issue. Now I'm getting that forces in the radial direction are equal to f-mgcosθ=Fc

However, if I set this equal to mv2/R, won't that be still solving for only one value of θ? Aren't I looking for a θ at which the mud falls off (an inequality)?

5. Feb 13, 2013

### tms

You're actually looking for a ≥ (or ≤ depending on how you set things up), so finding the = is good enough. That is, you're looking for the first point at which the clump falls off.

6. Feb 13, 2013

### haruspex

That assumes the force is initially less. I have a couple of concerns with this problem.
In general, the adhesive force cannot be purely radial. The resultant has to be radial, and the force of gravity is not, so the adhesion must supply a tangential force.
The adhesive force will be tested most when gravity opposes it. For the question to make sense, the mud should first appear high up on the wheel. The only reason a lot of mud can be thrown up in practice is that wet mud adheres very well to begin with but loses its grip as the mud deforms.

7. Feb 13, 2013

### porschedude

Thanks guys, I really appreciate the help. And at this point I'm resigned to say that this problem does not make sense. Given that the radius, velocity, and mass of the mud are constant, centripetal force (mv^2/R) will be constant. Given that all we're told about the adhesive force of the mud is that it's perpendicular to the wheel, I don't see how the problem can be solved. That is to say, that if the sum of the forces acting toward the center of the wheel are f-mgcos(theta), then only 2 values of theta will cause the sum of the forces acting toward the center of the wheel to be equal to mv^2/R (at every other theta the sum of the forces acting toward the center of the wheel will not be equal to mv^2/R)

8. Feb 13, 2013

### tms

I think the problem boils down to finding the point at which a tangential component to the adhesive force is necessary to keep the clump on the wheel. It is possible to get an answer that seems to make sense in that light.

9. Feb 13, 2013

### tms

In addition to the radial forces, look at the vertical forces; you can make a substitution that will help give a plausible answer. And remember that it doesn't matter if the radially inward force is greater than the outward force; that will just make the clump stick to the wheel.

10. Feb 13, 2013

### haruspex

Same result. That will happen the moment the clump leaves road level. At that point, the radial force required for adhesion is at its maximum: countering gravity + providing centripetal. An instant later, the gravitational force acquires a tangential component.

11. Feb 13, 2013

### tms

That was my first thought, but then I thought I must be wrong. Now ...

12. Feb 13, 2013

### haruspex

My guess is that the problem setter solved for equality of radial force and didn't stop to think whether it was increasing or decreasing.

13. Feb 15, 2013

### tms

I had a thought about this problem. If there is an implied frictional force between the mud and the tire that is large enough to prevent any movement in the tangential direction under any circumstances. That would allow the clump to stick until the vertical component of the adhesive force becomes less than gravity. Perhaps that is what the problem setter had in mind.

14. Feb 15, 2013

### haruspex

That just says to resolve in the radial direction, because there's an unknown and unlimited force in the tangential direction. So we have that the mud sticks as long as mg cos θ + mr ω2 < limit, where the wheel has rotated an angle θ since acquiring the mud. But mg cos θ is a maximum at θ = 0, so we're back in the same problem. Have I misunderstood your suggestion?

15. Feb 16, 2013

### tms

No, I'm the one doing the misunderstanding. I had thought that I had gotten the answer being asked for, but looking more closely it makes no sense.

It also seems that if the adhesive force is strong enough it will stick forever, even allowing tangential slipping.