Circular Motion, Gravity, Muddy Wheel

In summary, the problem presented involves calculating the angle (theta) at which a piece of mud will drop off a wheel with radius R, when a car is moving with constant velocity v. The mud has a mass of m and adheres to the wheel with an adhesive force of f perpendicular to the surface of the wheel. After considering various components and forces, the net force is found to be equal to the centripetal force, leading to the solution for theta. However, the problem may be unrealistic as it does not take into account the tangential component of the adhesive force and the behavior of wet mud.
  • #1
porschedude
11
0

Homework Statement


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.

Homework Equations


Fc=mv^2/R

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.
 
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  • #2
porschedude said:
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,
How can the resultant be towards the center of the wheel? The two components of [itex]f[/itex] by themselves point to the center. Add gravity in another direction, and the sum can't point to the center.
 
  • #3
tms said:
How can the resultant be towards the center of the wheel? The two components of [itex]f[/itex] by themselves point to the center. Add gravity in another direction, and the sum can't point to the center.

Ah, nice catch. Ok, so how would you recommend I determine the radial component of the net force
 
  • #4
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
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
tms said:
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.
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
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
haruspex said:
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.
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
porschedude said:
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)
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
tms said:
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.
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
haruspex said:
Same result. That will happen the moment the clump leaves road level.
That was my first thought, but then I thought I must be wrong. Now ...
 
  • #12
tms said:
That was my first thought, but then I thought I must be wrong. Now ...
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
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
tms said:
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.
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
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.
 

What is circular motion?

Circular motion is the movement of an object along a circular path. It occurs when an object travels at a constant speed around a fixed point or axis.

How is circular motion related to gravity?

Circular motion is related to gravity because gravity is the force that keeps an object moving in a circular path. It acts as a centripetal force, constantly pulling the object towards the center of the circle.

What is the muddy wheel analogy for circular motion?

The muddy wheel analogy is a way to understand circular motion by imagining a wheel rolling through a muddy surface. The mud on the outer edge of the wheel is moving faster than the mud near the center, just like how objects in circular motion have varying speeds.

What factors affect circular motion?

The two main factors that affect circular motion are the object's speed and the radius of the circular path. Increasing the speed or decreasing the radius will result in a faster circular motion.

How does inertia play a role in circular motion?

Inertia, the tendency of an object to resist changes in its motion, plays a role in circular motion by causing objects to continue moving in a circular path unless acted upon by an external force, such as gravity. This is why objects in space, where there is no significant air resistance, can continue moving in a circular orbit around a planet.

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