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Rolling Friction

  1. Jan 11, 2007 #1
    Consider a hollow cylinder rolling down an inclined plane, without slipping. I know that the rolling friction does no work on the point which is in contact with the cylinder, since that point must be at rest since we have rolling without slipping, but isn't the ENTIRE BODY still moving, so the friction force acts over this distance, so it must be doing work on the BODY. These seem to contradict. Furthermore, we know that forces acting anywhere on the body force the body to accelerate like a point mass at the center of mass, so in this case the friction force is acting opposite the acceleration of the center of mass, and hence doing work on the object. This seems to contradict the fact that rolling friction is supposed to do no work. So why does rolling friction do no work? What is wrong with the reasonings above?
     
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  3. Jan 11, 2007 #2

    Doc Al

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    You need to distinguish between "real" work and what is sometimes called "pseudowork" (though that last term never really caught on - pity!). "Real" work is the same work that appears in the first law of thermodynamics (conservation of energy) and depends on the details of how the force is applied and the movement of the point of contact. No real work is done by friction on the rolling cylinder, since the point of contact doesn't move. That's why mechanical energy is conserved as the cylinder rolls down the incline. (If friction did work, you'd get an increase in internal energy--"heat"--somewhere.)

    On the other hand, friction is most definitely exerting a force on the cylinder. If you take a force acting on an object (like friction) and multiply it by the displacement of the object's center of mass, you get a quantity that looks like a work term but is better called pseudowork (or "center of mass" work)--what it determines is not the real work done on the object, but the change in the KE of the center of mass of the object. This is usually called the "Work-Energy" theorem:
    [tex]F_{net}\Delta x_{cm}=\Delta (\frac{1}{2}m v_{cm}^2)[/tex]

    Despite the name, this is really a consequence of Newton's 2nd law, not a statement of energy conservation.
     
  4. Jan 11, 2007 #3
    So doesn't "pseudo-work" also have to cancel? I mean, I think that pseudo-work cancels because the work done by the force on the center of mass is exactly canceled by the work done on one point as it moves around the circle: for example, consider a point instantaneously in contact, then let the circle roll, the point is now somewhere higher, and even though the force is acting on a different point, it has the same effect as acting on the point in the direction of motion, since its a rigid body; since the motion of the point is the same as the motion of the center of mass, the pseudowork cancels.

    Is this valid?

    I don't quite believe what I said though, because it seems to me as if we might be "double counting" here: a force on the point in the direction of its motion would also effect the center of mass. I am just confused; help!
     
  5. Jan 11, 2007 #4

    Doc Al

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    Not sure I follow your reasoning, but pseudowork doesn't "cancel" since it acts directly on the center of mass. Since pseudowork (and the work-energy theorem) is just an application of Newton's law, you can use it to calculate the change in center of mass KE. (The terminology may be confusing: Pseudowork is quite real, but it's just not "work" in the conservation of energy sense.)

    Right. Any net force on an object will affect the acceleration of the center of mass. (You can either calculate the acceleration directly, or use the W-E theorem to find the change in translational KE.) The net force on the rolling cylinder will be:
    [tex]F_{net} = mg\sin\theta - F_f[/tex]

    Of course, you don't know the friction force ahead of time, but you can calculate it by applying Newton's 2nd law. Apply it to both translation and rotation, and then solve for the acceleration and friction.
     
  6. Jan 11, 2007 #5
    The way you are suggesting is T = I(alpha), F = ma method of solving the problem, and this makes a lot of sense to me. It's just that it seems that there should be a way to justify it using energy conservation besides saying that, work is simply due to the force being applied to the SAME point of contact over a distance. So the way above was one way I was thinking of justifying why the work that the friction force does to the translational movement is equal in magnitude and opposite in direction to the work that the friction does to a point moving around a circle (since the point travels the same distance as the CM). But in trying to justify it this way, it seemed like there was an apparent contradiction in overcounting. So essentially, my question is: what is another way of justifying that friction does no work, besides the "sketchy" (to me, because I always thought was just Fd, and since the object does move, its fine; is there another more rigorous definition?) that the force isn't acting on the same point?

    thanks for your help in advance
     
  7. Jan 11, 2007 #6

    Doc Al

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    That's just one way of solving the problem.

    Again, you are describing pseudowork, not real work. If friction did real work, the total mechanical energy of the cylinder would change--but it doesn't.

    To find the work (the actual work, as in conservation of energy) done by a force you must study the details of how the force is applied. The point of application must move, otherwise no work is done. As you know, since the cylinder rolls without slipping, the instantaneous speed of the point of contact is zero--it doesn't move.

    Don't confuse the fact that an applied force must accelerate an object--per Newton--with the thought that it must also transfer energy to the object (that's what real work means). Another example: An accelerating car. Friction of the road on the tires certainly accelerates the car, but it does no work on the car (for the same reason, since we assume no slipping). It's not the road that provides energy for the car, it's the gasoline and the engine! Of course, you can use pseudowork (if you know the friction force) to calculate the change in KE using the work-energy theorem described above.

    Here's another example: You jump in the air. Your KE obviously increased, but did the ground do work on you? No! Your point of contact with the ground did not move, so the ground did no work on you. So where did the energy come from? From the chemical energy in your muscles, of course, not from the ground. Of course, you can consider the average force that the ground exerts on you and use it to calculate your KE--but that's an application of pseudowork again, not real work.

    Let me know if this is helping at all.
     
  8. Jan 11, 2007 #7
    ok, i guess the examples made it clear. but i have one more question that's bothering me: essentially, what you have said implies that the work energy theorem holds iff we define work to be the force applied on the body at a contact point (dot product) the distance the contact point moves. So why is this true? I mean, why, mathematically and intuitively, would this be true? So does this imply that things like torque were basically defined because work energy holds only for single contact points. I'm just trying to really understand why we must use that definition of force, because then the fact that rolling friction does no work immediately follows.

    thanks
     
  9. Jan 11, 2007 #8

    Doc Al

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    I think you have it backwards. The "work" in the work-energy theorem is not real work, but pseudowork. (Only for the special case of a point particle will the pseudowork equal the real work.) Reread what I wrote in post #2. Note that in the work-energy theorem you take the net force (not the actual, individual forces) and multiply by the displacement of the center of mass (not the displacement of the contact point).

    Calculating the real work done on a system is often difficult, since it involves tracking individual forces and their points of contact. In the case of the rolling cylinder (and the other examples I gave) the calculation is easy since it's clear that the point of application does not move and thus the real work is zero. But that does tell us that the forces do not add energy to the system.
     
  10. Jan 11, 2007 #9
    ok, now I'm really confused; I thought that the reason rolling friction does no work is because when you consider a point on the ground, its at rest, and as soon as it starts moving, the fore is being applied at that point, so no work is done, since work is by definition the force times the distance the contact point moves WHEN the force is acting on it. also, I thought the entire point was that there was no real change in KE, because although the force slows done the velocity of the center of mass, it makes the circle spin around the center of mass, and that this effect canceled out. But Im not even sure why that is necessarily true, since you could be double counting...ugh...why is this so confusing?
     
  11. Jan 11, 2007 #10

    Doc Al

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    Exactly right! But this has nothing to do with the Work-Energy theorem. (That's what I was objecting to.)
    Exactly right: Since there is no loss of mechanical energy due to friction (it does no work) the total KE (rotational + translational) plus PE remains constant. The decrease in translational KE (due to the friction force) is exactly compensated for by an increase in rotational KE.

    Think about it: The same force that slows the cylinder's translational speed also produces a torque about its center. And since there's no slipping, the rotational speed is coupled to the translational speed via:
    [tex]V = \omega R[/tex]

    So the torque*theta (which will give you the increase in rotational KE) exactly equals force*distance (which gives you the decrease in translational KE). Perhaps this is what you've been saying all along--if so, perfectly correct!

    I suspect that you understand things better than you think you do. :wink:
     
  12. Jan 11, 2007 #11
    ok, so now that we are on the same page, I have a couple of questions: isn't energy conservation completely due to the work energy theorem; i.e., the reason you can ignore friction is because it does no work, so you just have to deal with gravity potential and kinetic energy. It is this specific DEFINITION of work that allows to make these claims. So why does work have to be defined as force times distance contact point moves while force is acting on it? As long as the force is acting on the body, isn't that ok? Doesn't the work energy theorem still hold.

    Sorry for nagging you, but I want to understand this subtlety...
     
  13. Jan 12, 2007 #12

    Doc Al

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    Don't confuse the "work"-energy theorem with conservation of energy. The W-E theorem, perhaps better called the "center of mass" theorem, applies even for systems that have internal degrees of freedom: objects that can roll, deform, heat up, have friction, etc. It is only equivalent to conservation of energy for simple point mass system with no internal energy.
    Again, the "work" used in the W-E theorem is really pseudowork, not an energy term.
    When you use real work, then yes, you can account for any energy mechanically transferred into a system.
    Realize that the "work" used in the work-energy theorem does not care about contact points or how they move. (Reread my earlier posts.) In applying that theorem, all you care about is the net force on an object and the movement of its center of mass (not the contact point). Yes, the work-energy theorem still holds since it is a direction consequence of Newton's 2nd law.

    Not nagging at all! And this is subtle stuff--realize that many professors of (elementary) physics confuse these ideas.

    The bottom line is this: Despite the name, the W-E theorem is not a statement about conservation of energy, but an application of Newton's laws. Conservation of energy goes beyond Newton's laws.

    (I'm happy to discuss this further, but I'll be away from PF for the next week.)
     
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