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seen Jul 20 at 20:36

Jul
20
comment Why friction is zero when wheel slip is zero?
I can't resist pointing out the amusing typo in my last comment. I meant "a real-world elastic tire", not "a wheel world elastic tire".
Jul
20
comment Why friction is zero when wheel slip is zero?
+1 Rick and thank you for the discussion on my answer above. For those who don't want to read our whole back-and-forth, the key point is for tires with elasticity (any real-world tire) there is compression and thus V is always a bit less than R omega. See here
Jul
20
comment Why friction is zero when wheel slip is zero?
@Rick So while what I said was true for a perfectly rigid wheel (and I do think the OP was perhaps confused about static friction since they mention a car standing still), the graph describes a wheel world elastic tire. Thank you for your patient clarification on this point.
Jul
20
comment Why friction is zero when wheel slip is zero?
@Rick Actually, on a re-read, maybe I get what you're saying. An idealized, perfectly elastic wheel would move due to static friction while experiencing zero slip, whereas a real-world elastic wheel would also move due to static friction (at low enough speeds that it wasn't sliding), but would have a non-zero slip at any speed. So for physical wheels the graph makes sense reading it as the full frictional force, not just a kinetic friction. My objection that S (eq. 1 in your PDF) could be zero and you still are moving won't apply if you consider the wheel to be elastic. I buy it.
Jul
20
comment Why friction is zero when wheel slip is zero?
I tried to be clear that I was simplifying by ignoring the elasticity of the wheel - which, I agree, is too big a simplification to explain the OP's graph, but not too big to explain the role of static friction, which is what the OP seemed to be missing. Do you disagree that a moving wheel can experience static friction (because the tangential velocity of the wheel at the point of contact with the road is zero)?
Jul
20
comment Why friction is zero when wheel slip is zero?
@Rick, thanks for the links, but I'm not sure where you disagree with me on. I agree with the answer you linked to, which says that you get slip at small forces because the tire is elastic, and also that unless the wheel is sliding across the surface, the car is accelerating from static friction.
Jul
16
comment Why friction is zero when wheel slip is zero?
While I'm pedantically correcting myself, I should say that you could have forward motion even in the zero-friction case if you were to get out and push. :) But if your engine is just spinning the wheels, then it's only because of friction between the wheels and the ground that this gets turned into forward motion. And if there's no slip, then it's static friction.
Jul
16
comment Why friction is zero when wheel slip is zero?
@Rick Actually even in the case of a moving vehicle on level ground, if there is no slip then there is static friction between the wheel and the road (because the point of contact has zero instantaneous velocity in the horizontal direction). It's this friction that converts the rotation of the wheels into forward motion. If there were no static friction, then you couldn't have forward motion without slip. But the OP asked about a car on a slope, where there's static friction even at rest to resist gravity.
Jul
16
comment Why friction is zero when wheel slip is zero?
@Rick The frictional force described in that PDF (given by eq. 2) is kinetic friction, which goes to zero when the rotation of the wheel keeps up with it's motion (i.e., when the numerator in eq. 1, and thus the slip, is zero). But the OP asked about the case of a car "standing still on a slope". In that case there is static friction. The PDF and the plot aren't wrong by any means, they're just describing a moving vehicle, not one at rest.
Jul
13
comment Why friction is zero when wheel slip is zero?
@Rick Yes, otherwise I don't see how the plot makes sense. It's certainly not the case that the static coefficient of friction is zero when there's zero slippage. As the questioner notes, you can have static friction even when standing completely still on a slope.
Jul
13
comment Why friction is zero when wheel slip is zero?
@Rick, I understood the question to be asking about the case where there is no slip. "Why there is no friction when there is no slip?" In that case (for an idealized, wheel, i.e., a circle rolling on a plane) there can be no kinetic friction due to the contact between the wheel and the road. There can still be static friction, as I note. There can also be kinetic friction at the axle, but not at the point of contact between the wheel and road (for idealized tires.) If there is slip, then there would be kinetic friction.
Nov
2
awarded  Yearling
Apr
27
comment Feynman diagram for annihilation
@Harold, You would need two photons to conserve 4-momentum while satisfying $E = pc$, but because it's a virtual photon (an internal line) you're allowed to violate $E = pc$.
Apr
25
comment Feynman diagram for annihilation
@Harold What physical process you're talking about is identified by the initial and final states. If in an experiment you have two electrons going in and two electrons going out, you can't look at it and say "the intermediate state is a single photon". That is just one diagram that contributes to the process. At tree-level you also have the diagram where a photon is exchanged (the t-channel), and there are higher order diagrams as well.
Apr
25
comment Why the name “optical phonon”?
@Seanny123 Yes, it's a very important fundamental concept in solid state physics. I'm not sure if I can do justice to it in a comment, perhaps you should post a separate question on that? (Or have a look at an introductory text on solid state.) The short version is that it's a cell of the reciprocal lattice (the lattice of k-vectors that arises from the Fourier transforms of the states in your original lattice). It is relevant both to the case where the primitive cell contains one atom or multiple atoms.
Apr
2
comment How can Magnets be used to pick up pieces of metal when the force from a magnetic field does no work?
This paper discusses the quantum mechanical case: academic.csuohio.edu/deissler/PhysRevE_77_036609.pdf
Apr
2
comment How can Magnets be used to pick up pieces of metal when the force from a magnetic field does no work?
As a simplified (classical) example, consider a loop of current-carrying wire dangling from a string (so it's free to rotate) in an external magnetic field. As the loop rotates to align with the field, it gains rotational kinetic energy, but at the same time there's an induced EMF due to the change in magnetic flux through the loop. This opposes the motion of the circulating charges in the loop, thus robbing them of some kinetic energy. I believe if you do the calculation you'll see the amount of energy removed is the same as the amount of energy added.
Apr
2
comment How can Magnets be used to pick up pieces of metal when the force from a magnetic field does no work?
This answer doesn't seem quite right to me. When a magnetic field gives kinetic energy to an object, it induces an EMF that cancels out an equal amount of internal energy of the object. So the net work done by the magnetic field is truly zero.
Mar
20
comment Capacitor with dielectric between the plates
@guru Sorry, I believe you are incorrect. You get an attraction from inducing dipoles in the dielectric. The negative charges in the dielectric are pulled closer to the positive plate, so their attraction to it outweighs the repulsion felt by the more distant positive charges. Likewise, the positive charges are pulled closer to the negative plate, so their attraction towards it outweighs the repulsion felt by the more distant negative charges.
Mar
20
comment Capacitor with dielectric between the plates
@DumpsterDoofus That sounds right to me. Why does it seem strange?