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Ideal Rolling: static friction and temperature

  1. Jul 28, 2005 #1
    Ideal rolling is rolling without slipping. Precisely, it is when the distance covered on the ground is equal to the arc length rotated through the circle during any time interval.

    Since rolling involves only static friction (as opposed to kinetic friction in rolling with slipping), the force between the tire and the road does no work. So we would expect that an ideal tire would not heat up.

    Real tires heat up a lot. It is easy to calculate the force of friction, and the work done (by measuring temperature of the rubber and pressure of the air before and after the test trip and acounting for the road heating the tire), and in this way calculate the total distance that your tire "slipped" during the trip.

    I am curious how close real tires are to the ideal of rolling without slipping.
     
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  3. Jul 28, 2005 #2

    brewnog

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    I think one of the key differences is a tyre's deformation. In real life, we don't have an infintessimally small contact area. A rolling tyre is constantly squishing inwards (radially) and outwards (axially) in order to provide a contact area with the surface it is rolling upon. This (rather than any dynamic friction) gives rise to most of the heat produced by the tyre, assuming that it's not skidding at all.
     
  4. Jul 29, 2005 #3

    arildno

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    Rolling is that the "rolling object" in actuality "swimming" through the ground, i.e, kinetic energy is extracted from the rolling object in deforming the ground as it rolls along. If this kinetic energy is not transferred back to the rolling object, we get what is known as "rolling friction" acting upon the rolling object.
     
  5. Jul 29, 2005 #4

    Meir Achuz

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    The energy goes into the tire, not the ground. The tire doesn't need to slip to heat up. The tire heats up because the energy put into the rubber as it is compressed is more than the energy the rubber gives back as it leaves the ground. This is called "hysteresis loss". It is due to the fact that the rubber is not completely elastic so that Hooke's law is not exact as it compresses and then comes back into shape.
     
  6. Jul 29, 2005 #5

    arildno

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    Have a look at the following link:
    http://webphysics.davidson.edu/faculty/dmb/PY430/Friction/rolling.html
    At no point did I deny that the wheel underwent inelastic deformations as well, but what I wrote was misleading, dumb and to some extent false all the same.
    Thanks for the correction.
     
    Last edited: Jul 29, 2005
  7. Jul 29, 2005 #6

    Stingray

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    Let
    [tex]\omega_{0} = v/r
    [/tex]
    be the angular velocity of an ideal nonslipping tire. In reality, tires always slip a little bit when exerting any force. It is customary to parametrize this by something called the longitudinal slip angle. If [tex]\omega[/tex] is the wheel's true angular velocity, then the slip angle is
    [tex]
    S = (\omega-\omega_{0})/\omega_{0} ,
    [/tex]
    for [tex]\omega<\omega_{0}[/tex] (braking), and otherwise
    [tex]
    S= (\omega - \omega_{0})/\omega .
    [/tex]

    The force exerted by the tire is then related to the slip angle by curves looking like http://code.eng.buffalo.edu/dat/sites/tire/img55.gif

    So you might get up to 10% slip or so under hard braking or acceleration. Lateral slip is probably more effective at heating up a tire, but neither is the dominant effect for regular driving (as others have mentioned).
     
  8. Jul 30, 2005 #7

    Meir Achuz

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    The trouble with that Davidson website is that anyone who looks at a car sitting on a concrete road can see the deformation of the tire, but there is no deformation of the concrete (not as shown in the diagram). The contact forces (unless you accelerate or panic stop fast enough to skid) are vertical (not as shown in the diagram). The pressure on the front end of the tire's contact area is larger than that at the back end, causing a negative torque. This is the same torque that causes a rolling rubber ball to stop.
     
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