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Bike tire momentum problem

  1. Apr 8, 2003 #1
    You have a bike tire with X mass and Y moment of inertia.
    You hold the tire on a vertical plane just slightly above the ground. You then spin it to give it Z amount of angular momentum.

    If you were to let the tire drop to the ground, what would its acceleration and final velocity be? (not taking into account drag)
  2. jcsd
  3. Apr 9, 2003 #2
  4. Apr 9, 2003 #3


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    It would depend on the coefficient of friction (both static and kinetic) between the tire and the ground. If the wheel bounces, you'd need the coefficient of restitution as well.

    It's not a simple problem (unless you make a bunch of simplifying assumptions)
  5. Apr 9, 2003 #4


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    The answer depends on the friction coefficients between the tire and ground. For example if there was no friction at all, the tire would keep on rotating around its axis with constant angular velocity (conservation of angular momentum relative to the center of the wheel)and at the most could bounce up and down. With "low" friction the tire would start rolling and sliding, energy would be dissipated continuously until pure rolling sets in. With "high" friction the tire would start rolling without sliding and no energy would be dissipated after landing.
    In the general case (with friction and considering an almost zero vertical landing speed just to disregard bouncing) what is conserved (immediately before and after touching the ground) is the total angular momentum with respect to landing point since all the interaction forces are bounded there and hence produce a null external momentum with respect to this pole.
    The system is also pretty unstable (a small angle with the vertical direction will create a totally different evolution).

    Hope this helps!

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