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Bike mechanics question

  1. Jul 12, 2009 #1
    I was discussing this with a friend:

    Say you are doing a tailwhip on a bike (bike spins around pivot at headtube)

    Simpify this by looking at it as a bar pivoted at one end, with a wheel mounted to the bar with the radius in the direction of the bar. We want to spin this around the pivot in the direction of the wheels axle. (The wheel is spinning, but I am not sure if this makes any difference?)

    Now, if we make the wheel's radius smaller, does the force required to spin the wheel/bar around the pivot decrease? The "back edge" of the wheel is closer to the pivot, so less torque is needed to turn it around the pivot, but the "front edge" of the wheel is further from it, so more torque is needed to turn it around the pivot?

    However, can the wheel not be though of a point mass, positioned at the axle? In this case, provided the wheel's mass does not change, the radius of the wheel is independent?

  2. jcsd
  3. Jul 12, 2009 #2
    I did not probably get it right (the movement part, i'm not really a bike monster, so i'm not sure about what a tailwhip is), but it looks like a spinning wheel doing O -> | -> O etc, and a bar doing something like \ -> | -> / with an overall movement like this one (3 photographs :P)
    \ - | - /
    O | O
    watching from the side of the bike.
    At this point you should have the torque of the wheel and the one of the bar. Reducing thw wheel's radius should reduce its torque leaving the same torque to the bar that remains the same as before.
    So i didn't get the whole

    At this point you should be able to see whether the change of radius is influent, depending on the density of mass of the wheel and the one of the bar, to see if the wheel influences the torque enough to make something change effectively.

    Hope that helps.

  4. Jul 13, 2009 #3

    What the question is aiming at is this: what if you switch your wheels for 24" dia. instead of 26"? Assume mass of wheel stays the same.

    The axle of the wheel is at the same position, so the wheel's mass is centred at the same point. But, the "back edge" of the wheel is 1" closer to the headtube (the pivot), and the "leading edge" of the wheel is 1 inch further away. But these effects don't balance, since they are different distances from the pivot?
    Last edited by a moderator: Sep 25, 2014
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