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Bicycle tires question.

  1. Dec 24, 2003 #1
    bicycle tires question.......

    ok.....heres a question that was given in the interview to a student who wanted to study physics at cabridge university.

    why when there is air loss in the tires of a bicycle(pressure decrease)....the bicycle becomes(heavier)?,meaning......harder to cycle.
     
  2. jcsd
  3. Dec 24, 2003 #2

    NateTG

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    There is a bunch of energy that goes into deforming the rubber of the tires and shifting the air around that doesn't get used if the tires are filled.
     
  4. Dec 24, 2003 #3
    thats true.......but its not by far the main reason of the huge difference in the effort u have to put in to continue cycling.

    pls try again!
     
  5. Dec 25, 2003 #4

    Integral

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    I have always assumed that it was the greater area of contact between the the tire and the road which caused greater rolling resistance thus low pressure tires are MUCH harder to push down the road.

    Is there more?
     
  6. Dec 25, 2003 #5
    the (point),(area...) of contact of the tire with the road is stationary....the tyre at that point does't move with respect to the road.....so there is no point of discusing resistance due to friction.(so the greater area of contact is irrelevant in that respect).

    TRUE there is greater resistance in the rolling of the (wheel) tire.....but what causes it?

    i have given u a strong hint now.
     
  7. Dec 26, 2003 #6

    Doc Al

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    I would expect that the major effect of soft tires is this: the tire is squashed forward and bunched up---essentially pulled away from the rim. In turn, the rubber pulls against the rim, creating a torque on the wheel that must be overcome to keep riding.

    Looked at in terms of external forces, the normal force (the ground pushing the tires up) has been redistributed. In a firm tire, the normal force is directly underneath the center of the wheel. In the soft tire case, the normal force is now squashing the rubber in front of center. Thus the normal force now exerts a torque on the wheel.
     
  8. Dec 26, 2003 #7
    you got it!
    good thinking!
    and good and complete explanation......not just a dry answer.
    you should apply for cabridge........:)
     
    Last edited: Dec 26, 2003
  9. Dec 26, 2003 #8

    Integral

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    Ahhh! Beautiful! It makes very good sense now.

    Thanks for completing a picture for me.
     
  10. Dec 26, 2003 #9
    I am not fully convinced. Wouldn't the normal force also act (equally) on the other side of the axis of rotation, that is, at the back end of the footprint?
     
  11. Dec 27, 2003 #10

    Doc Al

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    No, since the "footprint" is not symmetric. The footprint is distorted towards the front.
     
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