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B Gear Ratio in Bicycles using Rotational Motion

  1. Jul 23, 2015 #1
    When we change the gears of the bicycle we are riding, we change the the disc we are currently at (which are located at the place where we pedal) to some other disc. This means the radius of the circular disc we were pedaling/rotating changes. So that means if we were rotating the disc with angular velocity ##ω##, if ##r## changes (radius of the disc) ##rω## changes. And that means the speed with with the chain which rolls over the disc, i.e. ##v = rω## changes. But how does that make the bike move faster with the same angular velocity we were providing it as before? If we want it to move faster, then the velocity of COM of the rear/front tire should increase, but how does changing gears do that?
     
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  3. Jul 23, 2015 #2

    A.T.

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    What happens with the angular velocity of the back wheel, if the speed of the chain changes?
     
  4. Jul 23, 2015 #3
    Okay..it increases too. But I still didn't get how that increases the speed of the bike? I think it is because of the force imparted by the ground on the rear tire, which happens because the rear tire pushes the ground back..so the ground pushes the tire front..giving it some acceleration. Which increases the velocity of the rear tire and as a result of the entire bike.

    Also, if we stop pedaling why does the bike come to a complete halt after sometime? How can it be friction because friction is always acting in the direction of motion of the bike?
     
    Last edited: Jul 23, 2015
  5. Jul 23, 2015 #4
    Air resistance , friction between parts of the cycle , and other stuff too ( Not sure what that other stuff may be , though ) .
     
  6. Jul 23, 2015 #5
    How exactly does the rear tire apply force to the ground? It has a velocity at the bottom-most point..so how does it lead to applying a force?
     
  7. Jul 23, 2015 #6

    A.T.

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    No.
    Friction.
     
  8. Jul 23, 2015 #7
    Yes.
    Friction is applied by the ground not by the rear tire. What I am saying is friction is a reaction force. What was the action force?
     
  9. Jul 23, 2015 #8

    A.T.

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    Both. There are two equal but opposite forces between tire and ground.
    It doesn't matter which of the two forces you call "action" and which "reaction". They are both friction.
     
  10. Jul 23, 2015 #9
    Okay, so I should ask: "How" does the rear tire apply a force on the ground?
     
  11. Jul 23, 2015 #10
    How or Why ?
     
  12. Jul 23, 2015 #11
    Both
     
  13. Jul 23, 2015 #12

    A.T.

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  14. Jul 23, 2015 #13
    When angular velocity of the back wheel increases , the wheel gains a tendency to slip on the ground . Friction opposes this tendency of relative motion by increasing - this causes acceleration of the wheel's center of mass , causing increase in velocity of the Com , thus making net velocity of point in contact with ground zero , thus preventing relative motion , and in this process causing increase in bike's speed .
     
  15. Jul 23, 2015 #14

    A.T.

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    If there is no slippage, then angular velocity and COM's linear velocity change simultaneously. So I don't see how you can tell which change is causing the other change. Seem arbitrary to me.
     
  16. Jul 23, 2015 #15
    Well the angular velocity increases with time , and so it must be causing simultaneously , the velocity of Com to increase too .

    Take a rubber ball , spin it and drop on the ground . It would first slip and then eventually , start to roll . What would you say is the mechanism/cause for this ?
     
  17. Jul 23, 2015 #16

    A.T.

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    You could just as well say:

    The velocity of Com increases with time, and so it must be causing simultaneously , the angular velocity to increase too .

    But here the angular velocity would decrease, the opposite of the case above. And a bike isn't dropped with spinning wheels, so what does it have to do with an accelerating bike?
     
  18. Jul 23, 2015 #17
    So changing the gear doesn't cause a direct increase in angular velocity of the wheel ?

    My point here is that friction , in the ball's case is opposing the relative motion and therefore causing an increase in ball's horizontal velocity ; and the only difference in the case of the ball and wheel is that the ball had slipping in the start and hence friction opposed angular velocity , in the case of the wheel it would not , it would merely increase the Com's velocity .
     
  19. Jul 23, 2015 #18

    A.T.

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    Switching gears doesn't necessary change the bike speed, which (assuming no slippage) determines the angular velocity of the wheel.

    You can (and usually do) accelerate a bike without any relative motion at the contact (slippage).
     
  20. Jul 23, 2015 #19
    No - I don't think so either . But the case the OP talks about involves increasing the speed of the bike .

    That is what I have been saying all along . What I've been trying to do is give the reason for this .
     
  21. Jul 23, 2015 #20

    A.T.

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    Locally the "reason" is static friction. Globally the "reason" is simply that the equations of motion say so, given certain boundary conditions.
     
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