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Amount of Revolutions Based on Multiple Radii

  1. Oct 21, 2013 #1
    1. The problem statement, all variables and given/known data
    A chainwheel of a bicycle has a radius of .09 m, the radius of the smallest cogwheel (the “highest” gear) is .01 m, and the radius of the rear wheel is .35 m. At what rate (in revolutions per minute) must you be pedaling in order for the bicycle to have a forward speed of 8.33 m/s?

    v=8.33
    r1=.09
    r2=.01
    r3=.35


    2. Relevant equations

    v=[itex]\frac{2*pi*r}{T}[/itex]

    3. The attempt at a solution

    T=[itex]\frac{2*pi*r}{v}[/itex]

    I am unsure what to do when presented with three radii. Am I suppose to average them in order to solve for T?
     
  2. jcsd
  3. Oct 21, 2013 #2

    SteamKing

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    Hardly. Don't you understand how bicycle gears work?

    The bicycle has a forward speed of 8.33 m/s and the radius of the rear wheel is 0.35 m. How fast must this wheel turn in order for the bike to travel at 8.33 m/s?

    The cogwheel, radius 0.01 m, is also attached to the rear wheel of the bike. The chain goes around the cogwheel and the chainwheel. The pedals are attached to the chainwheel. In order to turn the rear wheel, the rider turns the chainwheel with the pedals.

    Knowing the angular velocity of the rear wheel, how many RPMs must the chainwheel turn?
     
  4. Oct 21, 2013 #3
    So am I trying to solve for the RPMs of the chain wheel which is equivalent to the speed of the bike traveling at 8.33 m/s? If so, the chain wheel would turn 0.0679 RPM.
     
  5. Oct 22, 2013 #4

    SteamKing

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    No, as I said in my previous post, you start with the rear wheel and figure out the RPMs it is turning to give a velocity of 8.33 m/s, then you work forward to the pedals. The cogwheel turns at the same RPM as the rear wheel.

    BTW, your answer of 0.0679 RPM means that it would take almost 15 minutes to make 1 revolution of the pedals. Clearly, this is an unrealistic answer.
     
  6. Oct 22, 2013 #5
    If I am understanding the question the same way you are, this is what I get.

    Rear Wheel
    T=[itex]\frac{8.33}{2*pi*.35}[/itex]=3.79 RPM

    Cog Wheel
    v=[itex]\frac{2*pi*.01}{3.79}[/itex]=0.0166 m/s

    Pedal

    T=[itex]\frac{2*pi*.09}{0.0166}[/itex]=34.1 RPM

    Also, is the units RPM (revolutions per minute) right since the speed that I am using is m/s?
     
  7. Oct 22, 2013 #6

    SteamKing

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    You haven't paid attention to your units.

    If the bike V = 8.33 m/s, then the number of revs calculated must occur in one second.

    For the cog wheel, you don't care what the velocity would be. You know that the cog wheel must spin the same number of revolutions as the back wheel of the bike (the two are rigidly connected together).

    The relative sizes of the cog wheel and the chain wheel determine the ratio of angular speeds of the two wheels.

    Try again.
     
  8. Oct 22, 2013 #7
    Okay, so...

    V=8.33m/s * 60s/min = 499.8 m/min

    Rear Wheel
    T=[itex]\frac{499.8}{2*pi*.35}[/itex]=227.273 RPM

    Cog Wheel RPM = Rear wheel RPM

    Pedal

    [itex]\frac{2*pi*.01}{227.273}[/itex]=[itex]\frac{2*pi*.09}{T}[/itex]

    T=2045.46 RPM

    Does this make sense now? Also, when I do a ratio of the speed between the cog wheel and pedal, isn't that essentially solving for the speed of the cog wheel and setting it equal to the pedal?
     
  9. Oct 22, 2013 #8

    SteamKing

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    2000 RPM is what a car engine turns at. The gears on a bike are supposed to make it easier for the rider to pedal the bike at high speed.

    In your proportion to calculate the pedal speed, the 2pi is not required. You are trying to find the number of revolutions per minute, not the number of radians per minute.

    Look at it this way: The pedal wheel is nine times bigger than the cog wheel. For each complete turn of the cog wheel, the pedal wheel only spins one-ninth of a turn.
     
  10. Oct 22, 2013 #9
    Okay, so the purpose of gears are to reduce the the effort needed to spin another gear.

    Using your explanation, the RPM of the pedal would be 25.25 RPM.
     
  11. Oct 23, 2013 #10

    SteamKing

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    That's much more reasonable.
     
  12. Oct 23, 2013 #11
    The purpose of other gears are not to reduce the effort to needed to spin another gear, but to reduce the amount of time to spin another gear. Does this make sense?
     
  13. Oct 23, 2013 #12

    SteamKing

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    Having a smaller gear attached to the rear wheel and a large chain gear attached to the pedals allows the rider to build up a reasonable speed on the bike with reduced effort at pedaling. In this example, the ratio is 9:1, so for every turn of the pedals, the rear wheel spins 9 times, including the cog wheel.
     
  14. Oct 23, 2013 #13

    haruspex

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    It depends what you mean by effort. In the mechanics of levers it means the applied force, which is minimised by the lowest gear http://en.wikipedia.org/wiki/Lever#Classes_of_levers. If you mean work, the work done on the pedals is the same regardless of the gear.
    The reason for having gears is a matter of physiology. There is an energy cost in flexing the muscles at a high rate, favouring selection of the higher gear; but generating a large force is also expensive even when little useful work is done. So for any given combination of current speed and target acceleration there is an optimal gear. If you watch the top cyclists tackling a hill, it's interesting that even among the leaders there is often a considerable difference in their choices of gear. There are spinners and there are crankers.
     
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