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Motion of 2 wheels

  1. Jul 25, 2016 #1
    1. The problem statement, all variables and given/known data
    Math is a tool which support for Physics. But In some case Physics idea can use to solve problems easily. Now, there is an example that Physics's result is used to solve hard integral.
    A man is cycling which has 2 wheels. Smear of 2 wheels paint 2 curves in ground and they arent intersect.
    Prove that annular area painted by 2 wheel is constant.
    Use that to find that integral:
    $$I=\int_0^∞(\frac{2.e^{-t}}{1+e^{-2t}}-\frac{8e^{-3t}}{(1+e^{-2t})^3})dt$$

    2. Relevant equations
    I think we use some gyroscope 's equations.

    3. The attempt at a solution
    Dont have any idea :(
    Thanks for helping
     
  2. jcsd
  3. Jul 25, 2016 #2

    haruspex

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    I am not at all sure what this means. Is there a diagram?
    A possible interpretation is that the bicycle is performing some loop, not necessarily circular, and the annulus between the two tracks is independent of the exact path.
     
  4. Jul 25, 2016 #3
    It is area of curve out - area of curve in
     

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  5. Jul 25, 2016 #4

    haruspex

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    Yes, I understand what an annulus is. What I don't understand is the reference to its being constant. Constant with respect to what variable?
     
  6. Jul 25, 2016 #5

    haruspex

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    By the way, the integral looks quite easy to me.
     
  7. Jul 25, 2016 #6
    ah it doesnt respect with the man, and how he run. It respect swith the bike. Example mass m,....
     
    Last edited: Jul 25, 2016
  8. Jul 25, 2016 #7

    berkeman

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    Staff: Mentor

    So the inner circle is traced by the back wheel, and the outside circle is traced by the front wheel? And the area is constant with respect to the speed of the bicycle? As you go faster, the circles get bigger, and the difference between them gets smaller?

    Or I could be misinterpreting your question altogether... :smile:
     
  9. Jul 26, 2016 #8

    haruspex

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    No idea what you mean by that.
    Are the tracks supposed to be circular, or can they be any shape as long as they do not cross?
    If circular, does the constancy mean that it is independent of the inner radius, i.e., if the inner radius is made larger then the tracks will be closer together, and the annulus stays the same area?
    If the tracks need not be circular, it is going to be very interesting proving the annulus area is independent of the shape.
     
  10. Jul 26, 2016 #9
    Yes.
    Yes, that it.The problem says that:" The area doesnt respect the man, how he run, he strong or weak. Just respects the bike example m, radius of wheels,... " I dont remember that when i translate the problem
     
    Last edited: Jul 26, 2016
  11. Jul 26, 2016 #10

    Delta²

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    Don't understand the problem very good, but the integral is not very hard, using substitution ##x=e^{-t}## the first term is straightforward, while the second term
    is reducible to integrals of the form ##\int \frac{1}{(1+x^2)^n}dx## n=1,2,3 which aren't so easy but not that hard either.
     
  12. Jul 26, 2016 #11
    .
     
  13. Jul 26, 2016 #12

    haruspex

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    There's an even easier route if you recognise the functions being expressed as exponentials.
     
  14. Jul 26, 2016 #13

    haruspex

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    It clearly does not depend on the mass of the bike or radius of its wheels either. Yet we are asked to show the area is constant with respect to something, presumably something which one might have thought would affect the area. What?!
    Does the original question indicate whether the tracks are circular?
     
  15. Jul 26, 2016 #14
    No, problem dont say. I think the bike motion as a gyroscope. And we find the equation of motion of gyroscope. And I think 2 parts of intergral is 2 area of 2 curve .maybe Intergral is equal ##I=S2-S1##
     
  16. Jul 26, 2016 #15

    haruspex

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    Gyroscopes cannot have anything to do with it. We do not care how the bicycle moved in regard to time.
    All that matters is the geometry of a bicycle, namely, that (approximately) the point of contact of the front wheel with the road lies in the plane of the rear wheel. Let the distance between the points of contact with the road is L. If you pick a point on the trace of the rear wheel (the inner trace) and construct the tangent forward from there it will intersect the outer trace a distance L from the point.

    By the way, I suspect the result is true for arbitrary paths of the rear wheel. I considered two families of case: circle radius R and square side S. In each case, the annulus has the same area, regardless of size parameter.
     
  17. Jul 26, 2016 #16
    I just finded the solution. It has in
    Mark Levi
    The Mathematical Mechanic: Using Physical Reasoning to Solve Problems.
    Thanks for all helping^^
     
  18. Jul 26, 2016 #17
  19. Jul 26, 2016 #18

    haruspex

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    You beat me to it. It's 5am here. I got up to write this post because I just figured out the geometric/calculus argument that proves the general case for the annulus area.
    However, I still have no idea how this connects with the integral, have you? Can you solve the integral?
     
  20. Jul 26, 2016 #19
    That intergral equal ##\frac{\pi}{4}##. But the connects with integral seems not natural. Use the first problem and we have new problem.
    The bike seems as a model with 2 wheels AB and they connect by hard rod L. A move along x-axis and B is fetterless. They move in xy-plane
    That intergral using: ##e^{-t}=tan{\frac{\alpha}{2}}## We have the motion equation of B. That intergral is area which is B-graph and x-axis. It equals 1/4 circle.
    This problem isnt interesting .-.
     
    Last edited: Jul 26, 2016
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