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Velocity of cyclist undergoing air resistance

  1. Mar 3, 2013 #1
    1. The problem statement, all variables and given/known data
    I'm attempting to solve the differential equation,

    [itex]\frac{dv}{dt}=\frac{P}{mv}-\frac{1}{2m}C\rho Av^{2}[/itex]​
    where [itex]P, \rho, m, A, C[/itex] are constants.

    The differential equation is used to approximate the velocity of a cyclist undergoing air resistance.

    It's actually presented as a numerical problem but I'm wondering if it's possible to solve it analytically.

    2. Relevant equations

    [itex]\frac{dv}{dt}=\frac{P}{mv}-\frac{1}{2m}C\rho Av^{2}[/itex]​
    [itex]P, \rho, m, A, C[/itex] are constants

    3. The attempt at a solution
    I'm not sure how to classify the ODE. It's not separable, linear nor exact. Not sure how I could use change of variables either.
     
    Last edited: Mar 3, 2013
  2. jcsd
  3. Mar 3, 2013 #2

    mfb

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    A first step: Simplify the equation to ##\frac{dv}{dt}=\frac{c}{v}-bv^2## with new constants c=P/m and b=CρA/2.

    The limit velocity is given by ##V=\sqrt[3]{\frac{c}{b}}## and we can construct a typical time ##T=\sqrt[3]{\frac{1}{cb^2}}##, write ##v=w V## and ##t=T \tau## to get $$\frac{dw}{d\tau}=\frac{1}{w}-w^2$$

    WolframAlpha does not find a solution, so it looks like a hard problem.
     
  4. Mar 3, 2013 #3
    Just wondering how you determined the limit velocity to be ##V=\sqrt[3]{\frac{c}{b}}##.
     
  5. Mar 4, 2013 #4

    mfb

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    Setting dv/dt=0 gives 0=c/V - bV^2, with the posted solution.
     
  6. Mar 4, 2013 #5
    Ah OK.

    Thanks, mfb.
     
  7. Mar 4, 2013 #6
    So I've gone through the problem numerically and the results that I'm getting are identical, regardless of whether or not I include the viscous term ## -\frac{\eta Av}{mh} ## in the differential equation.

    So, without the viscous term,
    [itex]\frac{dv}{dt}=\frac{P}{mv}-\frac{1}{2m}C\rho Av^{2}[/itex]​

    While with the viscous term,
    [itex]\frac{dv}{dt}=\frac{P}{mv}-\frac{1}{2m}C\rho Av^{2} -\frac{\eta}{mh}Av [/itex]​

    I've gone through a couple of scenarios:

    1) Cyclist in air with the following constants:
    ## P = 400 W, m = 70kg, C = 0.5, \rho = 1.225 kg/m^{2}, A = 0.33 m^{2}, \eta = 2*10^{-5} Pas, h = 1.5 m, v_{0} = 4 m/s^{2} ##

    Here, my calculated limiting velocity is ## 15.818174 m/s^{2} ## for both viscous and non-viscous considerations.

    oEnOGaB.gif


    2) Cyclist in water with the following constants:
    ## P = 400 W, m = 70kg, C = 0.5, \rho = 1000 kg/m^{2}, A = 0.33 m^{2}, \eta = 1*10^{-3} Pas, h = 1.5 m, v_{0} = 4 m/s^{2} ##.

    In this case, the limiting velocity due without viscous drag is ## 1.692526 m/s^{2} ##, while with viscous drag, it is: ## 1.692525 m/s^{2} ##.

    grnptqC.gif

    I've worked out the limiting velocities as you pointed out by hand, mfb, and everything seems to check out...

    Would it be correct to consider the Reynold's number in these scenarios? Since the Reynold's number in both of these scenarios is quite large, the viscous drag is negligeable.
     
    Last edited: Mar 4, 2013
  8. Mar 5, 2013 #7

    mfb

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    Right.
     
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