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ODE problem

  1. Jan 27, 2008 #1
    [SOLVED] ODE problem

    1. The problem statement, all variables and given/known data
    A situation in which the air resistance is proportional to the velocity of an object squared.

    Object dropped off of a building with height 100m.

    F = -mg + Fr Fr=.5*cw*p*A*v^2

    cw=.5 p = air density = 1?



    2. Relevant equations

    I need to come about with a solution for y(t)

    3. The attempt at a solution

    I've used Mathcad to help me with a problem when the velocity is proportional to only the velocity linearly, using equations derived analytically.

    I just don't know where to begin with this one. I can't derive it analytically to obtain an expression for v(t) so that I can throw that into my runga-kutta solver along with the force expression as the two time derivatives.

    I have tried using ODE solver, but it fails to produce anything. It works for two iterations, up to t=2, then quits on me when I try to obtain more results.

    I believe my best approach is to solve this analytically, but I can't do it, my prof. did something strange with letting the variable be v^2, but the exercize he did this in was a few weeks ago, and foggy in my mind. Thanks for any help!
     
  2. jcsd
  3. Jan 27, 2008 #2
    From Newton's 2nd Law

    [tex]m\,\frac{d^2\,y}{d\,t^2}=m\,g-k\,v^2\Rightarrow \frac{d^2\,y}{d\,t^2}=g-\frac{k}{m}\,v^2 \quad (1)[/tex]

    where [itex]k[/itex]=constant. You can write the accerelation as

    [tex]\frac{d^2\,y}{d\,t^2}=\frac{d\,v}{d\,t}=\frac{d\,y}{d\,t}\frac{d\,v}{d\,y}=v\,\frac{d\,v}{d\,y}[/tex]

    thus (1) reads

    [tex]v\,\frac{d\,v}{d\,y}=g-\frac{k}{m}\,v^2\Rightarrow \frac{v\,d\,v}{g-\frac{k}{m}\,v^2}=d\,y[/tex]

    Integrate the above equation to obtain [itex]v=v(y)[/itex]. Then integrate
    [tex]v=\frac{d\,y}{d\,t}[/tex]
    to obtain [itex]y(t)[/itex].
     
  4. Jan 27, 2008 #3
    ahh... thank you very much. Didn't think about the change of base.
     
  5. Jan 27, 2008 #4
    Integrating
    [tex]\frac{v\,d\,v}{g-\frac{k}{m}\,v^2}=d\,y[/tex]
    yields

    [tex]v=-\sqrt{-\frac{m\,g}{k}+c_1\,e^{-2k\,y/m}}\Rightarrow \int\frac{d\,y}{\sqrt{-\frac{m\,g}{k}+c_1\,e^{-2k\,y/m}}}=-\int d\,t[/tex]

    For the first integral make the change of variables

    [tex]\sqrt{-\frac{m\,g}{k}+c_1\,e^{-2k\,y/m}}=z,\, d\,y=-\frac{z}{g+\frac{k}{m}\,z}\,d\,z[/tex]
     
    Last edited: Jan 27, 2008
  6. Jan 27, 2008 #5
    you are a lifesaver. Thanks a ton.
     
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