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Nonlinear Retarding force

  1. Feb 15, 2007 #1
    1. The problem statement, all variables and given/known data
    A particle moves in a medium under the influence of a retarding force equal to [tex]mk(v^3 + a^2 v)[/tex], where k and a are constants. Show that for any value of the initial speed the particle will never move a distance greater than pi/2ka and that the particle comes to rest only for [tex]t \rightarrow \infty[/tex]

    2. Relevant equations
    Legrangian seems overkill, so I used Newton's.

    3. The attempt at a solution
    [tex]\frac{dv}{dt} = k (v^3 + a^2 v)[/tex]

    then seperate the ODE and integrate

    [tex] \frac{lnv}{a^2} - \frac{ln(a^2 + v^2)}{2a^2} = kt + C[/tex]

    multiply by 2a^2

    [tex] 2lnv - ln(a^2 + v^2) = 2kta^2 + C[/tex]

    use log properties and combine the natural logs

    [tex]ln ( \frac{v^2}{a^2 + v^2} ) = 2kta^2 + C[/tex]

    exponentiate and carry constant down

    [tex]\frac{v^2}{a^2 + v^2} = Ce^{2kta^2}[/tex]

    add and subtract a^2 in numerator to simplify, and then subtract the one

    [tex] \frac{-a^2}{v^2+a^2} = Ce^{2kta^2}[/tex]

    use algebra to isolate v

    [tex] v^2 = a^2 (1 - \frac{1}{1-Ce^{2kta^2}})[/tex]

    Now I am at the point where I can solve for the integration constant, but I don't see it giving me anything close to revealing a max distance of pi/2ka. Also, in my solution as t approaches infinity v = a, and not zero. Maybe I made an algebraic mistake?
    Last edited: Feb 16, 2007
  2. jcsd
  3. Feb 16, 2007 #2


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    The error is clearly in the integration of the ODE.
  4. Feb 16, 2007 #3


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    Presumably, though you didn't show it, you used partial fractions to write
    [tex]\frac{dx}{k(v^3+ a^2v)}= \frac{A}{v}+ \frac{Bv+C}{v^2+ a^2}[/tex]

    How did you integrate [tex]\frac{Bv+C}{v^2+ a^2}[/tex]?
  5. Feb 16, 2007 #4
    I used mathematica, though partial fractions is the obviously the way to do it by hand.
  6. Feb 16, 2007 #5


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    The first error is integrating the wrong ODE.

    [tex]\frac{dv}{dt} = - k (v^3 + a^2 v)[/tex] would be better, since the form of the answer implies k is positive.
  7. Feb 16, 2007 #6
    Yep, that was actually the problem. I went through it again and figured out what I did wrong. You end up with v = +/- a/sqrt(-1-Ce^(2kta^2).
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