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Classical: Quadratic Drag and Gravity

  1. Oct 17, 2011 #1
    1. The problem statement, all variables and given/known data
    A particle of mass "m" whose motion start with downard velocity V0 in a constant gravitational field. The drag force is quadratic and proportional to kmv2. What is the distance s through which the particle falls in accelerating from v0 to v1. Give your expression for s in terms of k, g, v0, v1


    2. Relevant equations
    F = m[itex]\frac{dv}{dt}[/itex]=mg - kmv2


    3. The attempt at a solution

    My attempt:
    [STRIKE]m[/STRIKE](g-kv2) = [STRIKE]m[/STRIKE][itex]\frac{dv}{dt}[/itex]
    ergo:
    [itex]\frac{dv}{dt}[/itex] = (g-kv2)
    Separating:

    [itex]\int[/itex]dv/(kv2-g) = [itex]\int-kdt[/itex]

    Yields:
    √k/√g*arctan(√k*V/√g) = -kt + C

    Pretty much stuck at this point, and am not even sure this is the proper way.
     
  2. jcsd
  3. Oct 18, 2011 #2
    your first step is correct. After that you need to find the terminal speed:
    Vter=sqrt(g/k)
    use that to substitute for k in your second equation:
    (dv/dt)=g(1-(v^2)/(v_ter^2))
    Then separate the variables & THEN integrate.

    g*integral (0->t) of dt= integral(v0->vt) of(dv/(1-(v^2)/(v_ter^2)))

    and this is where I'm stuck -__-
     
  4. Oct 18, 2011 #3
    You have a mistake at the last step of integration. The expression you give with arctan is for the integral of [tex] \frac{1}{kv^2+g} [/tex] but we have to find integral of [tex]\frac{1}{kv^2-g}[/tex] which can be found by breaking the fraction into two fractions [tex]\frac{A}{v-\sqrt{\frac{g}{k}}}+\frac{B}{v+\sqrt{\frac{g}{k}}}[/tex].

    Also there shouldnt be any k in the right hand side of your integral equation. It sould just be [tex] \int-1dt=-t+C[/tex].

    After you integrate and find the expresion for v susbstitue for t=0, v=v0 to find the constant C.
     
    Last edited: Oct 18, 2011
  5. Oct 18, 2011 #4
    Last edited by a moderator: Apr 26, 2017
  6. Oct 18, 2011 #5
    Here's what I get:
    t + C = 1/2*[itex]\sqrt{k/g}[/itex][ln[itex]\frac{v-√g/k}{v+√g/k}[/itex]]

    When I work out the rest nothing simplifies.
    I've also tried it using your first suggestion stuckAgain.

    Using some info in this book (taylor classical)
    I got v = [itex]\sqrt{g/k}[/itex] tanh([itex]\sqrt{g}t[/itex]/[itex]\sqrt{k}[/itex]) but this is the general equation?
     
    Last edited: Oct 18, 2011
  7. Oct 18, 2011 #6

    vela

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    You don't want time in the equation. You want the displacement instead. Use the fact that
    [tex]\frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt} = v\frac{dv}{dx}[/tex]which is simply the chain rule, to get a differential equation relating v and x.
     
  8. Oct 18, 2011 #7
    How do you derive dv/dx and dx/dt from equations with neither of these terms?

    ie.
    dv/dt = g-mv

    dy/dt grabbing the already solved equation (y would be the falling problem)
    yields
    dy/dt = [itex]\sqrt{g/k}[/itex]tanh([itex]\sqrt{g/k}t[/itex]
    dv/dy = 0?
    I see no way to convert unless I'm missing key steps here
    if you sub in and solve for v(y) = you get an absolutely insane equation

    This class is Classical Mechanics II (senior level) hence why it was placed in the advanced section:
    The only other thing I can see ever working (haven't worked it out 100%)

    v(t)=-9.8t+Vo and y(t) = -4.9t2+V0t + X0 (X0 can be assumed to be 0)

    But this does not look at all promising
     
    Last edited: Oct 18, 2011
  9. Oct 18, 2011 #8

    vela

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    OK, use
    [tex]\frac{dv}{dt} = \frac{dv}{dy}\frac{dy}{dt} = v\frac{dv}{dy}[/tex]If your confusion is where this equation comes from, I say again: it's the chain rule from calculus. Look it up in your calculus book if you don't believe me. Or simply differentiate v=v(x) with respect to time.

    Your other alternative is to integrate v=dy/dt and solve for y(t). Then you'll need to do a bunch of algebra to eliminate t between the two equations you have to get v in terms of y.
     
  10. Oct 18, 2011 #9

    vela

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    I thought that might have been the case, but this is a common problem in many lower-division classes so I decided to get the thread moved.
    These equations don't apply at all. They're only valid when acceleration is constant, which it clearly isn't in this case.
     
  11. Oct 18, 2011 #10

    vela

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    You don't want v(y), you want y(v). The answer literally pops right out when you integrate.
     
  12. Oct 18, 2011 #11
    I'm not at all confused by the chain rule, I'm confused by how to employ it from the starting conditions.
    I've successfully solved the equation, which looks like 1/k*ln[(cosh(1/k*arctanh(v/√gk))] evaluated from v0 to v1
    I cannot imagine this is what the prof is looking for though (even though he made up this question and as such I'm sure the answer makes little sense to begin with)

    [itex]\frac{dv}{dy}[/itex] is where I'm having issues deriving anything without integrating and then going back and re-deriving, which does not seem helpful

    Also v itself has t within it, so I'm unsure how this solves the problem
    I get that you can solve dv/dy as a partial, but you do not have the V equation

    How can you solve for dv/dy without first having v(y)?

    Are you suggesting v([itex]\frac{g}{v}[/itex]-k) breaking it up into partials and then integrating?
    Which yields:

    y(v) = g-kv (I sincerely hope this isn't the answer based upon how much work I've tossed into this)
     
    Last edited: Oct 18, 2011
  13. Oct 18, 2011 #12

    vela

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    The differential equation you get is
    [tex]v \frac{dv}{dy} = g-kv^2[/tex]When you separate, you get
    [tex]dy = \frac{v\,dv}{g-kv^2}[/tex]You can integrate the righthand side with the simple substitution u=g-kv2.
     
  14. Oct 18, 2011 #13
    alright thanks,
    definitely did not understand this one at all... I appreciate your help.
    I was able to get the answer I believe.
     
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