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Are drag coefficients negative?

  1. Feb 29, 2008 #1
    1. The problem statement, all variables and given/known data

    Find an equation describing a body in free-fall in air. This is not really homework. I'm in a differential equations class write now, and I have fun finding real world applications for such things.

    2. Relevant equations

    [tex]m\frac{d^{2}y}{dt^{2}} = B\frac{dy}{dt}-mg[/tex]

    3. The attempt at a solution

    Rewriting the above:

    [tex]\frac{d^{2}y}{dt^{2}} - \frac{b}{m}\frac{dy}{dt} = -g[/tex]

    The corresponding homogeneous equation is:

    [tex]\frac{d^{2}y}{dt^{2}} - \frac{b}{m}\frac{dy}{dt} = 0[/tex]

    Two solutions to the homogeneous equations are:

    [tex]y_{1} = C_{1}[/tex] and [tex]y_{2} = C_{2}e^{\frac{b}{m}t}[/tex]

    And as a particular solution:

    [tex]y_{p} = \frac{gm}{b}t[/tex]

    Therefore by the superposition principle, we have a general equation as follows:

    [tex]y(t) = y_{1}(t) + y_{2}(t) + y_{p}(t) = C_{1} + C_{2}e^{\frac{b}{m}t} + \frac{gm}{b}t[/tex]

    So here's where my question arises. I believe my assumption (the initial differential equation) should be a reasonable approximation of the real world. And I plugged my solution into the original differential equation. I seem to have answered it correctly. If both of these assumptions are true, then this equation doesn't make sense. It asserts that we are falling up! The only we I can think that this equation can still hold is if B were negative. I think after that, everything should make sense. Is this a correct assumption, or did I just make a mistake in the mathematics above?

    Thanks in advance for your time and any help.
    Last edited: Feb 29, 2008
  2. jcsd
  3. Feb 29, 2008 #2


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    The drag force is [tex]- B\frac{dy}{dt} [/tex] since it is always opposite to th emotion (where B is assumed to be a positive constant).
    Your particular solution is incorrect, it should be proportional to t^2
    (and of course, if you set B=0 you should recover the usual free fall equation)
  4. Feb 29, 2008 #3
    Sorry for the confusion, I'm using g as 9.8 rather than -9.8. I should have clarified. But I believe that should clear up the confusion with the drag force sign. Basically I set the drag force to be positive whereas the gravitational force is negative.

    I'm not sure why my particular solution does not work.

    [tex]\frac{d^{2}y_{p}}{dt^{2}} = 0[/tex]

    [tex]\frac{dy_{p}}{dt} = \frac{gm}{b}[/tex]

    Plugging this into the original differential equation I have

    [tex]-\frac{b}{m}\frac{gm}{b} = -g[/tex] which seems to be what I need. I must be missing something.
  5. Feb 29, 2008 #4


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    But it does not matter whether you use +9.8 or -9.8 (i.e. whether your y axis points upward or downward) because the drag force is always opposite to the motion so no matter what your choice is for your direction of the y axis, the force will be -B dy/dt
  6. Feb 29, 2008 #5
    Ahh, yes! How could I have missed that.

    Thanks a lot. That clears up a lot.
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