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Free fall with air drag force

  1. Nov 6, 2004 #1
    free fall with drag force

    If the equation F=-bv^2 describes the drag force of an object...then the differential equation for the object's motion would be:

    dv/dt= -g+bv/m

    or is it...

    dv/dt= g-bv/m

    After solving the equation, should I get...
    V=mg/b[1-e^(-bt/m)]

    Also, does this look like the position v time graph of the object?
    http://home.earthlink.net/~urban-xrisis/phy001.gif
     
    Last edited: Nov 6, 2004
  2. jcsd
  3. Nov 6, 2004 #2

    Pyrrhus

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    Ok i'll do the analysis.

    Forces acting on a body falling.
    Down positive (Using resistive force proportional to the speed)

    We got a first order DE

    [tex] mg - bv = m \frac{dv}{dt} [/tex]

    To not get into much detail, this type of DE

    [tex] \frac{dy}{dt} = ay - b [/tex]

    Has the following solution

    [tex] y = \frac{b}{a} + ce^{at} [/tex]

    For an initial value, to find C.

    [tex] y = \frac{b}{a} + [y_{o} - \frac{b}{a}]e^{at} [/tex]

    Thus for our case the solution is

    [tex] v = \frac{mg}{b} + [v_{o} - \frac{mg}{b}]e^{-\frac{bt}{m}} [/tex]

    If we arrange the terms and [itex] v_{o} = 0 [/itex]

    [tex] v = \frac{mg}{b} - \frac{mg}{b}e^{-\frac{bt}{m}} [/tex]

    [tex] v = \frac{mg}{b}(1 - e^{-\frac{bt}{m}}) [/tex]
     
    Last edited: Nov 6, 2004
  4. Nov 6, 2004 #3
    instead of F=-bv^2, what if F=-2v? The question tells me to ignore the effects of gravity in this problem. Would the DE then be dv/dt= 2v/m?

    Since f=bv...and b=-2

    dv/dt= g-bv/m
    dv/dt= g-(-2)v/m
    the question asks to ignore gravity...
    dv/dt= -(-2)v/m
    dv/dt= 2v/m
     
  5. Nov 6, 2004 #4

    Pyrrhus

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    If we ignore gravity, then indeed our Newtonian analysis will be

    The Object is falling, and the air drag is in the opposite direction
    Down positive.

    [tex] m\frac{dv}{dt} = -bv [/tex]

    [tex] \frac{dv}{dt} = \frac{-bv}{m} [/tex]

    Solving this:

    [tex] \frac{dv}{dt} = \frac{-bv}{m} [/tex]

    [tex] \frac{dv}{v} = \frac{-b}{m}dt [/tex]

    [tex] \int_{v_{o}}^{v} \frac{dv}{v} = \int_{0}^{t} \frac{-b}{m}dt [/tex]

    [tex] \ln |v|]_{v_{o}}^{v} = \frac{-b}{m}t]_{0}^{t} [/tex]

    For [itex] v_{o} = 0 [/itex]

    [tex] ln |v| = \frac{-b}{m}t [/tex]

    [tex] v = e^{\frac{-b}{m}t} [/tex]
     
    Last edited: Nov 6, 2004
  6. Nov 6, 2004 #5
    Does this look correct?
     
  7. Nov 6, 2004 #6
    However, the retarding force is not F=bv, it is F=-2v as stated above. Correct me if I’m wrong, but isn’t [tex] \frac{dv}{dt} = \frac{-bv}{m} [/tex] for when F=bv? Should it be [tex] \frac{dv}{dt} = \frac{2v}{m} [/tex] ?
     
  8. Nov 6, 2004 #7

    Pyrrhus

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    Sure, just substitute b with 2, and for the graphic, It looks rather odd to me, but maybe someone else can find it the physical meaning.
     
    Last edited: Nov 6, 2004
  9. Nov 6, 2004 #8
    Last edited: Nov 6, 2004
  10. Nov 6, 2004 #9

    Pyrrhus

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    the graphic looks better now, and i added the extra solve steps for neglecting gravity.
     
  11. Feb 10, 2010 #10
    what is b in this equation
     
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