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Experimental Derivation of the Drag Force

  1. Aug 30, 2014 #1
    I am trying to experimentally derive the drag equation. I have dropped coffee filters with varying masses and determined their resulting terminal velocity. I plotted the data and found that the mass of the coffee filter is proportional to the terminal velocity squared. I was now wondering how I can use this to derive the drag equation, which states: F_D = .5CpAv2
     
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  3. Aug 30, 2014 #2

    Nugatory

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    At terminal velocity the known constant gravitational force is equal to the unknown drag force; equate them and do a bit of algebra and you'll have the ##v^2## dependency. You will also have to do some measurements with a constant mass and different surface areas to trade out the dependency on area.
     
  4. Aug 30, 2014 #3
    Would you be so kind to explain this a bit further. The graph I obtained is attached.
     

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  5. Aug 30, 2014 #4

    CWatters

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    At terminal velocity the filter is not accelerating. So the forces acting on it must sum to zero. You know what the two forces are.
     
  6. Aug 30, 2014 #5
    That's obvious but from the graph that I have provided can I derive that mg=dv^2 where d is the drag constant? Or can I show that the proportionality constant is equal to g/d somehow?

    The graph shows the following:
    $$ {v_{\textrm{T}}}^{2} \propto m$$
    and from this I want to show that:
    $$\Sigma F=mg-D{v_{\textrm{T}}}^{2}=0$$
    So that the drag force can be defined as following:
    $$F_{drag}=Dv^{2}$$
     
  7. Aug 30, 2014 #6

    CWatters

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    So...

    mg = .5CpAv2

    Your graph is for

    v2 = mg/.5CpA

    The standard equation for a straight line is..

    y = slope*x + constant

    but constant = 0

    So in this case

    y = v2
    x = mass m

    and the slope of the graph

    = g/.5CpA

    You know the slope.
    You know the area A
    You know g

    That just leaves Cp.

    Sorry had to edit a few typos in this post.
     
  8. Aug 30, 2014 #7

    Redbelly98

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    I don't see how that would follow using only the graph. It follows from Newton's second law, and that ∑F=0 at a constant velocity.

    See C. Watters post #6. Cp is the only unknown in the expression for the slope, so it can be calculated.
     
  9. Aug 30, 2014 #8
    Ya that makes sense. Would it be correct to infer from the graph that the drag force is equal to: $F_drag=Dv^2$, where D is just a general drag constant?
     
  10. Aug 30, 2014 #9

    CWatters

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    I don't think you can call D a "general" drag constant because D includes A (the area of the object). So it's specific to that size of filter cone rather than cones in general.

    D = .5CpA

    Had to edit for typo again.
     
  11. Sep 1, 2014 #10

    Redbelly98

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    And wouldn't Cp include the density of air as well? That has to be in the drag force expression somewhere.

    From http://en.wikipedia.org/wiki/Drag_(physics)#Drag_at_high_velocity
    we have
    [tex]F_{drag}=\frac{1}{2} \ \rho_{air} \ C_d \ v^2[/tex]
    where Cd is dimensionless -- and I'm pretty sure depends on the object's shape.
     
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