Experimental Derivation of the Drag Force

  • Thread starter koko122
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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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  • #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.
 
  • #3
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Would you be so kind to explain this a bit further. The graph I obtained is attached.
 

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  • #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.
 
  • #5
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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:
$$F_{drag}=Dv^{2}$$
 
  • #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.
 
  • #7
Redbelly98
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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:
$$\Sigma F=mg-D{v_{\textrm{T}}}^{2}=0$$
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.
 
  • #8
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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?
 
  • #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.
 
  • #10
Redbelly98
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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.
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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