Understanding the Drag Force Derivation: How Does the Process Work?

[sponsoredwalk]

Hello PF'ers, I'm having a bit of conceptual trouble with understanding the drag force
derivation.

I take the drag force as being $$D \ = \ \frac{1}{4} \ A \ v^2$$.

I want to imagine dropping an object from the highest point at which the equation w = mg
will be a satisfactory approximation for the weight.

My problem is understanding how the process works, imagine dropping the object straight
down from the highest point and we set up a free-body diagram as so:

$$w_{weight} \ - \ f_{drag} \ = \ ma$$

$$mg \ - \ \frac{1}{4} \ A \ v^2 \ = \ m\frac{dv}{dt}$$

$$\ m\frac{dv}{dt} \ = \ mg \ - \ \frac{1}{4} \ A \ v^2$$

$$\ \frac{dv}{dt} \ = \ g \ - \ \frac{1}{4m} \ A \ v^2$$

$$\ dv \ = \ ( \ g \ - \ \frac{1}{4m} \ A \ v^2 \ ) \ dt$$

$$\ \int_{v_initial}^{v_{final}} \ dv \ = \ \int_{v_initial}^{v_{final}} \ ( \ g \ - \ \frac{1}{4m} \ A \ v^2 \ ) \ dt$$


$$v_{final} = v_{initial} \ + \ ( \ gt \ - \ \frac{1}{12m} \ A \ v^3 \ ) |^{v_{final}}_{v_{initial}}$$

I'm thinking that may be wrong, unless the velocity is a function of time which I don't
think it is in the set up I've given, right? I should have done:

$$v_{final} = v_{initial} \ + \ ( \ gt \ - \ \frac{1}{4m} \ A \ v^2 \ t \ ) |^{final}_{initial}$$

I think so!

Anyway, my other concern is with the terminal speed, when the forces balance I can write:

$$w_{weight} \ - \ f_{drag} \ = \ m \ \cdot \ 0$$

$$mg \ - \ \frac{1}{4} \ A \ v^2 \ = \ 0$$

$$mg \ = \ \frac{1}{4} \ A \ v^2$$

$$v \ = \ \sqrt{ \frac{4mg}{A} }$$

Going off what I've done so far, which has the very real potential of being wrong, I
could write that the velocity I just calculated should also equal the integral I did
above seeing as they both give the velocity you'll be falling with from this point on.

$$\sqrt{ \frac{4mg}{A} } \ = \ v_{initial} \ + \ ( \ gt \ - \ \frac{1}{4m} \ A \ v^2 \ t \ ) |^{v_{final}}_{v_{initial}}$$

The initial velocity is zero in this situation.

I'm just thinking, what is the set up, You do an integral to find say the height fallen
from the initial point to the place where terminal velocity kicks in and then you can use
the equations of constant acceleration to find the time/height etc... from the ground.

Just trying to get it all together, any help would be greatly appreciated!

 

[zhermes]

I take the drag force as being $$D \ = \ \frac{1}{4} \ A \ v^2$$.

If A is a constant then yeah; if A is supposed to be the area of the object then you need a few other constants (e.g. the density of air, and some efficiency factor).

I'm thinking that may be wrong, unless the velocity is a function of time which I don't
think it is in the set up I've given, right?

The velocity is definitely(!) a function of time. I.e. when you release the object, its velocity is zero, and presumably it increases until terminal. Thus your solution to the differential equation is incorrect, collect the v and dv terms on one side, and the dt on the other side... then integrate with v=v(t).

$$\sqrt{ \frac{4mg}{A} } \ = \ v_{initial} \ + \ ( \ gt \ - \ \frac{1}{4m} \ A \ v^2 \ t \ ) |^{v_{final}}_{v_{initial}}$$

v_final is the terminal velocity.

 

[PhanthomJay]

The force of Drag equation is D =(1/2)(k)(v^2)(A), where k is a function of the density of the air and the shape factor (Drag coefficient). In your equation , k = 0.5, using SI units, where the drag coefficient is 0.4 (about the shape of a sphere), and the density of air is about 1.27kg/m^3. This equation cannot be used for different shapes or different densities of the medium through which the object falls.

The solution to differential equation is not as you have listed. I'm no expert in the calculus, but you can look it up somewhere. The speed of the object is a function of its weight, shape, density of the medium, and time. It's one of those e^(bt) solutions, where the terminal velocity is approached, but never quite reached. But in calculating terminal velocity, you can just set the weight = to the drag force, to solve for the terminal velocity.

 

[sponsoredwalk]

Hi thanks for the reply! Yes the A is supposed to be the area, this is an intro college physics
text and obviously isn't too rigorous on this point so I'll just imagine A to factor all of
these things in for now.

So obviously I haven't done diff eq's properly yet but we'll see how things go, I'm kind of
copying my book so bear with me

$$m \ \frac{dv}{dt} \ = \ m \ g \ - \ \frac{1}{4} \ A \ v^2$$

If I set $$\frac{1}{4} \ A \ = \ c$$

$$\frac{m}{c} \ \frac{dv}{dt} \ = \ \frac{m \ g}{c} \ - \ \ v^2$$

If I set $$\frac{m \ g}{c} \ = \ v_t$$ I can work with it as the book does.
Honestly, I'm not sure if this is supposed to be a velocity...

$$\frac{m}{c} \ \frac{dv}{dt} \ = \ v_t \ - \ \ v^2$$

$$\frac{1}{v_t \ - \ v^2} \ \cdot \ \frac{m}{c} \ \frac{dv}{dt} \ = \ 1$$

$$\frac{1}{v_t \ - \ v^2} \ \cdot \ \ \frac{dv}{dt} \ = \ \frac{c}{m}$$

$$\frac{dv}{v_t \ - \ v^2} \ \cdot \ \ \ = \ \frac{c}{m} \ dt$$

$$\int_{0}^{v_f} \ \frac{dv}{v_t \ - \ v^2} \ \cdot \ \ \ = \ \int_{0}^{t} \ \frac{c}{m} \ dt$$

I honestly don't know what the deal with that v_t is, I'm sure I'll learn about it
when I do proper differential equations. I suppose I could get to the root of my question
without faking it through any more math

My question is, once I calculate this, I'll have a final velocity, I suppose that gives you
the terminal speed then, the calculation I did where $$w \ - \ f_{drag} \ = \ m \cdot 0$$
doesn't enter into this calculation at all, no? I have the feeling that this calculation should
give you the same terminal velocity that the strange integral above does because they both
account for the same forces, would I be missing something?

If I'm doing a calculation like the above integral, I could integrate this again to find the
height you've moved and then the rest of the fall will become an equation where you'll
use the equations of constant acceleration to find time it'll hit the ground from this point
to there etc... Would that be correct conceptually?

 

[zhermes]

$$\frac{dv}{v_t \ - \ v^2} \ \cdot \ \ \ = \ \frac{c}{m} \ dt$$

Yeah! That looks great.

$$\int_{0}^{v_f} \ \frac{dv}{v_t \ - \ v^2} \ \cdot \ \ \ = \ \int_{0}^{t} \ \frac{c}{m} \ dt$$

Remember, you already know what the final velocity is; this equation could tell you the time (t) it takes to get to it... What you want to do is an indefinite integral (no bounds on your integral) which will give you a general relation between v and t, i.e. a function v(t).

I honestly don't know what the deal with that v_t is

I think it should have been $$v_t \equiv \sqrt{ \frac{mg}{c} }$$ which is again the terminal velocity. Its just a natural (convenient) constant to use, especially in anticipation of the final answer (relation between v and t).

One you have v(t) you could integrate that to get the distance traveling for whatever parameters.

 

[PhanthomJay]

Solving the 2nd order differential equation for v as a function of t will yield the terminal velocity by setting t=near infinity, which is the time it takes to reach terminal velocity (which is never reached). You will get the same result as using mg = Drag force. Don't ask me to do the math..it involves a hyperbolic function.

 

[zhermes]

Its not just an arctan?

 

[PhanthomJay]

Its not just an arctan?

No. Since, for quadratic air drag, F-net = mg - kv^2 = mdv/dt, then, dv/dt = (g-kv^2)/m, which is a non linear differential equation whose solution involves the tanh hyperbolic function. Some can do it easily using a numerical solution , or the calculus, but I've long since forgotten how to derive the solution for v as a function of time. So I looked it up. You will see that terminal velocity is approached, but not reached , at large values of t. Many falling objects reach 99% of terminal velocity in just a few seconds.