Air Drag in 1D

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  • #1
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Main Question or Discussion Point

I'm trying to derive the equations of motion for a particle falling in a uniform gravitational fill with air drag proportional to the square of velocity. However I'm getting the velocity as a complex number. Here is what I've done

The force of friction is [tex]F=-k\left(\dfrac{dx}{dt}\right)^2\dfrac{\textbf{v}}{||\textbf{v}||}=-k\left(\dfrac{dx}{dt}\right)^2[/tex]

We put the particle stationary at [tex]x=0[/tex]

The Lagrangian is [tex]\mathcal{L}=\dfrac{1}{2}m\left(\dfrac{dx}{dt}\right)^2-mgx[/tex]

[tex]\dfrac{d}{dt}\left(\dfrac{\partial \mathcal{L}}{\partial \dot{x}}\right)-\dfrac{\partial \mathcal{L}}{\partial x}=-k\left(\dfrac{dx}{dt}\right)^2[/tex]

[tex]m\dfrac{d^2 x}{dt^2}+mg=-k\left(\dfrac{dx}{dt}\right)^2[/tex]

If we put [tex]c=\dfrac{k}{m}[/tex]

[tex]\dfrac{d^2x}{dt^2}+c\left(\dfrac{dx}{dt}\right)^2+g=0[/tex]

We put [tex]p=\dfrac{dx}{dt}[/tex]

We have [tex]\dfrac{d^2x}{dt^2}=\dfrac{dp}{dt}=\dfrac{dp}{dx}\dfrac{dx}{dt}=\dfrac{dp}{dx}p[/tex]

The differential equation becomes

[tex]\dfrac{dp}{dx}p+cp^2+g=0[/tex]

[tex]\dfrac{p}{cp^2+g}\dfrac{dp}{dx}=-1[/tex]

[tex]\int\dfrac{p}{cp^2+g}\,dp=-x[/tex]

To solve the integral we put [tex]u=cp^2+g[/tex]

[tex]\dfrac{du}{dp}=2cp[/tex]

[tex]p=\dfrac{1}{2c}\dfrac{du}{dp}[/tex]

[tex]\dfrac{1}{2c}\int \dfrac{1}{u}\,du=-x[/tex]

[tex]\int \dfrac{1}{u}\,du=\ln |u|=\ln u[/tex] because [tex]u>0[/tex]

[tex]\dfrac{1}{2c}\ln (cp^2+g)+A=-x[/tex]

[tex]A(cp^2+g)=e^{-2cx}[/tex]

At [tex]t=0,x=0,p=0[/tex]

[tex]Ag=1[/tex]

[tex]A=\dfrac{1}{g}[/tex]

[tex]\dfrac{c}{g}p^2+1=e^{-2cx}[/tex]

[tex]p^2=\dfrac{g(e^{-2cx}-1)}{c}[/tex]

Now the sign of [tex]\dfrac{g(e^{-2cx}-1)}{c}[/tex] is determined by [tex]e^{-2cx}-1[/tex] which is not necessary positive definite.

In fact if we put [tex]c=1,x=1[/tex] we get [tex]e^{-2}-1<0[/tex] which means that [tex]p\in \mathbb{C}[/tex]

But [tex]p=\dfrac{dx}{dt}[/tex] which makes no sense at all. Did I do something wrong?
 

Answers and Replies

  • #2
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[tex]\dfrac{d}{dt}\left(\dfrac{\partial \mathcal{L}}{\partial \dot{x}}\right)-\dfrac{\partial \mathcal{L}}{\partial x}=-k\left(\dfrac{dx}{dt}\right)^2[/tex]
The sign of the drag force in the above equation is wrong.
By the way, I don't see why we have to use Lagrangian here, since Newtonian method is much more simple.

EDIT: Actually you may change either the sign of the drag force or the sign of the potential energy, but only one of them. Both yield the same correct equation.
 
Last edited:
  • #3
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[tex]m\dfrac{d^2 x}{dt^2}+mg=-k\left(\dfrac{dx}{dt}\right)^2[/tex]
I think you have a sign issue here. Shouldn't it be:

[tex]m\dfrac{d^2 x}{dt^2}=mg-k\left(\dfrac{dx}{dt}\right)^2[/tex]

Mass*acceleration equals the sum of the forces. The net force is gravity minus drag.
 
  • #4
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The sign of the drag force in the above equation is wrong.
By the way, I don't see why we have to use Lagrangian here, since Newtonian method is much more simple.

EDIT: Actually you may change either the sign of the drag force or the sign of the potential energy, but only one of them. Both yield the same correct equation.
Yes I see that now. Can you explain where I went wrong in my derivation of the friction.This is what I thought

[tex]F=-k\dot{x}^2 \hat{\textbf{v}}[/tex]

[tex]\hat{\textbf{v}}=\dfrac{\textbf{v}}{||\textbf{v}||}=\dfrac{(\dot{x})}{\dot{x}}=(1)[/tex]

Where the brackets denote a vector in the x direction.

Now obviously [tex]\hat{\textbf{v}}=(1)[/tex] is wrong, because the particle accelerates in the negative direction, but why?

Edit
Never mind. I find out on my own.
 
Last edited:
  • #5
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There is nothing wrong with [tex]\vec{F}=-kv^2\hat{v}[/tex]. But you must be careful when jotting down [tex]F = -kv^2[/tex] (*) (that means F<0). Since the ball is falling, the force must act upwards. So when you write (*), that means the positive direction of x axis is downward.
Now in order that potential energy V = mgx, the x axis must point upwards, which is inconsistent with (*) as explained above.
 

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