Air resistance differential equation

  • #1
Hello all, I want to say thank you in advance for any and all advice on my question. My classical mechanics textbook (Marion Thornton) has been taking me through motion for a particle with retarding forces.

The example it keeps giving is:

m dv/dt = -kmv

which can be solved for:
v = v0e-kt and
x = v0/k(1-e-kt)

But out of curiosity I tried using the actual drag force equation "1/2ρCAv2" instead of "kmv." But I can't figure out how to solve the differential:
##\ddot x ## + 1/2ρCA##\dot x ##2 = 0

How do you solve this thing? I'm stuck since it's not the standard
x'' + ax' + bx = 0

My solution yielded:

$$ \int\frac{\mathrm{d}\dot x }{ \dot x^2} = \frac{1}{2m}\rho CA\int \mathrm{dt} $$

which just gives some weird thing:
$$t = e^{\frac{1}{2m}\rho CAx} $$
which can't be right.
 
Last edited:

Answers and Replies

  • #2
Orodruin
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Having a quadratic air resistance term, your differential equation is no longer linear so you will need to apply other methods. As you have noted, the differential equation you get for ##v## is separable, i.e.,
$$
\frac{dv}{v^2} = K\, dt.
$$
However, your integration of this does not seem correct to me. Please show the details of your computation.
 
  • #3
I treated ##
\int {\frac {\mathrm {d \dot{x}}} {\mathrm {\dot{x^2}}}}## the same as ##\int \frac {\mathrm{dx}} {x^2} = \int x^{-2} \mathrm{dx} = -x^{-1}##.

So then I got ##- \dot x^{-1} = \frac {\mathrm {dt}}{dx} = \frac {1}{2m}\rho CA t##.

Then multiplied and divided: ##\frac {\mathrm {dt}}{t} = \frac {1}{2m}\rho CA \mathrm {dx}##.

Then integrated ##\int {\frac {\mathrm {dt}}{t}} = \frac {1}{2m}\rho CA \int {\mathrm {dx}}##.

So I got ##\ln {t} = \frac {1}{2m}\rho CAx##, ##t = e^{\frac {1}{2m}\rho CAx}##
 
  • #4
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I treated ##
\int {\frac {\mathrm {d \dot{x}}} {\mathrm {\dot{x^2}}}}## the same as ##\int \frac {\mathrm{dx}} {x^2} = \int x^{-2} \mathrm{dx} = -x^{-1}##.

So then I got ##- \dot x^{-1} = \frac {\mathrm {dt}}{dx} = \frac {1}{2m}\rho CA t##.

Then multiplied and divided: ##\frac {\mathrm {dt}}{t} = \frac {1}{2m}\rho CA \mathrm {dx}##.

Then integrated ##\int {\frac {\mathrm {dt}}{t}} = \frac {1}{2m}\rho CA \int {\mathrm {dx}}##.

So I got ##\ln {t} = \frac {1}{2m}\rho CAx##, ##t = e^{\frac {1}{2m}\rho CAx}##
What about taking into account the initial condition in your integration to get v? You omitted the constant of integration.
 
  • #5
Thank you by the way Orodruin for mentioning that it's nonlinear. I'm doing some research on "homogeneous first-order nonlinear ordinary differential equations." I guess that's the kind of equation this is.
 
  • #6
What about taking into account the initial condition in your integration to get v? You omitted the constant of integration.
Would that make a big difference tho? I could always just set up the constant to be trivial.
 
  • #7
Ray Vickson
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Would that make a big difference tho? I could always just set up the constant to be trivial.
Try it for yourself: with a constant of integration, and without it. Do you get mathematically equivalent ##x(t)## formulas?
 
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  • #8
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Would that make a big difference tho? I could always just set up the constant to be trivial.
Ask yourself this: Do you think I would have responded the way I did if I didn't already know that it would alleviate your difficulty?
 

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