Air resistance differential equation

[David Koufos]

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.

 

[Orodruin]

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.

 

[David Koufos]

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 $\ln {t} = \frac {1}{2m}\rho CAx$, $t = e^{\frac {1}{2m}\rho CAx}$

 

[Chestermiller]

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.

 

[David Koufos]

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.

 

[David Koufos]

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.

 

[Ray Vickson]

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?

 

[Chestermiller]

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?