Solving Nonlinear Differential Equations for Air Drag in One Dimension

[FrogPad]

I'm working on this project that involves air drag. The model for the air drag is given as:

$$\vec F_d = \frac{1}{2} C \rho A v^2$$

I'm using Newton's Second law in relation to this force and gravity (in one dimension) which yields:

$$a = \frac{1}{m} \left( -mg + \frac{1}{2} C \rho A v^2 \right)$$

I'm in the middle of an ODE course, so I have not dealt with anything nonlinear... so this is where my question is. If I convert everything to the differential form:

$$a = \frac{d^2x}{dt^2}$$
$$v = \frac{dx}{dt}$$

So what does $$v$$ become in the $$\vec F_d$$ equation?

It is $$\left( \frac{dx}{dt} \right)^2$$. I've just never encountered this. Does it become?
$$\frac{dx^2}{dt}$$

thanks in advance

 

[saltydog]

Frog, you have an equation of the form:

$$y^{''}=k_1+k_2\left(y^{'}\right)^2$$

(it's just the square of the derivative that all).

This is an equation in which the independent variable is missing. The standard approach is to let:

$$y^{'}=p$$

then:

$$y^{''}=p\frac{dp}{dy}$$

Can you figure that one out?

Substituting this into the original equation yields:

$$p\frac{dp}{dy}=k_1+k_2p^2$$

Can you finish it? That is, separate variables, integrate, then convert back to y.

 

[arildno]

It seems you are dealing with a one-dimensional air drag model.
In that case, it is certainly true that the MAGNITUDE of the drag force is given by [itex]\frac{1}{2}C\rho{A}v^{2}=\frac{1}{2}C\rho{A}(\frac{dx}{dt})^{2}[/tex], however that is NOT the correct expression of the (directed) force.<br /> The force is always anti-parallell to the direction of the velocity, thus you have that the drag force is given by:<br /> $$F_{d}=-\frac{C}{2}\rho{A}|\frac{dx}{dt}|\frac{dx}{dt}$$[/itex]

 

[saltydog]

It seems you are dealing with a one-dimensional air drag model.
In that case, it is certainly true that the MAGNITUDE of the drag force is given by [itex]\frac{1}{2}C\rho{A}v^{2}=\frac{1}{2}C\rho{A}(\frac{dx}{dt})^{2}[/tex], however that is NOT the correct expression of the (directed) force.<br /> The force is always anti-parallell to the direction of the velocity, thus you have that the drag force is given by:<br /> $$F_{d}=-\frac{1}{2}\rho{A}|\frac{dx}{dt}|\frac{dx}{dt}$$[/itex]

$Well that should make it interesting Arildno. I'll try working with that one too.<br /> <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f642.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":smile:" title="Smile :smile:" data-smilie="1"data-shortname=":smile:" />$

 

[FrogPad]

Ok. I'm getting confused here. I've just never had any 'training' with equations like this. So I kind of feel like I'm guessing here, and not understanding what to do.

Let me clarify one thing first. This model is for an intro engineering course, so we are actually going to solve this equation computationally with Euler's method. However, I thought I might actually be able to solve this part of the equation. (Which I still hope to do).

From:
$$p\frac{dp}{dy}=k_1+k_2p^2$$

I would do the following?

$$p\frac{dp}{dy}=k_1+k_2p^2$$

$$\int p\,dp=\int (k_1 +k_2p^2)dy$$

$$\frac{p^2}{2}=k_1y+k_2 p^2 y + c$$

So yay (if I did this correctly) I'm left with: $$\frac{p^2}{2}=k_1y+k_2 p^2 y + c\,\,\,\,\,(1)$$Now I should solve $(1)$ for $p$ and substitute $y'=p$ into the expression? Then solve for $y$ ?