Solving ODE Problem: Air Resistance Proportional to Velocity Squared

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
4 replies · 3K views
bndnchrs
Messages
26
Reaction score
0
[SOLVED] ODE problem

Homework Statement


A situation in which the air resistance is proportional to the velocity of an object squared.

Object dropped off of a building with height 100m.

F = -mg + Fr Fr=.5*cw*p*A*v^2

cw=.5 p = air density = 1?



Homework Equations



I need to come about with a solution for y(t)

The Attempt at a Solution



I've used Mathcad to help me with a problem when the velocity is proportional to only the velocity linearly, using equations derived analytically.

I just don't know where to begin with this one. I can't derive it analytically to obtain an expression for v(t) so that I can throw that into my runga-kutta solver along with the force expression as the two time derivatives.

I have tried using ODE solver, but it fails to produce anything. It works for two iterations, up to t=2, then quits on me when I try to obtain more results.

I believe my best approach is to solve this analytically, but I can't do it, my prof. did something strange with letting the variable be v^2, but the exercize he did this in was a few weeks ago, and foggy in my mind. Thanks for any help!
 
Physics news on Phys.org
From Newton's 2nd Law

[tex]m\,\frac{d^2\,y}{d\,t^2}=m\,g-k\,v^2\Rightarrow \frac{d^2\,y}{d\,t^2}=g-\frac{k}{m}\,v^2 \quad (1)[/tex]

where [itex]k[/itex]=constant. You can write the accerelation as

[tex]\frac{d^2\,y}{d\,t^2}=\frac{d\,v}{d\,t}=\frac{d\,y}{d\,t}\frac{d\,v}{d\,y}=v\,\frac{d\,v}{d\,y}[/tex]

thus (1) reads

[tex]v\,\frac{d\,v}{d\,y}=g-\frac{k}{m}\,v^2\Rightarrow \frac{v\,d\,v}{g-\frac{k}{m}\,v^2}=d\,y[/tex]

Integrate the above equation to obtain [itex]v=v(y)[/itex]. Then integrate
[tex]v=\frac{d\,y}{d\,t}[/tex]
to obtain [itex]y(t)[/itex].
 
ahh... thank you very much. Didn't think about the change of base.
 
Integrating
[tex]\frac{v\,d\,v}{g-\frac{k}{m}\,v^2}=d\,y[/tex]
yields

[tex]v=-\sqrt{-\frac{m\,g}{k}+c_1\,e^{-2k\,y/m}}\Rightarrow \int\frac{d\,y}{\sqrt{-\frac{m\,g}{k}+c_1\,e^{-2k\,y/m}}}=-\int d\,t[/tex]

For the first integral make the change of variables

[tex]\sqrt{-\frac{m\,g}{k}+c_1\,e^{-2k\,y/m}}=z,\, d\,y=-\frac{z}{g+\frac{k}{m}\,z}\,d\,z[/tex]
 
Last edited:
you are a lifesaver. Thanks a ton.