Solving ODE Problem: Air Resistance Proportional to Velocity Squared

Click For Summary

Homework Help Overview

The discussion revolves around solving an ordinary differential equation (ODE) related to an object dropped from a height of 100 meters, where air resistance is proportional to the square of the velocity. The original poster expresses difficulty in deriving a solution for the velocity as a function of time, particularly in the context of integrating the forces acting on the object.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of Newton's second law and the transformation of the acceleration equation into a form suitable for integration. There are attempts to derive expressions for velocity and position through integration, with some participants questioning the original poster's understanding of the problem setup.

Discussion Status

Some participants have provided guidance on integrating the equations and transforming variables, while others have expressed gratitude for the insights shared. The discussion appears to be progressing with various interpretations and approaches being explored, but no consensus has been reached regarding a complete solution.

Contextual Notes

The original poster mentions challenges with using an ODE solver and recalls a previous class example that may not be clear in their memory. There is also mention of a constant related to air resistance, which may require further clarification.

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

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)

where k=constant. You can write the accerelation as

\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}

thus (1) reads

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

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

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

For the first integral make the change of variables

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

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 7 ·
Replies
7
Views
5K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
3
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K