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Old Feb 20th 2010, 12:00 PM   #1
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Free falling body and resistance of air. A problem with equations of motion.

Hello everybody, I'm new to this forum and I have some difficulties when I try to solve a problem of free falling body when it is affected by gravitation and air resistance, which is equal to k*v^2/2 (k*y'^2/2). I need to find the equation of motion y(t) and the final velocity when t approaches infinity. I'm not that good at physics and differential equations, so I need your help.

I have tried to solve this by myself, but it was an intuition-like solving and the results I got aren't quite clear (I guess they aren't right too, hehe). So look what I got here:

http://img.photobucket.com/albums/v4...a/P2190001.jpg
http://img.photobucket.com/albums/v4...a/P2190002.jpg

I got that a velocity when t approaches infinity is sqrt(2mg/k) and y(t) is some function you see in those pictures. As I don't have initial conditions, I can't determine those coefficients (it is said that a body is just falling).

I hope I wrote those exercises clearly and I stated this problem clearly too. I'm looking forward to seeing some answers / thoughts about that from you.

Last edited by stingy; Feb 20th 2010 at 12:49 PM.
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Old Feb 20th 2010, 04:16 PM   #2
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You should be able to use any initial conditions, because it'll reach terminal velocity regardless of how fast or where it started.

It might also be easier to use velocity instead of displacement (y) as your variable. Then you can start out with just a 1st order DE. Once you solve it for v(t) you can just integrate to get y(t). If you define y=0 at the starting point this gives you the constant of integration.
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Old Feb 21st 2010, 04:15 AM   #3
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Yes, I did just the way you wrote, but anyway, I'm not sure about that constant c1 which I got after integrating the first order differential equation. I got that v(t) = y'(t) = 2m/k(t+c1) + sqrt(2mg/k). How can I determine that c1 coefficient? If we assume that at the moment t = 0 the velocity v = 0, it just looks weird. I mean, if we wrote t = 0 to the equation v(t), we would have that v goes to infinity. How come?
I read this article Free fall - Wikipedia, the free encyclopedia and it has v(t), though I have no idea how did they get it.

Last edited by stingy; Feb 21st 2010 at 04:38 AM.
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Old Feb 21st 2010, 02:19 PM   #4
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I could find c1 of your solution OK. And putting t=0 into your equation for v(t) doesn't make v=infinity.

You can get a bit more confidence in your solution from the terminal velocity. If you put t=infinity into it, you get

v = sqrt(2 m g / k)

which is of the same form as Wikipedia's Terminal Velocity page's

v = sqrt(2 m g / (rho A Cd))
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Old Feb 21st 2010, 02:29 PM   #5
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To make things simple, we can write that v(t) =~ 1/t + constant. So when t = 0, 1/t -> infinity and so v(t) -> infinity, because that constant (terminal velocity) doesn't matter. And yes, when t -> infinity, 1/t = 0 and v(t) = constant (terminal velocity). Mhm, I'm still confused, I will try to solve this problem with my friends, though I don't think that we will come to success.
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Old Feb 21st 2010, 02:38 PM   #6
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Originally Posted by stingy View Post
v(t) =~ 1/t + constant
This isn't the form of v(t) that you gave in your last message. There was + c1 in the denominator there, which protected it from blowing up at t=0.
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Old Feb 22nd 2010, 08:38 AM   #7
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Oh, I made a huge mistake when I tried to create a Lagrangian equation of motion so the following differential equation was bad and solving it was almost useless too. And one more mistake I made when I tried to determine the coefficient c1 (even though the differential equation was wrong) is that I took a limit of that v(t) when t approaches infinity instead of just writing t = 0 into that equation. I'm really sorry for these mistakes, shame on me.

Frictional force is a derivative of the dissipation function with a respect to velocity with minus sign. That's the mistake I made in the beginning. If you are interested, I got the right answer here:

http://img.photobucket.com/albums/v491/anncia/mmm.jpg

Thanks for your patience, empiler1, and sorry once again for mistakes I made and time I took from you.

Last edited by stingy; Mar 6th 2010 at 07:29 AM.
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Old Feb 22nd 2010, 04:34 PM   #8
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Awesome. I never quite understood how to do Lagrangians, but hey!
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