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Mechanics problem involving a Differential Equation

  • Thread starter darrenhb
  • Start date
  • #1
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Homework Statement



Fnet = Ffriction + Fairresistance = μmg + cv2 = ma

Essentially I just need to get this in a v(x) form. Prof said that we need to go about it by doing this:

F = m[itex]\ddot{x}[/itex] = m [itex]\dot{x}[/itex] * d[itex]\dot{x}[/itex]/dx

and then integrating to get v(x)

The Attempt at a Solution



so...

μmg + cv2 = m [itex]\dot{x}[/itex] * d[itex]\dot{x}[/itex]/dx

And this is where I just can't figure it out. I'm terrible with DEs.

I thought of moving the 1/dx to the other side and then integrating both sides, but I get stuck, especially since v2 can be written as [itex]\dot{x}[/itex]2, right? And that messes with my sleep-deprived brain a little, lol. Any nudge in the right direction would be greatly appreciated!
 

Answers and Replies

  • #2
Hi!
If I recall correctly, the v^2 proportionality law(of drag) applies to high speeds, only; Are you sure that you need to have v^2 there? Stoke's law, in this case, I suppose, is a better fit(i.e k*v);
How did you arrive at the following relation:?[itex] {\frac{d}{dx}\dot{x}} = 0 [/itex], I can't see its use here...
We'll work on it,
Daniel
 
  • #3
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The question unfortunately doesn't give much of a hint as to how fast this might be traveling, so I'm not sure I could cancel out the v2. I also can't see where I typed d[itex]\dot{x}[/itex]/dx = 0. :S
 
  • #4
It's not so much that you typed it, it's what it is... :)
Meaning:
[itex] \frac{d}{dx}{dx}{dt} = \frac{d}{dt}\cdot\frac{dx}{dx} = 0 [/itex] Meaning that it can't be used...
The first thing to do, in my view would be to rewrite it, such that:
[itex] \Large mx''(t) = mk(x'(t))^2 + C [/itex] Where C is a constant carrying your constant friction force.
Then one would lean towards differentiating again(to get rid of the C) with respect to t:
[itex] \large mx'''(t) = 2mkx'(t)x''(t) [/itex] and dividing by m, and substituting z(t) = x'(t), and for simplicity, taking k=1:
[itex] \large z'' = zz', \frac{d^2z}{dt^2} = z\frac{dz}{dt} [/itex].
Multiplying by dt:
[itex] \large zdz = (\frac{d^2z}{dt^2})dt [/itex].
Now one must integrate, and continue the process further.
Try it,
Hope that works for you,
Daniel
P.S
Here's some "cheating"(:D), just to verify your steps:
http://www.wolframalpha.com/input/?i=-x''+==+x*x'"
 
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