Mechanics problem involving a Differential Equation

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Homework Help Overview

The discussion revolves around a mechanics problem involving a differential equation related to forces acting on an object, specifically friction and air resistance. The original poster seeks to express the equation in terms of velocity as a function of position.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to manipulate the equation involving net force and acceleration to derive a velocity function. They express confusion regarding the integration process and the implications of the drag force's dependence on velocity squared.
  • Some participants question the appropriateness of using the v² term for drag at varying speeds and suggest alternative approaches, such as Stokes' law.
  • There is a discussion about the validity of certain mathematical manipulations and the interpretation of derivatives in the context of the problem.

Discussion Status

The conversation is ongoing, with participants exploring different interpretations of the problem and offering guidance on potential steps forward. There is no explicit consensus, but suggestions for rewriting the equation and integrating are present.

Contextual Notes

The original poster notes a lack of information regarding the speed of the object, which complicates the analysis of the drag force. There is also mention of confusion regarding the mathematical expressions used in the discussion.

darrenhb
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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!
 
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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
 
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
 
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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