Mechanics problem involving a Differential Equation

In summary, Daniel is trying to find a way to solve a problem that he is having with integrating drag forces. He is having trouble with the v^2 proportionality law. His steps were to rewrite it such that mx''(t) = mk(x'(t))^2 + C. He differentiated it with respect to t and found z''=zzz'. Multiplying by dt, and integrating he arrived at zdz.
  • #1
darrenhb
7
0

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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  • #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
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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  • #5




I understand your confusion with this problem involving a differential equation. It can be difficult to work with equations that involve derivatives and integrals, but with practice and a clear understanding of the concepts, it can become easier.

In this problem, we have a force equation that involves the net force (Fnet), friction force (Ffriction), and air resistance force (Fairresistance). The goal is to find the velocity (v) as a function of position (x).

To begin, we can rewrite the equation as:

Fnet = μmg + cv^2 = m\ddot{x}

Next, we can use the chain rule to rewrite the second term on the left side as:

cv^2 = m\dot{x} * d\dot{x}/dx

Now, we can rearrange the equation to isolate the velocity term:

cv^2 = m\ddot{x} - μmg

v^2 = (m\ddot{x} - μmg)/c

Finally, we can take the square root of both sides to get the velocity as a function of position:

v(x) = √[(m\ddot{x} - μmg)/c]

This is the solution for the velocity as a function of position, which was the goal of the problem. Keep in mind that this is just one possible solution, and there may be other ways to approach the problem and derive the same result. I hope this helps to clarify the process for solving this type of mechanics problem involving differential equations. Keep practicing and you will become more comfortable with them!
 

1. What is a differential equation?

A differential equation is an equation that relates a function with its derivatives. It describes how a function changes over time or space in terms of its rate of change.

2. How is a differential equation used in mechanics problems?

In mechanics problems, differential equations are used to model and analyze the motion of objects. By using equations that describe the relationship between position, velocity, and acceleration, we can solve for the behavior of a system over time.

3. What are the types of differential equations commonly used in mechanics problems?

The two types of differential equations commonly used in mechanics problems are ordinary differential equations (ODEs) and partial differential equations (PDEs). ODEs involve a single independent variable, while PDEs involve multiple independent variables.

4. Can differential equations be solved analytically or numerically?

Differential equations can be solved using both analytical and numerical methods. Analytical solutions involve finding a closed-form expression for the function, while numerical solutions involve approximating the function using numerical methods. The choice of method depends on the complexity of the equation and the desired level of accuracy.

5. How are initial conditions and boundary conditions used in solving differential equations?

Initial conditions and boundary conditions are used to determine the constants of integration in a differential equation. Initial conditions specify the values of the function and its derivatives at a specific point in time, while boundary conditions specify the behavior of the function at the boundaries of the problem domain. These conditions help to obtain a unique solution to the differential equation.

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