Linear Differential Equations in Kinematics

[P3X-018]

Linear DE in Kinematics

Hey
I would like to know, how it is possible to solve the following
differentialequations

[tex]\ddot{x}(t)+\alpha \dot{x}(t)^2=0[/tex]

and the one that really gives troubles

[tex]\ddot{y}(t)+\alpha\dot{y}(t)^2=-g[/tex]

when given that [tex]x(0)=0[/tex], [tex]y(0)=0[/tex], [tex]\dot{y}(0)=v_0\sin \theta[/tex] and [tex]\dot{x}(0)=v_0\cos \theta[/tex]. Where [tex]\alpha=k/m[/tex]. The problem in equation 2 is, that i can't even solve the integral

[tex]\int \frac{1}{-g-\alpha \dot{y}^2}\:d\dot{y}[/tex]

Doing the substitution here doesn't get get me anywhere. Maybe someone have made this problem before, it's motion in 2-dimensions, with
airresistance. The 2 equations comes from

[tex]m\vec{a}=\vec{F}_g - \vec{F}_{airr}[/tex]

 

[dextercioby]

HINT:not [\tex],but [/tex]

Daniel.

 

[dextercioby]

1.Your equations are not linear.
2.Write them using the velocity.E.g.
[tex] m\frac{dv(t)}{dt}-kv^{2}(t)=-mg [/tex]

Can u solve it,now??

Daniel.

 

[P3X-018]

That doesn't help, I started by writing it in that way, but then rewrited it here. The problem is the integral

[tex]\int \frac{1}{-g-\alpha \dot{y}^2}\:d\dot{y}[/tex]

How can you solve this. By using the substitution, you end up with an even more complicated integral, and partial integration doesn't help. Can you that problem? I hope you can help me. Thanks in advance.

 

[dextercioby]

Can't u solve this integral??

[tex] \int \frac{dv}{-mg+kv^{2}} [/tex]

Daniel.

PS.If u can't,what are u doing solving ODE-s??

 

[arildno]

Check up an arcus tangens and artanh (inverse of hyperbolic tangens)

 

[dextercioby]

Check up an arcus tangens and artanh (inverse of hyperbolic tangens)

Let's not go there,Arildno... Maybe separation into simple fractions...??

Daniel.

 

[P3X-018]

No I can't solve this integral:

[tex] \int \frac{dv}{-mg+kv^{2}} [/tex]

Because when I substitude let's say

[tex]u=-mg+kv^{2}[/tex]

then

[tex]v=\sqrt{\frac{u+mg}{k}}[/tex]

and

[tex]\frac{du}{dv}=2kv \Leftrightarrow dv=\frac{du}{2kv}[/tex]

But when I use the partial integration technique, I end up with a even more complicated integral... Try solve it and you will see. Perhaps I should check up an arcus tangens and artanh?

 

[dextercioby]

Do u have any experience doing integrals??In this case,u can decompose the fraction/integrand into 2 simple fractions.

Daniel.

PS.Do you know this integration technique??

 

[Nylex]

No I can't solve this integral:

[tex] \int \frac{dv}{-mg+kv^{2}} [/tex]

Because when I substitude let's say

[tex]u=-mg+kv^{2}[/tex]

then

If you want to use an inverse hyperbolic/trig integral, don't subsitute like this. What you need to do is make the denominator into the form 1 - av^2 (where a is a constant (can you see how to do that?), then you can look up what this integral is. It's not arctan, as d/dx (arctan x) = 1/(1 + x^2), but I can't remember which one it is.

 

[urble]

looking up integrals ? just draw a right triangle... make pythagoras proud [/hint]

if you use trig substitution, you'll get to [tex]\int\sec\theta\ d\theta = \ln | \sec\theta + \tan\theta |+C[/tex] which will be the same as if you were to go by seperation.

 

[P3X-018]

Well, I havn't been online for 2 days now, but I have solved the ODE. I just jused arctan Nylex says.

 

[dextercioby]

Well, I havn't been online for 2 days now, but I have solved the ODE. I just jused arctan Nylex says.

WHAT??U mean "arctanh" as from "arcus tangens hyperbolicus",right?
As i said,partial fraction would have dunnit much easier.

Daniel.

 

[P3X-018]

No, I used arctan;
[tex]\frac{d}{dx}\arctan x=\frac{1}{1+x^2}[/tex]
And yes I sepretated into simple fractions.

 

[dextercioby]

Then your result was WRONG,WRONG!You should have used "arctanh".

Daniel.

PS.Are u sure it was the integral discussed above,the one with one minus at the denominator??

 

[dextercioby]

No, I used arctan;
[tex]\frac{d}{dx}\arctan x=\frac{1}{1+x^2}[/tex]
And yes I sepretated into simple fractions.

Please,separate into simple fractions
[tex] \frac{1}{1+x^{2}} [/tex]

Daniel.

 

[P3X-018]

I have the calculations in Mathcad, how can I upload a file in here, so that you can see them..

 

[dextercioby]

If it's not more than 50KB,you can post them as an attachement.OOPS,can u transfer them into another format??Attachment manager doesn't support "mathcad/matlab" files,sorry.

Try to post the general idea,at least.

Valid extensions:
"Valid file extensions: bmp doc gif jpe jpeg jpg pdf png psd txt zip"

Daniel.

 

[P3X-018]

Here is how i solve it;

[tex]m\frac{d}{dt}v_y(t)+kv_y(t)^2=-mg[/tex]

[tex]\frac{d}{dt}v=-g-\alpha v^2[/tex]

where \alpha = k/m, then

[tex]\frac{dv}{-g-\alpha v ^2}=dt [/tex]

[tex]\int\frac{1}{-g-\alpha v^2}dv=t + c[/tex]

[tex]\frac{1}{-g}\int\frac{1}{1+\frac{\alpha}{g}v^2}dv=t + c[/tex]

[tex]\frac{1}{-g}\int\frac{1}{1+\left(\sqrt{\frac{\alpha}{g}}v\right)^2}dv=t + c[/tex]

Then i substitude

[tex] u=\sqrt{\frac{\alpha}{g}}v[/tex]

[tex] \frac{1}{-g\sqrt{\frac{\alpha}{g}}}\int\frac{1}{1+u^2}du[/tex]

therefor

[tex] \frac{-1}{\sqrt{g\alpha}}\arctan\left(\sqrt{\frac{\alpha}{g}}v\right)=t+c[/tex]

Where does it go wrong :(?

 

[dextercioby]

I thought so.The initial equation is wrong.Gravity and aerodynamic force have opposite signs...


Daniel.

P.S.Redo calculations.

 

[P3X-018]

Hmm isn't it;

[tex] \vec{F}_{res}=\vec{F}_g-\vec{F}_{air}[/tex]

[tex] m\begin{pmatrix} v_x'(t)\\v_y'(t) \end{pmatrix} = \begin{pmatrix} 0\\-mg \end{pmatrix} -k\begin{pmatrix} v_x(t)\\v_y(t) \end{pmatrix}^2 [/tex]

 

[arildno]

Hmm isn't it;

[tex] \vec{F}_{res}=\vec{F}_g-\vec{F}_{air}[/tex]

[tex] m\begin{pmatrix} v_x'(t)\\v_y'(t) \end{pmatrix} = \begin{pmatrix} 0\\-mg \end{pmatrix} -k\begin{pmatrix} v_x(t)\\v_y(t) \end{pmatrix}^2 [/tex]

Nope; you should use:
[tex]m\frac{d\vec{v}}{dt}=-m\vec{g}-k||\vec{v}||\vec{v}[/tex]
Think about it..
(I am assuming zero air velocity here)

 

[P3X-018]

What does [tex]||\vec{v}||[/tex] means, it's not magnetude is it??

 

[arildno]

Sure it's magnitude.

 

[P3X-018]

But what is the difference between;

[tex]m\frac{d\vec{v}}{dt}=-m\vec{g}-k||\vec{v}||\vec{v}[/tex]

And my expression in #21?

 

[dextercioby]

Arildno,are u saying that gravity and aerodynamic forces have the same sense wrt the Oy axis??

Daniel.

 

[arildno]

That would depend upon the direction of the velocity.
I haven't checked all your calculations yet, but here's the answer, for unidirectional vertical motion:
[tex]v>0:[/tex]
[tex]m\frac{dv}{dt}=-mg-kv^{2}[/tex]
[tex]v<0:[/tex]
[tex]m\frac{dv}{dt}=-mg+kv^{2}[/tex]

That is, you must be careful about the SIGN of your velocity!

Daniel:
If the projectile moves UPWARDS, then the air resistance works downwards; if the projectile moves downwards, the the air resistance works upwards.
Air resistance opposes the direction of motion at all times.

 

[dextercioby]

The body was in free fall,Arildno.So there's no point in considering the other case.You mislead and confused him...

Daniel.

 

[P3X-018]

Havn't the gravity and aerodynamic forces same signs, when the projectil is on it's way to the top, and on the way down, they have opposite signs?

 

[P3X-018]

No it's not freefall.. And your right arildno in #27.

 

[dextercioby]

Havn't the gravity and aerodynamic forces same signs, when the projectil is on it's way to the top, and on the way down, they have opposite signs?

That's something totally different.I was considering free fall.Yes,the aerodynamics force changes the sign...

Daniel.

 

[arildno]

Havn't the gravity and aerodynamic forces same signs, when the projectil is on it's way to the top, and on the way down, they have opposite signs?

Precisely!
However, you didn't really make it clear from the start that you were considering BOTH cases; I well understand Daniel's objection (he has worked under the most natural assumption given your info).

 

[P3X-018]

Yea I know it was my fault. I didn't consider the 2 cases, but in the second equation then you got to use arctanh.

 

[arildno]

I didn't consider the 2 cases, but in the second equation then you got to use arctanh.

Since partial fraction decomposition disgusts me, go for artanh.
(I guess Daniel has a different view on this..)

 

[dextercioby]

Take both methods... You'll get the expression for "arctanh" in terms of "ln"...A useful result...A nice proof to it,also...

Daniel.

P.S.Arildno,i LOVE HYPERBOLIC FUNCTIONS... :!)