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

$$\ddot{x}(t)+\alpha \dot{x}(t)^2=0$$

and the one that really gives troubles

$$\ddot{y}(t)+\alpha\dot{y}(t)^2=-g$$

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

$$\int \frac{1}{-g-\alpha \dot{y}^2}\:d\dot{y}$$

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

$$m\vec{a}=\vec{F}_g - \vec{F}_{airr}$$

 

[dextercioby]

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

Daniel.

 

[dextercioby]

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

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

$$\int \frac{1}{-g-\alpha \dot{y}^2}\:d\dot{y}$$

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??

$$\int \frac{dv}{-mg+kv^{2}}$$

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:

$$\int \frac{dv}{-mg+kv^{2}}$$

Because when I substitude let's say

$$u=-mg+kv^{2}$$

then

$$v=\sqrt{\frac{u+mg}{k}}$$

and

$$\frac{du}{dv}=2kv \Leftrightarrow dv=\frac{du}{2kv}$$

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:

$$\int \frac{dv}{-mg+kv^{2}}$$

Because when I substitude let's say

$$u=-mg+kv^{2}$$

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 $$\int\sec\theta\ d\theta = \ln | \sec\theta + \tan\theta |+C$$ which will be the same as if you were to go by separation.

 

[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;
$$\frac{d}{dx}\arctan x=\frac{1}{1+x^2}$$
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;
$$\frac{d}{dx}\arctan x=\frac{1}{1+x^2}$$
And yes I sepretated into simple fractions.

Please,separate into simple fractions
$$\frac{1}{1+x^{2}}$$

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;

$$m\frac{d}{dt}v_y(t)+kv_y(t)^2=-mg$$

$$\frac{d}{dt}v=-g-\alpha v^2$$

where \alpha = k/m, then

$$\frac{dv}{-g-\alpha v ^2}=dt$$

$$\int\frac{1}{-g-\alpha v^2}dv=t + c$$

$$\frac{1}{-g}\int\frac{1}{1+\frac{\alpha}{g}v^2}dv=t + c$$

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

Then i substitude

$$u=\sqrt{\frac{\alpha}{g}}v$$

$$\frac{1}{-g\sqrt{\frac{\alpha}{g}}}\int\frac{1}{1+u^2}du$$

therefor

$$\frac{-1}{\sqrt{g\alpha}}\arctan\left(\sqrt{\frac{\alpha}{g}}v\right)=t+c$$

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;

$$\vec{F}_{res}=\vec{F}_g-\vec{F}_{air}$$

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

 

[arildno]

Hmm isn't it;

$$\vec{F}_{res}=\vec{F}_g-\vec{F}_{air}$$

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

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

 

[P3X-018]

What does $$||\vec{v}||$$ means, it's not magnetude is it??

 

[arildno]

Sure it's magnitude.

 

[P3X-018]

But what is the difference between;

$$m\frac{d\vec{v}}{dt}=-m\vec{g}-k||\vec{v}||\vec{v}$$

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:
$$v>0:$$
$$m\frac{dv}{dt}=-mg-kv^{2}$$
$$v<0:$$
$$m\frac{dv}{dt}=-mg+kv^{2}$$

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.