# Making a differential equation of a dynamical system

1. Dec 25, 2008

### eddiechai2003

1. The problem statement, all variables and given/known data
A stick, whose mass is neglected, with length $${l}$$ is fixed at point $${O}$$ on a vertical axis. $$\theta$$ is the angle formed between the stick and the vertical axis. A spiral spring is attached to the vertical axis and the stick in a manner that when at rest, the stick is in vertical position ($$\theta{=0}$$) and the vector $$\stackrel{\rightarrow}{OM}$$ is parallel to the vertical axis. The spiral spring, with the spring constant $${k}$$, applies a force whose driving moment is proportional to the angle $$\theta$$ on the stick.

At the other end of the stick in $${M}$$, there is a mass $${m}$$. Assume that $$\mu{=}\frac{k}{ml}$$and $$\gamma{=}\frac{g}{l}$$, find the differential equation in this form: $$\ddot{\theta}$$$${=f(}\theta$$,µ,$$\gamma{)}$$

2. Relevant equations

3. The attempt at a solution
I am having difficulties with the driving moment. How to etablish a differential equation by taking into account the driving moment and the friction?

Do you have any suggestions?

Thanks.

Last edited: Dec 25, 2008
2. Dec 25, 2008

### HallsofIvy

Staff Emeritus
I have moved this to "Calculus and Beyond" because I cannot imagine doing differential equation PRE-calculus!

But it seems to me that you will need to know at what point on bar the force is applied. Or are you using "driving moment" to mean "force times distance"?

3. Dec 25, 2008

### eddiechai2003

Hello HallsofIvy,

Actually, the original question was not in english. I translated it from french version and I am sure it is slightly different from the original question since I didn't totally understand the notion of moment or driving moment (according to wikipedia). Maybe you know better than me, then you could clarify this part for me.

Thanks.

4. Dec 25, 2008

### mohlam12

Hey !

I'm fluent in french and have done some mechanics in my first year. Maybe I can help you if you write the original question in french ?

5. Dec 25, 2008

### eddiechai2003

Hello mohlam12,

The french version of the question:

Une tige sans masse de longueur $${l}$$, est fixée en $${O}$$ sur un axe. On note $$\theta$$ l'angle formé par la tige et la verticale montante. Un ressort en spiral est fixé entre l'axe et la tige de telle façon qu'au repos ($$\theta{=0}$$) la tige soit verticale, le vecteur $$\stackrel{\rightarrow}{OM}$$ étant orienté vers le haut. Le ressort exerce sur la tige une force de rappel dont le moment est proportionnel à l'angle $$\theta$$ de coefficient de raideur $${k}$$. On suppose qu'au bout de la tige en $${M}$$, on fixe une masse $${m}$$.

On note $$\mu{=}\frac{k}{ml}$$, $$\gamma{=}\frac{g}{l}$$. Montrer que l'angle $$\theta$$ est une solution d'une équation différentielle de la forme:
$$\ddot{\theta}$$$${=f(\theta;\mu,\gamma)}$$.

6. Dec 26, 2008

### mohlam12

Sorry everyone, I'll give explanation in french so he can solve it quickly.

Alors on a trois forces, P le poids, R les frottements, et F une certaine force de rappel appliquée par le ressort. En appliquant le principe fondamental de la dynamique de rotation, on a, et en choisissant un sens de rotation approprié, on a
M(P) + M(R) - Mc = J_delta.(teta 2 points) (désolé j'ai du mal avec latex)

Mc étant le moment d'inertie de rappel, qui est égal à k(Teta)
M(P) étant le moment d'inertie de la force P par rapport à l'axe, et qui est égal à mglsin(teta)
M(R) étant le moment d'nertie de la force R par rapport à l'axe, et qui est nul.
J_delta étant le moment du système par rapport à l'axe, et qui est égal à ml²

l'équation devient alors : mglsin(teta) - k(teta) = ml²(teta 2 points)

En remplacant dans cette équation mu et gamma, tu obtiens un truc du genre :

(teta 2 points) + (mu/l).teta - gamma.(sinteta) = 0 qui est l'équation différentielle demandée.

PS : si tu pouvais réécrire mes équations en latex je t'en serais reconnaissant merci !

7. Dec 26, 2008

### eddiechai2003

Hello mohlam12,

$$\ddot{\theta}{+\frac{\mu}{l}.\theta-\gamma.sin(\theta)=0}$$

It seems to me correct but I have doubt since the dimension is not homogeneous.

$${[\frac{\mu}{l}]\neq[\gamma]}$$ ($$\theta$$ and $${sin(\theta)}$$ are dimensionless)

I will recheck your solution tonight as I need to go now.

By the way, how can I take into account the force of friction as demanded later in the question : On suppose maintenant que le système est soumis à une force de frottement opposée à la vitesse de coefficient $$\alpha>0$$.

I am bad in mechanics since I am studying Electrical and Electronics course. My major difficulty is to obtain the corresponding differential equation, after that I can solve the remaining sub questions by my own using matlab.

Last edited: Dec 26, 2008
8. Dec 29, 2008

### eddiechai2003

Hello,

After working for a few hours, I come out with a solution which is "dimensionally" correct. However, I am still not very convinced about the exactitude of the solution since I don't know much about the spiral spring and I couldn't find much information about it on the internet. Below is my solution:

$$\ddot{\theta}\pm\gamma.\sin\theta+\mu.l.\theta+\alpha.\dot{\theta}=0$$

By assuming the "moment of inertia" or "moment induced by the spiral spring"= $$k.l^2.\theta$$ and $$\alpha$$ is the coefficient of friction.

Is it a plus or a minus in front of $$\gamma.\sin\theta$$? It is confusing since the english version and the french version of the wikipedia about the inverted pendulum have different equation.

I want to thank HallsofIvy and mohlam12 for your helps and if you have comments on the equation please don't hesitate write them here.

Thanks.

Last edited: Dec 29, 2008
9. Dec 29, 2008

### eddiechai2003

It seems like I have messed somethings up.

There are 2 different kinds of spring constant. Say $$k_L$$ for linear springs (compression and extension) and, say $$k_T$$, for torsion springs.

$$[k_L]=[\frac{Force}{Length}]$$ while $$[k_T]=[\frac{Force\times Length}{Radian}]$$.

So, mohlam12, your solution was correct. It was my fault to not verify things carefully.

Here again, the solution is:

$$\ddot{\theta}-\gamma.\sin\theta+\frac{\mu}{l}.\theta+\alpha.\dot{\theta}=0$$

Am I right?