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I'm trying to model a basic ball and beam system using Euler-Lagrange Equation. My system looks something like this:

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I have come up with this final Euler-Lagrange Equation:

$$\left(\frac{J_B}{r^2}+ m \right) \ddot{r}_B + mg\beta\theta - mr_B \dot{\theta}^2 = 0.$$

Where $J_B$ is the ball's moment of inertia, $r$ is the radius of the ball, $m$ is the mass of the ball, $g$ is the acceleration constant, $\beta$ is the ratio $d/L$, $r_B$ is the position of the ball along the beam, and finally $\theta$ is the gear angle.

The Euler-Lagrange Equation was acquired after finding the partial derivatives with respect to $r_B$.

My question is: How do I proceed with finding the transfer function? I have seen two research papers straight away cancelling the term with $\dot{\theta}$ from the equation, changing the equation to laplace domain, and finding $r_B$ to $\theta$. I'm assuming that this was done due to an assumption, but I'm unable to figure out what this assumption is. Is this a correct way to do it?

Alternatively, can I proceed with leaving the $\dot{\theta}$ term and changing it to the laplace domain, and again find the transfer function from there?

Also, the research paper has proceeded with finding the transfer function from the Euler-Lagrange equation taken by finding the partial derivatives with respect to $r_B$. What about finding it with respect to $\theta$?

I'm a bit confused, so I'd appreciate some clarification.

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  • $\begingroup$ Hi, welcome to Physics Stackexchange! Would you be able to reference the papers please? I would be intrested to have a look. $\endgroup$
    – Chris Long
    Commented Jul 22, 2021 at 10:29
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    $\begingroup$ @ChrisLong I have done that to the picture I added, but for some reason it isn't shown. The name of the paper is "Modelling and Control of Ball and Beam System using Coefficient Diagram Method (CDM) based PID controller" $\endgroup$
    – Zelreedy
    Commented Jul 23, 2021 at 12:08

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In the paper you referenced the $\dot\alpha$ term which will become the $\dot\theta$ term is dropped when they linearise the equation as the $\dot\alpha$ term is quadratic and is assumed to be negligible. Obviously, retaining this term will increase accuracy though.

The Euler-Lagrange equation with respect to $r_B$ is used because $r_B$ is the only degree of freedom. $\theta$ and $\dot\theta$ are not free but fixed like $d,L,J_B,m$ (albeit a fixed function of $t$) as they are the driving terms. If the wheel was not driven then you would also consider the Euler-Lagrange equations with respect to $\theta$ but then I am not sure the transfer function would make much sense to calculate as then you have no driving "force".

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  • $\begingroup$ I understand the first part, thank you. With regards to the second part: is θ considered driven as its value is determined by the value of rB? Because θ represents the servo motor angle, which rotates to balance the ball (Or is my understanding of driven wrong?) $\endgroup$
    – Zelreedy
    Commented Jul 23, 2021 at 23:19
  • $\begingroup$ I haven't read the paper in its entirety so do let me know if I have assumed something I should not have. That said it looks like $\theta$ is changed by some controller not by the physics of the system in the diagram in the OP. In order to use the Lagrangian formulation with $\theta$ being a degree of freedom we would need to somehow expand the Lagrangian to encapsulate the feedback loop. However, it is likely easier to treat $\theta$ as a driving force and compute the driving force required based on $r_B$. Effectively we are deciding the equation of motion for $\theta$, not the phsyics. $\endgroup$
    – Chris Long
    Commented Jul 23, 2021 at 23:51

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