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I am working through the book Geometric Control of Mechanical Systems by Bullo and Lewis https://www.amazon.com/gp/product/0387221956/ and I am stuck on a problem, E4-18. The problem was evidently at one point the subject of a research paper, https://pdfs.semanticscholar.org/387d/4bb1c336aa0da87ab1d3a59f53532a2c74d2.pdf . I am trying to reproduce what the authors did in that paper so that I may solve the Bullo and Lewis problem. Taking the paper's authors' kinetic energy function as correct, including their "kinetic energy Riemannian metric", I am trying to reproduce their equations of motion from the Lagrangian.

Taking the derivative of the Lagrangian to produce the equations of motion with a kinetic energy Riemannian metric evidently includes using Christoffel symbols, per Bullo and Lewis. I computed the 4 Christoffel symbols with $\theta$ "in the top", to be used in the first equation of motion for $\ddot{\theta}$, and I get that 2 of them, the ones with $z\theta$ and $\theta z$ in the bottom, are both non-zero and the same, leading to a $-2mz\dot{z}\dot{\theta}$ term in the first equation of motion for $\ddot{\theta}$. I also get that $\Gamma^{\theta}_{zz} = 0$.

However, I get a non-zero value for $\Gamma^{\theta}_{\theta\theta}$, which should lead to an $mlz\dot{\theta}^2$ term in the equation of motion for $\ddot{\theta}$, but the paper's authors don't have said term in their equation of motion for $\ddot{\theta}$.

Can someone help me figure out what I am doing wrong? Thanks much in advance.

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  • $\begingroup$ I can't access the pages of the textbook, so it would be helpful to know what the problem is. I'm guessing you are computing the Christoffel symbols $\Gamma^{a}_{\ bc}$ for cylindrical coordinates $(x^1,x^2,x^3)=(r,\theta,z)$? $\endgroup$
    – QuantumEyedea
    Commented Aug 1, 2018 at 15:37
  • $\begingroup$ why you don't use the Lagrangian formalism to drive the EOM $$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q_i}}\right) -\frac{\partial L}{\partial q_i}=0\,\quad q_1=\vartheta\,,q_2=z$$ $\endgroup$
    – Eli
    Commented Aug 1, 2018 at 17:18
  • $\begingroup$ @Greg.Paul Thanks so much for replying! The coordinates are just $(\theta,z)$: $\theta$ is the angle of the platform; $l$ is the perpendicular distance of the channel from the point of rotation (a parameter, not a variable); and $z$ is the distance from the center of the channel. The problem is (a) to compute the kinetic energy and the "kinetic energy Riemannian metric" (b) to compute the potential energy (c) to compute the Christoffel symbols (of the kinetic energy Riemannian metric) and (d) to compute the equations of motion. $\endgroup$
    – Jeffrey Rolland
    Commented Aug 1, 2018 at 18:45
  • $\begingroup$ @Eli Thanks so much for replying! Just using what you wrote down with the Lagrangian in the paper doesn't reproduce the equations of motion in the paper; one evidently needs to include the Christoffel symbols. I've never computed the EOM from a Lagrangian with Christoffel symbols before, and I'm evidently doing it wrong now. $\endgroup$
    – Jeffrey Rolland
    Commented Aug 1, 2018 at 18:49

1 Answer 1

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I Think you can drive the EOM's from the Lagrangian?

\begin{align*} &\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q_i}}\right) -\frac{\partial L}{\partial q_i}=0\,\quad q_1=\vartheta\,,q_2=z\\\\ &L=\left(m\left(z^2+l^2\right)+\frac{1}{2}I\right)\dot{\vartheta}^2 -2\,m\,l\dot{\vartheta}\dot{z}+m\,\dot{z}^2\\ &\left(\frac{\partial L}{\partial \dot{\vartheta}}\right)= 2\left(m\left(z^2+l^2\right)+\frac{1}{2}I\right)\dot{\vartheta}- 2\,m\,l\dot{z}\\ &\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\vartheta}}\right)= 2\left(m\left(z^2+l^2\right)+\frac{1}{2}I\right)\ddot{\vartheta}- 2\,m\,l\ddot{z}+4\,m\,\dot{z}\,\dot{\vartheta}\\ &\frac{\partial L}{\partial \vartheta}=0\\\\ &\text{EOM for $\vartheta$}\\ &\boxed{\left(m\left(z^2+l^2\right)+\frac{1}{2}I\right)\ddot{\vartheta}- \,m\,l\ddot{z}+2\,m\,\dot{z}\dot{\vartheta}=0}\quad \surd ?? \end{align*}

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