I have a question about a fixed coordinate system as limit of rotating system. Consider for example a pendulum. The Lagrangian in the rotating frame is given by \begin{equation} L(\mathbf{r}, \mathbf{\dot{r}}) = \frac{m}{2} \vert \mathbf{\dot{r}} + \boldsymbol{\omega} \times \mathbf{r} \vert^2 - U(\mathbf{r}) = \vert \mathbf{\dot{r}} \vert^2 + 2\boldsymbol{\omega} \cdot (\mathbf{r} \times \mathbf{\dot{r}}) + \left( \omega^2 r^2 - (\boldsymbol{\omega} \cdot \mathbf{r})^2 \right) - U(\mathbf{r}) \end{equation} Now I am confused how to find the limit of a fixed coordinate frame. What is confusing me is that for a non-rotating system $\boldsymbol{\omega}$ should vanish. However, to end up with the pendulum equations I should keep the centrifugal force and state that it is compensated by centripetal force. How is this possible? Furthermore, if I would use \begin{equation} \boldsymbol{\omega} \propto \mathbf{r} \times \mathbf{v} \end{equation} then a constant factor in the second term would be still in the energy. It vanishes in the equations of motion, but is in the energy. What does this mean?

  • $\begingroup$ There should be a $\boldsymbol{\omega}$ for the pendulum and a separate $\boldsymbol{\Omega}$ for the frame. I think you are missing some terms. $\endgroup$ Jun 10, 2019 at 19:55
  • $\begingroup$ Why your U does not show potential energy of pendulum? $\endgroup$
    – SAKhan
    Jun 10, 2019 at 20:21
  • $\begingroup$ Okay this is something I also thought about. What do you mean with the potential energy? $\endgroup$
    – Q.stion
    Jun 11, 2019 at 16:08
  • $\begingroup$ “ keep the centrifugal force ” the energy has not the unit of force, there is no centrifugal force in your equation. \begin{equation} \boldsymbol{\omega} \propto \mathbf{r} \times \mathbf{v} \end{equation} . This is wrong the angular velocity of the rotation frame is known function of t $\endgroup$
    – Eli
    Jul 7, 2023 at 7:04


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