1

Take a simple example like the pendulum with a moving support (But only considering the pendulum mass $m$, not the support mass $M$): https://en.wikipedia.org/wiki/Lagrangian_mechanics#/media/File:PendulumWithMovableSupport.svg

From my understanding I could formulate the Lagrangian for the moving reference frame from the perspective of an observer sitting on the support. Using the pendulum angle $\theta$ as the only coordinate and considering only the velocity relative to the moving reference frame. $r$ is the length of the pendulum. With $L=T-V$ ($T$ for kinetic, $V$ for potential energy, $Q$ for generalized forces not included in the energy) I end up with:

$$ \frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial L}{\partial \dot{\theta}} - \frac{\partial L}{\partial \theta} - Q = m r^2 \ddot{\theta} + m g r \sin \theta - Q = 0 $$ With the acceleration of the frame $\ddot{x}$ and the position vector of m, $\mathbf{r}$ I can add the only additional force: $$ Q = \frac{\partial \mathbf{r}}{\partial \theta} \cdot \ddot{x} (-1, 0) = - m \ddot{x} r \cos \theta $$ So I can eliminate $m$ and $r$ and get: $$ r \ddot{\theta} + g \sin \theta + \ddot{x} \cos \theta = 0 $$ Which is the same solution as wikipedia gives for the moving pendulum.

So far, all examples I've seen considered a moving reference frame by adding the relative speed to an inertial reference frame in the system's kinetic energy. As I understand it, I can just as well look at all movements relative to the moving frame and view the acceleration from 'outside' as an extra force. Which in some cases might make the algebraic stuff easier, I think.

Am I just lucky in this case and making wrong assumptions or is this a valid way to get to the equations of motion?

Qmechanic
  • 201,751
bobribar
  • 11

2 Answers2

0

enter image description here

Start with the position vector to the mass

$$\vec R=\left[ \begin {array}{c} x \left( \tau \right) +l\sin \left( \varphi \right) \\ l\cos \left( \varphi \right) \end {array} \right] $$

where $~x(\tau)~$ is known.

from here you obtain the kinetic and potential energy and with EL the equation of motion

$$ \ddot\varphi-{\frac {-g\sin \left( \varphi \right) + \left( {\frac {d^{2}}{d{\tau }^{2}}}x \left( \tau \right) \right) \cos \left( \varphi \right) }{l }} =0$$

where dot is $~\frac{d}{d\tau}~$

the kinetic energy $~T~$ is a function of $~\dot x~,\dot\varphi~,\varphi$

Eli
  • 11,878
  • Yes of course, in the resting reference frame that is clear. But I want to look at the energy in the moving reference frame. – bobribar Jul 02 '23 at 16:37
  • @bobribar the energy is $E=T+U~$ is not conserved – Eli Jul 02 '23 at 17:35
  • ofc, it's an accelerated frame. My only question is: Do I have to consider the acceleration at the kinetic energy from the start or can I add it later as a generalized force (which adds energy to the system). In the given example I get to the same result either way but I am not sure if that's always the case. – bobribar Jul 02 '23 at 18:53
  • @bobribar $~T=\frac 12,m \left( \left( l\cos \left( \varphi \right) \dot\varphi +{\frac { d}{d\tau}}x \left( \tau \right) \right) ^{2}+{l}^{2} \left( \sin \left( \varphi \right) \right) ^{2}{\dot\varphi }^{2} \right) ~$ and $~U=m,g,l\cos \left( \varphi \right)~$ so the energy is $~E=T+U~$ – Eli Jul 02 '23 at 19:00
  • @bobribar I didn't use the generalized force in my equation and force has not the unit of energy – Eli Jul 02 '23 at 19:06
  • I know what you did. I want to know whether I can use the way I described (which results the same for the given example). – bobribar Jul 02 '23 at 20:25
  • We have equal equation of motion, but where is your energy equation? – Eli Jul 02 '23 at 20:43
  • As I wrote, I built the energy equations in the moving reference frame, without $\dot{x}$ – bobribar Jul 03 '23 at 05:35
0

Seems like I didn't make a mistake, it works out for other examples too and this link https://galileoandeinstein.phys.virginia.edu/7010/CM_28_Non_Inertial_Frame_Coriolis.pdf answers the question also this way. It's much easier in a lot of examples to add the fictitious forces later and to not consider the movement of the frame in the kinetic energy.

bobribar
  • 11
  • 1
    As it’s currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. – Community Jul 04 '23 at 11:26