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I am given the following Lagrangian:

$$L = \frac{ml^2}{2}\left((\dot\theta)^2 + \sin^2{(\theta)}\dot\phi^2\right) + \frac{I\dot\phi^2}{2} + mgl\cos{\theta}$$

which is meant to represent a simple pendulum mounted on a rotating turntable, and $m, g, l$, and $I$ are some physical constants. The task is to obtain the equations of motion, which I think I did, but we were given no solutions, so I can't know for sure if I am correct or not.

To solve it I applied the Euler-Lagrange equation (ELE), separately on both the variables $\theta$ and $\phi$:

0 = $\frac{\partial L}{\partial \theta} - \frac{d}{dt}\left(\frac{\partial L}{\partial\dot\theta} \right) = \frac{ml^2}{2}\dot\phi^2\sin{2\theta}-mgl\sin{\theta} + ml^2 \ddot\theta$

from which the first equation of motion is: $\ddot\theta = \frac{g}{l}\sin{\theta}-\frac{1}{2}\dot\phi^2\sin{\theta}$.

Then I applied the ELE for $\phi$: 0 = $\frac{\partial L}{\partial \phi} - \frac{d}{dt}\left(\frac{\partial L}{\partial\dot\phi} \right) = (-ml^2\sin^2{\theta}-I)\ddot\phi = 0$.

This is where I sort of ran into trouble. My next reasoning was that since the term in the brackets isn't always zero, the only way for the equation to be satisfied is if $\ddot\phi=0$, but I don't know if this is appropriate in Lagrangian mechanics.

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  • $\begingroup$ There is a sign error in your equation of motion for $\theta$. In the equation of motion for $\phi$, the time derivative in the Euler-Lagrange equation also acts on $\sin^2(\theta)$ giving a contribution in $\dot \theta$. This term allows for $\ddot \phi \neq 0$. $\endgroup$ May 26, 2022 at 14:16

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With the Lagrangian $$L = \frac{ml^2}{2}\left(\dot \theta^2 + \sin^2(\theta) \dot \phi^2\right) +\frac{I}{2} \dot\phi^2 + mgl\cos(\theta)$$

the Euler-Lagrange equations are : \begin{align} ml^2\ddot \theta &= ml^2\cos(\theta)\sin(\theta)\dot\phi^2 -mgl\sin(\theta) \\ \frac{d}{dt}\left( ml^2\sin^2(\theta)\dot \phi+ I\dot \phi\right) &= 0 \end{align}

This means there is a constant $C$ such that : $$\dot \phi = \frac{C}{I + ml^2\sin^2(\theta)}$$

and the equation for $\theta$ is : $$\ddot \theta =\frac{C^2\cos \theta\sin\theta}{(I+ml^2\sin^2(\theta))^2}-\frac{g}{l}\sin(\theta)$$

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  • $\begingroup$ Thank you, this helps a lot! $\endgroup$
    – NX37B
    May 26, 2022 at 14:45

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