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Another worked Landau pendulum problem

Posted by peeterjoot on July 14, 2012

Question: Pendulum with support moving in line

This problem like the last, but with the point of suspension moving in a horizontal line $x = a \cos\gamma t$.

Our mass point has coordinates

\begin{aligned}p &= a \cos\gamma t + l i e^{-i\phi} \\ &= a \cos \gamma t + l i ( \cos \phi - i \sin \phi ) \\ &= ( a \cos \gamma t + l \sin \phi, l \cos \phi ),\end{aligned} \hspace{\stretch{1}}(1.10)

so that the velocity is

\begin{aligned}\dot{p} = ( -a \gamma \sin \gamma t + l \dot{\phi} \cos \phi, -l \dot{\phi} \sin \phi ).\end{aligned} \hspace{\stretch{1}}(1.11)

Our squared velocity is

\begin{aligned}\dot{p}^2 &= a^2 \gamma^2 \sin^2 \gamma t + l^2 \dot{\phi}^2 - 2 a \gamma l \dot{\phi} \sin\gamma t \cos \phi \\ &= \frac{1}{{2}} a^2 \gamma^2 \frac{d{{}}}{dt}\left( t - \frac{1}{{2 \gamma}} \sin 2 \gamma t \right) + l^2 \dot{\phi}^2 - a \gamma l \dot{\phi} ( \sin( \gamma t + \phi) + \sin(\gamma t - \phi)).\end{aligned} \hspace{\stretch{1}}(1.12)

In the last term, we can reduce the sum of sines, finding a total derivative term and a remainder as in the previous problem. That is

\begin{aligned}\dot{\phi} (\sin( \gamma t + \phi) + \sin(\gamma t - \phi)) &= (\dot{\phi} + \gamma)\sin(\gamma t + \phi) - \gamma \sin(\gamma t + \phi)+(\dot{\phi} - \gamma)\sin(\gamma t - \phi) + \gamma \sin(\gamma t - \phi) \\ &= \frac{d{{}}}{dt} \left( -\cos(\gamma t + \phi) + \cos(\gamma t - \phi) \right)+ \gamma ( \sin(\gamma t - \phi) - \sin(\gamma t + \phi) ) \\ &= \frac{d{{}}}{dt} \left( -\cos(\gamma t + \phi) + \cos(\gamma t - \phi) \right)- 2 \gamma \cos \gamma t \sin\phi.\end{aligned} \hspace{\stretch{1}}(1.13)

Putting all the pieces together and dropping the total derivatives we have the stated solution

\begin{aligned}\mathcal{L} = \frac{1}{{2}} m \left( l^2 \dot{\phi}^2 + 2 a \gamma^2 l \cos \gamma t \sin\phi \right) + m g l \cos\phi\end{aligned} \hspace{\stretch{1}}(1.14)

References

[1] LD Landau and EM Lifshitz. Mechanics, vol. 1. 1976.