(adsbygoogle = window.adsbygoogle || []).push({}); Basic exercise for finding a Lagrangian from the Landau's "Mechanics"

Hello everyone!

1. The problem statement, all variables and given/known data

I've just started preparing for the classical mechanics course using only Landau & Lifgarbagez, so I'm doing everything according to their formulation.

And so I solved an exercise from the first chapter, but I'm not sure if I made that one correct (the answer is different, but a Lagrangian is ambiguous, isn't it?). The statement is:

Find the Lagrange function of a simple pendulum of length [itex] l [/itex], mass [itex] m [/itex], which pivot rotates along a circle of radius [itex] a [/itex] with a constant frequency [itex] f [/itex] in a homogeneous gravitational field with acceleration [itex] g [/itex].

2. Relevant equations

Formula for a Lagrangian of an isolated, one-particle system in 2 dimensions:

[tex]L=\frac{1}{2}m(\dot{x}^2+\dot{y}^2)+\textbf{F}\cdot \textbf{r}[/tex]

3. The attempt at a solution

I first formulated a more general case:

- the pivot moves around a circle satisfying [itex] \alpha(t) [/itex], ([itex]\alpha[/itex] is the angle between the pivot's radius vector and the y-axis),

- the angle between the pendulum and y-axis is [itex]\phi(t)[/itex],

- I've chosen the [itex]\alpha[/itex] and [itex]\phi[/itex] for the generalized coordinates of this system.

- the Cartesian coordinates of the mass are:

[tex]x=l \sin{\phi}+a \sin{\alpha}[/tex]

[tex]y=l \cos{\phi}+a \cos{\alpha}[/tex]

- the velocities:

[tex]\dot{x}=l \dot{\phi} \cos{\phi}+a \dot\alpha} \cos{\alpha}[/tex]

[tex]\dot{y}=-l \dot\phi} \sin{\phi}-a \dot\alpha} \sin{\alpha}[/tex]

U=[itex]-\textbf{F}\cdot \textbf{r}=-mgy=-mg(l \cos{\phi}+a \cos{\alpha})[/itex]

[tex]T=\frac{1}{2}m(\dot{x}^2+\dot{y}^2)=\frac{1}{2}m(l^2 \dot{\phi}^2 \textrm{cos}^2 \phi+a^2 \dot{\alpha}^2 \textrm{cos}^2\alpha+2 l a \dot{\phi} \dot{\alpha} \cos{\alpha}

\cos{\phi}+l^2 \dot{\phi}^2 \textrm{sin}^2 \phi+a^2 \dot{\alpha}^2

\textrm{sin}^2 \alpha+2 l a \dot{\phi} \dot{\alpha} \sin{\alpha} \sin{\phi}) [/tex]

[tex]L=T-U=m \left(\frac{1}{2}(l^2 \dot{\phi}^2+a^2 \dot{\alpha}^2)+l a \dot{\phi} \dot{\alpha} \cos{(\alpha-\phi)}+g(l \cos{\phi}+a \cos{\alpha}) \right)[/tex]

-And now I substitute [itex]\alpha=\alpha(t)=f t[/itex], and my final solution is:

[tex]L=m \left(\frac{1}{2}(l^2 \dot{\phi}^2+a^2 f^2)+l a \dot{\phi} f \cos{(f t-\phi)}+g(l \cos{\phi}+a \cos{f t}) \right)[/tex]

The answer in Landau & Lifgarbagez is:

[tex]L=\frac{1}{2}m l^2 \dot{\phi}^2+m l a f^2 \sin{(\phi-f t)}+m g l \cos{\phi}[/tex]

So is my solution correct and if not, what mistakes am I making? I will be grateful for every answer. I will just add that I don't fully understand the idea of the "total derivative" which, as I suspect is crucial for comparing this two solutions.

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# Homework Help: Basic exercise for finding a Lagrangian from the Landau's Mechanics

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