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I'm supposed to use the Lagrangian approach in the solution. I've just recently started classical mechanics course and it's still somewhat confusing to me.

3 pulley atwood machine

The problem I am working on is as follows: "2 masses $m_1$ and $m_2$ are connected with a line across 3 pulleys of which the middle one has mass $M$ and can move freely in the up-down direction. The line is frictionless and massless. Find the acceleration of $m_1$ and $m_2$."

What I've figured on my own so far:

velocities

  1. I've marked velocities of the blocks and the middle pulley and I've figured out that $$ \dot x_3 = \frac{\dot x_1 + \dot x_2}{2} $$ The reason is that if the section of the string from which $m_1$ is suspended shortens by $x_1$ , and that from which $m_2$ is suspended by $x_2$ , then the central loop lengthens by $x_1+x_2$ . Since this loop consists of two vertical sections (and a semicircular one around the pulley whose length stays constant), this means that the pulley descends a distance of $\frac{x_1+x_2}{2}$ .

  2. I wrote down the kinetic and the potential energy: $$T=\frac{1}{2} m_1 \dot x_1^2 + \frac{1}{2} m_2 \dot x_2^2 + \frac{1}{2} M (\frac{\dot x_1 + \dot x_2}{2})^2$$ $$U=g(m_1x_1-M\frac{\dot x_1 + \dot x_2}{2} + m_2x_2)$$

  3. At this point I could write down the Lagrangian as $$L=T-U$$

This is where I'm somewhat stuck, as I don't think I can write down the Lagrange's equation yet as I've got both $x_1$ and $x_2$ in there, and I would expect to only have one $x$ and it's derivative.

As I've said, I'm new to lagrangian mechanics and I've got zero intuition on how to approach such problems so I'd be more than welcome for any clue.

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  • $\begingroup$ You may want to double check the middle term in the potential energy $U$ involving $M$, possibly a typo. In particular, the term $M$ is multiplied by the average velocity: $\frac{\dot{x}_{1}+\dot{x}_{2}}{2}$, where I believe you should have the average position: $\frac{x_{1}+x_{2}}{2}$. I hope this helps. $\endgroup$ – ad2004 Nov 21 '20 at 21:28
  • $\begingroup$ Yes, my bad, thank you! $\endgroup$ – Misha Nov 21 '20 at 23:07
  • $\begingroup$ You learn by trying, making mistakes, going back again. If you say "I don't know what to do next," go look at some example problems. Don't expect to get this stuff correct the first time you try, or even the 10th. Hint for this: the string length is constant. $\endgroup$ – Bill N Nov 24 '20 at 18:41
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There is one Lagrangian for both $x$'s, but there is a separate Euler-Lagrange equation for each of $x_1$, $x_2$.

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