I'm just beginning to learn about Lagrangian mechanics, and I am asked to find the kinetic energy of this Atwood machine (See figure).

Atwood machine

I am told, that the kinetic energy should be:


I am also told, that the movement is so slow, that the spring is always stretched, and $x$ and $y$ are my generalized coordinates.

So what to do? My first though was, that since the spring is always stretched, all masses are dependent of each other, and I could maybe look at it as a whole string with one mass on the left, and two on the right. But that didn't give me the right answer. So I was hoping someone could give me a hint or something?

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    $\begingroup$ Have you tried writing an expression for the potential energy (since you already have kinetic energy), and then using the Euler-Lagrange equation? $\endgroup$ Mar 20, 2014 at 13:06
  • $\begingroup$ Nope, but the next question in this assignment is to show that the potential energy is some expression. So I'm guessing that I'm not allowed to do workarounds :) $\endgroup$ Mar 20, 2014 at 13:13
  • $\begingroup$ Ah, I get it. Normally for these problems you write $L = T - V$ and use Euler-Lagrange, but this is just part 1. I've given you a gist of how to do it in my answer, while trying not to give away the whole thing. $\endgroup$ Mar 20, 2014 at 13:23
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    $\begingroup$ I think when it says the spring is always stretched, it does not mean the spring length is constant, it just means the spring length never becomes zero. $\endgroup$ Mar 20, 2014 at 14:57

1 Answer 1


I think you should first express the kinetic energy of each block, using $\frac{1}{2} m v^2$, where $v$ is the velocity of the block. Then just sum these up. It looks like for two of the blocks, the velocity is $\dot{x}$, and for one of them the velocity is $\dot{x} - \dot{y}$. Be careful to remember that for one of the blocks the mass is $2m$.

  • $\begingroup$ Okay, I think I may have it now. The big bob is $m\dot{x}^{2}$, the small, and first bob is $\frac{1}{2}m\dot{x}^{2}$ since $x$ and $y$ is dependent on each other in this case. But, I'm not sure I totally understand why the last bob is $\frac{1}{2}m(\dot{x}-\dot{y})^{2}$ ? Or maybe that's because the last bob actually have $\frac{1}{2}m\dot{x}^{2}$ and the second bob has the above ? That would indeed make more sense, but maybe I'm mistaken ? $\endgroup$ Mar 20, 2014 at 13:41
  • $\begingroup$ The last bob is $\frac{1}{2}m(\dot{x}-\dot{y})^{2}$ because if the big bob moves $\Delta x$, then the string moves $\Delta x$, but the spring can stretch by $\Delta y$ to make the last bob move less. Maybe it will be more clear if you imagine what would happen if $x$ and $y$ increased at the same rate: the right-most bob should be stationary then, right? $\endgroup$ Mar 20, 2014 at 14:32
  • $\begingroup$ I think I have it now. I might have been confused by the "string is always stretched" statement, where I just though it was constant all the time. But the gentleman above made it a bit clearer. Thank you both :) $\endgroup$ Mar 20, 2014 at 15:00
  • $\begingroup$ Yeah, that line was a bit misleading. $\endgroup$ Mar 20, 2014 at 15:13

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