# Atwood machine with spring

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

I am told, that the kinetic energy should be:

$$T=2m\dot{x}^{2}+\frac{1}{2}m\dot{y}^{2}-m\dot{x}\dot{y}$$

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?

• Have you tried writing an expression for the potential energy (since you already have kinetic energy), and then using the Euler-Lagrange equation? Mar 20, 2014 at 13:06
• 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 :) Mar 20, 2014 at 13:13
• 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. Mar 20, 2014 at 13:23
• 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. Mar 20, 2014 at 14:57

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 $2m$.
• 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 ? Mar 20, 2014 at 13:41
• 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? Mar 20, 2014 at 14:32