Lagrange EQ for a Double Atwood Machine

[kq6up]

Homework Statement

7.27 ** Consider a double Atwood machine constructed as follows: A mass 4m is suspended from a string that passes over a massless pulley on frictionless bearings. The other end of this string supports a second similar pulley, over which passes a second string supporting a mass of 3m at one end and m at the other. Using two suitable generalized coordinates, set up the Lagrangian and use the Lagrange equations to find the acceleration of the mass 4m when the system is released. Explain why the top pulley rotates even though it carries equal weights on each side.

Homework Equations

$\frac { \partial L }{ \partial q } =\frac { d }{ dt } \frac { \partial L }{ \partial \dot { q } }$

The Attempt at a Solution

Looking at the solution I did all the steps after setting up the Lagrange e.q. correctly. My lagrange is incorrect, but I am not sure where I went awry. Could you folks give a hint?

Here is my Lagrange E.Q.:

$L=4m{ \dot { q } }_{ 1 }^{ 2 }+2m{ \dot { q } }_{ 2 }^{ 2 }-2m\dot { { q }_{ 1 } } \dot { { q }_{ 2 } } +2mg{ q }_{ 2 }$

edit: This is the resultant equation after I expanded all the terms and added cross terms. The algebra is good, the initial setup is different than what the solution manual had. However, I am having trouble seeing how our initial equations are not different equations stating the same problem. The SM yielded: $L=4m{ \dot { q } }_{ 1 }^{ 2 }+2m{ \dot { q } }_{ 2 }^{ 2 }+2m\dot { { q }_{ 1 } } \dot { { q }_{ 2 } } -2mg{ q }_{ 2 }$ after the algebraic smoke cleared.

Thanks,
Chris

 

[Orodruin]

Without you providing details, including what generalized coordinates you chose, it will be difficult to see where you went wrong. Just based on you lagrangian and that of the solution manual, I would say it seems you have defined q2 in the opposite direction from the SM as you will recover the same lagrangian if you do the transformation q2 -> -q2.

 

[kq6up]

Without you providing details, including what generalized coordinates you chose, it will be difficult to see where you went wrong. Just based on you lagrangian and that of the solution manual, I would say it seems you have defined q2 in the opposite direction from the SM as you will recover the same lagrangian if you do the transformation q2 -> -q2.

Would that not give the same answer?

Here is my original lagrangian without expanding and simplifying $L=2m{ \dot { q } }_{ 1 }^{ 2 }+\frac { 3 }{ 2 } m{ \left( { \dot { q } }_{ 2 }-{ \dot { q } }_{ 1 } \right) }^{ 2 }+\frac { 1 }{ 2 } m{ \left( { \dot { q } }_{ 2 }+{ \dot { q } }_{ 1 } \right) }^{ 2 }+4mg{ q }_{ 1 }-3mg\left( { q }_{ 1 }-{ q }_{ 2 } \right) -mg\left( { q }_{ 1 }{ +q }_{ 2 } \right)$

The SM lagrangian before expansion/simplification: $L=2m{ \dot { q } }_{ 1 }^{ 2 }+\frac { 1 }{ 2 } m{ \left( { \dot { q } }_{ 2 }-{ \dot { q } }_{ 1 } \right) }^{ 2 }+\frac { 3 }{ 2 } m{ \left( { -\dot { q } }_{ 1 }-{ \dot { q } }_{ 2 } \right) }^{ 2 },\quad U=-4mg{ q }_{ 1 }-mg\left( { { l }_{ 1 }-q }_{ 1 }{ +q }_{ 2 } \right) -3mg\left( { { l }_{ 1 }-q }_{ 1 }+{ l }_{ 2 }-{ q }_{ 2 } \right)$

Thanks,
Chris

 

[Orodruin]

No, it would not give the same answer. It would change the sign of the terms that are linear in $q_2$ and $\dot q_2$, just as you would get a different sign for a particle in a gravitational field depending on how you define the vertical coordinate (positive in the up or down direction). The terms quadratic in $q_2$ and $\dot q_2$ will be multiplied by a factor $(-1)^2 = 1$.

 

[kq6up]

I see that the equations are not the same. However, I am having a hard time understanding why choosing the direction of q2 should matter from a physical perspective. In my mind it seems like the choice of direction should be arbitrary. I do see from the math it is not. I am just having a hard time reconciling the two.

Edit:

One thing that does not make sense to me is how they have written their equations. It makes sense that the q2 will be falling towards the 3m side. When the pulley that q2 is wrapped around pulled up towards the q1 side, the fall of 3m should slow, and the rise of m should speed up by adding the velocity of rope q1. This is reflected in my equation not theirs as far as I can tell.

Edit II:

Ok, after some head scratching I came to the conclusion the two paragraphs above are wrong. I will leave them there just for a record.

If you take the q2 as running in the reverse direction for the SM version (which it is $2\ddot { y } +\ddot { x } =-g$ from the SM) the quantity L should be invariant. The first two terms don't matter as they are squared. The q2 being negative adds a negative in the 3rd term, and it cancels the negative in front of the fourth term (because it is running backwards in the SM). Ok I feel a little better now as it seems that they are equivalent. I will double check my math from that point on. If you are still convinced they are different, then I have absolutely no clue how. I spent a good hour thinking very carefully about it. I am open to being convinced of my er though.

Thanks,
Chris

 

[kq6up]

Finally figured it out. Both of the equations above give the correct answer (the SM and my lagrangian both work as they should). After I plugged it into the Euler-Lagrange (for the principle of stationary action) I made a simple algebraic mistake. Once I found that little mistake, I got g/7 just like the solution manual states. Your suggestion really threw me as I could not see how those lagrangians should be different other than the direction that q2 is defined. They weren't, but I really had to scrutinize everything to be 100% sure of that.

I should have carefully checked my algebra before posting -- lesson learned.

Thanks,
Chris

 

[Orodruin]

If you take the q2 as running in the reverse direction for the SM version (which it is $2\ddot { y } +\ddot { x } =-g$ from the SM) the quantity L should be invariant. The first two terms don't matter as they are squared. The q2 being negative adds a negative in the 3rd term, and it cancels the negative in front of the fourth term (because it is running backwards in the SM). Ok I feel a little better now as it seems that they are equivalent. I will double check my math from that point on. If you are still convinced they are different, then I have absolutely no clue how. I spent a good hour thinking very carefully about it. I am open to being convinced of my er though.

Yes, this is basically what I said. By "answer" above I am referring to the form of the Lagrangian, not "mass A will go up and B down". This will of course be invariant regardless of how you parametrize your problem.

L is not necessarily invariant under any transformation. However, regardless of what transformation you make you should end up with the same physical result, you just need to apply the inverse transformation to the end result.

 

[vanhees71]

I'd say it would help the OP very much, if he/she could send us a full description of the problem, perhaps attaching a little drawing. That may even help him/her to solve the problem himself/herself!

 

[kq6up]

Yes, I will post that in the future. The problem on this one was that I missed a tiny little algebra mistake that crept in at the very end of the problem. I was meticulous about thinking through the initial set up, so I had assumed I had no math errors. I guess I got sloppy toward the end of the problem.

Chris