# Lagrangian of System:

1. Jan 27, 2007

### HalfManHalfAmazing

1. The problem statement, all variables and given/known data
Two masses $$m_1$$ and $$m_2$$ are connected by a string passing through a hole on a smooth table so that $$m_1$$ rests on the table surface and $$m_2$$ hangs suspended. Assuming $$m_2$$ moves only in a vertical line determine the generalized coordinates for the system. Find the Lagrangian and find the Lagrangian Equations.

3. The attempt at a solution
I began by defining $$l$$ to the length of the string. I also assumed $$m_1$$ moved only along the x-axis. Thus the coordinates of $$m_1$$ are simply $$l_1$$ and the coordinates of $$m_2$$ are $$l_2$$ where $$l = l_1 + l_2$$. Are these the correct generalized coords? Determining the velocities (if these are the correct coords) is trivial and thus the lagrangian is also easy to calculate. My difficulty is with determing the coordinates. Thanks for your help!

2. Jan 27, 2007

### HalfManHalfAmazing

Also just to confirm if i did everything right, my equations of motion are $$m_1\ddot{l_1} = 0$$ and $$m_2\ddot{l_2} + m_2g = 0$$

3. Jan 28, 2007

### AlephZero

The way I would interpret the words "generalized coordinates" you should only have ONE generalized coordinate in this problem because there is only one degree of freedom.

In other words let the generalized coordinate by q, then displacement of m1 is x=q horizontally, and the displacement of m2 is y=-q vertically (positive upwards) because the string is inextensible.

Your equations of motion as you wrote them are not right, since m_1 l"_1 = 0 implies mass m_1 has no acceleration. Your equation don't contain the fact that there is a string joining the masses, because you haven't used the contraint l1 = (plus or minus) l2 in your two equations.

The point of using generalized coordinates is to formulate the problem in terms of a small number of degrees of freedom right from the start by writing the energies in terms of the generalized variables only.

You should get just one equation of motion involving q, m1, m2 and g.

The correct equations motion for the two masses are m_1l"_1 = P and m_2l"_2 + m2_g = -P where P is the tension in the string - but doing it that way is not solving the problem using generalized coordinates!

Last edited: Jan 28, 2007
4. Jan 28, 2007

### HalfManHalfAmazing

Thanks AlephZero! I'll post my next 'try'.

5. Jan 28, 2007

### HalfManHalfAmazing

Man. I can't produce the proper generalized coordinates. if l = length of the string and q = the distance of m_1 along the x-axis (or q = x). then l - q = y ?

6. Jan 28, 2007

### HalfManHalfAmazing

if that's the case should $$\dot{y} = -\dot{q}$$

7. Jan 28, 2007

### AlephZero

You seem to be getting there. There is more than one way of imagining this problem, so questions like "is this equation right" can't be answered without a picture.

This is how I'm thinking of it. If you imagined m1 on the left of m2, or had q positive in the opposite direction, the Lagrangian equation would be different but when you tranform back to physical X-Y coordinates the answer would be the same.

I chose the origin of the X-Y axes to line up with the initial position of the masses.

I drew it so that x = q, and y = -q
therefore x' = q' and y' = -q'

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Last edited: Jan 28, 2007
8. Jan 28, 2007

### HalfManHalfAmazing

Shouldn't y = l - q? How can the location of the second mass be always the same from the origin as the first mass? Or am I misunderstanding still?

9. Jan 28, 2007

### AlephZero

You can choose to put the origin of the axes anywhere you like. If you prefer to put the origin of the X-Y plane at the point where m1 is, then yes you would have y = l-q (where l is length of the vertical part of the string, not the length of the whole string as you originally said).

I chose to put the origin underneath m1 and at the same level as m2. It will work out right whichever way you do it.

One of the nice things about the Lagrange method is, since you only need to work out the PE and KE for each object in terms of the generalized coordinates, you can use a different physical X-Y-Z coordinate systems for each object if it makes things simpler.

10. Jan 29, 2007

### HalfManHalfAmazing

Thanks so much AlephZero! I'll post my last attempt (I say last because I feel very close to the solution)!

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