System of Masses - Atwood Machine

AI Thread Summary
The discussion focuses on verifying the formulation of the Lagrangian for a system involving pulleys and masses. The user initially encounters unexpected results for the acceleration of one mass as they approach a limit where another mass approaches zero. After reviewing their calculations, they realize their error was in the interpretation of the results rather than an algebraic mistake. The final equations derived show that as one mass approaches zero, the acceleration of the other mass does not equal gravitational acceleration but rather indicates that the system's dynamics remain consistent. The user expresses satisfaction with the verification process and the clarity gained through the discussion.
erobz
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Would anyone verify whether or not I've formulated the proper Lagrangian here for the system above (the pulleys are massless, inextensible ropes of length ##L## and ##S##):

$$\mathcal{L} = T - U $$

$$ \mathcal{L} = \frac{1}{2} m_1 { \dot l_1 }^2+ \frac{1}{2} m_2 \left( { \dot l_2}- {\dot l_1 } \right)^2 + \frac{1}{2} M \left( {\dot l_1}+ {\dot l_2} \right) ^2 + m_1 g l_1 + m_2 g ( L - l_1 + l_2) + M g ( L - l_1 + S - l_2 )$$

If I find out this step looks ok, I'll post the remainder. I'm getting some unexpected result at the end for ## \ddot{l_2}## when ##m_1 \to 0##, and I'm wondering if I'm making an algebra error in the middle, or I have made a mistake right from the get-go.

This is an extension of a recent HW problem that I proposed (and go figure... turns out I can't solve it myself). I figured I 'd start a new thread since this is a different method of analyzing it.

Thanks for any help.
 
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The Lagrangian looks good to me
 
Dale said:
The Lagrangian looks good to me
Thanks Dale! I'm going to piece this together so it's not too much to process. I'm a newbie to using this machinery.

Next step:

$$ \frac{ \partial \mathcal{L} }{ \partial l_1 } = ( m_1 - m_2 -M ) g $$

$$ \frac{ \partial \mathcal{L} }{ \partial l_2 } = ( m_2 -M ) g $$

$$ \frac{ \partial \mathcal{L} }{ \partial \dot{l_1} } = m_1 \dot l_1 - m_2\left( \dot{l_2} - \dot{l_1} \right) + M \left( \dot{l_2} + \dot{l_1} \right) $$

$$ \frac{ \partial \mathcal{L} }{ \partial \dot{l_2} } = m_2 \left( \dot{l_2} - \dot{l_1} \right) + M \left( \dot{l_1} + \dot{l_2} \right) $$
 
And:

$$ \frac{d}{dt} \frac{ \partial \mathcal{L} }{ \partial \dot{l_1} } = \left ( m_1 + m_2+M \right) \ddot {l_1} + \left( M - m_2\right) \ddot {l_2} $$

$$ \frac{d}{dt} \frac{ \partial \mathcal{L} }{ \partial \dot{l_2} } = \left( M - m_2 \right) \ddot{l_1} + \left( M+m_2 \right) \ddot{l_2}$$
 
Yes, I get the same
 
This is not someone’s physics homework, why did I get a warning?
 
You didn't get a point. I just moved it. Even though it isn't homework it is homework-like so it belongs here. I removed the points from the warning before moving it. It won't affect your membership at all
 
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Are you going to solve the equations of motion or leave it at that?
 
kuruman said:
Are you going to solve the equations of motion or leave it at that?
I’m planning on it, but since all that appears correct I’m thinking I have an algebra error to find…
 
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  • #10
I just had a lightbulb moment. It took many times putting into practice Einstein's definition of insanity... I didn't have an algebra mistake, but rather an interpretation error.

$$ \frac{ \partial \mathcal{L} }{ \partial l_1 } = \frac{d}{dt} \frac{ \partial \mathcal{L} }{ \partial \dot{l_1} }$$

$$ ( m_1 - m_2 -M ) g = \left ( m_1 + m_2+M \right) \ddot {l_1} + \left( M - m_2\right) \ddot {l_2}$$

$$ \ddot{l_1} = \frac{(m_1 -m_2 -M)g - (M-m_2) \ddot{l_2}}{m_1 + m_2 + M} \tag{1}$$

$$ \frac{ \partial \mathcal{L} }{ \partial l_2 } = \frac{d}{dt} \frac{ \partial \mathcal{L} }{ \partial \dot{l_2} }$$

$$( m_2 -M ) g = \left( M - m_2 \right) \ddot{l_1} + \left( M+m_2 \right) \ddot{l_2} \tag{2}$$

Substitute ##(1) \to (2)##:

$$ \ddot{l_2} = \frac{ (m_1 + m_2 + M)( m_2 - M ) - ( M - m_2)(m_1 -m_2 -M) }{ ( m_1 + m_2 + M )(M+m_2) -(M-m_2)^2 } g \tag{3}$$

I had expected the masses to be in freefall if ##m_1 \to 0##, I reached the conclusion that ##m_1 \to 0, \ddot{l_2} = g ##. That conclusion was reached in error.

I had found what I was looking for, but I just hadn't realized it ( because I was thinking in terms of what I had done in the HW problem trying to use Newtons Laws )

Instead, I see that as ##m_1 \to 0, \ddot{l_2} \to 0 g ## meaning the length of the rope is not changing w.r.t. the pulley its attached to. They are both accelerating at ## - \ddot{l_1} = g## which is found by substituting ##m_1 = 0, \ddot{l_2} = 0## into ##(1)##.

Thats very good news! (I think?)

P.S. I realize ##(3)## simplifies greatly...I was getting tired.

$$\ddot{l_2} = \frac{ 2m_1(m_2 - M) }{m_1M + m_1 m_2 + 2 M m_2}g \tag{3}$$
 
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  • #11
What is not so comforting about arriving at this solution is my failure in applying Newtons Laws to solve this problem as I have tried in

Problem with 2 pulleys and 3 masses

Can anyone figure out what is wrong about that approach?
 
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  • #12
post 10 looks good too
 
  • #13
Dale said:
post 10 looks good too
I found an error in this one! That means the system is consistent

erobz said:
$$ \ddot{l_2} = \frac{ 2m_1(m_2 - M) }{m_1M + m_1 m_2 + 2 M m_2}g \tag{3}$$
$$ \ddot{l_2} = \frac{ 2m_1(m_2 - M) }{m_1M + m_1 m_2 + \boldsymbol{4} M m_2}g \tag{3}$$
 
  • #14
Thanks for all the verification along the way @Dale.
 
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