System of Masses - Atwood Machine

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  • #1
erobz
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Misplaced Homework Thread, no template
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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.
 

Answers and Replies

  • #3
erobz
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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:

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

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

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

$$ \frac{ \partial \mathcal{L} }{ \partial \dot{l_2} } = m_2 \left( \dot{l_2} - \dot{l_1} \right) + M \left( \dot{l_1} + \dot{l_2} \right) $$
 
  • #4
erobz
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And:

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

$$ \frac{d}{dt} \frac{ \partial \mathcal{L} }{ \partial \dot{l_2} } = \left( M - m_2 \right) \ddot{l_1} + \left( M+m_2 \right) \ddot{l_2}$$
 
  • #6
erobz
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This is not someone’s physics homework, why did I get a warning?
 
  • #7
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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
 
  • #8
kuruman
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Are you going to solve the equations of motion or leave it at that?
 
  • #9
erobz
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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
erobz
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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
erobz
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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?
 
Last edited:
  • #13
erobz
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post 10 looks good too
I found an error in this one! That means the system is consistent

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

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