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Centre of mass

  1. Dec 29, 2005 #1
    I was looking over my notes for centre of mass for a system and it says:

    [tex] c = \frac {1} {M} \sum_{i} m_i\ddot{r}_i = \sum_{i}(E_i + \sum_{j \neq i}F_i_j) [/tex]

    where M is the total mass of the system.

    Then it considers the centre of Mass in motion:

    [tex] M \ddot{c} = \sum_i m_i \ddot{r}_i = \sum_{i}(E_i + \sum_{j \neq i}F_i_j) [/tex]

    [tex] = \sum_{i}E_i + \sum_{j \neq i}F_i_j = E + \sum_{i < j}(F_j_i + F_j_j)[/tex]

    [tex] = E [/tex]

    The thing is, I don't understand the line:

    [tex]E + \sum_{i < j}(F_j_i + F_j_j)[/tex]

    and how it comes about (especially the i < j) part.

    Any help would be grateful!
     
    Last edited: Dec 29, 2005
  2. jcsd
  3. Dec 29, 2005 #2

    Doc Al

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    Staff: Mentor

    Could there be a typo? I'd think that:
    [tex]\sum_{j \neq i}F_i_j = \sum_{i < j}(F_j_i + F_i_j)[/tex]

    (And this term disappears due to Newton's 3rd law.)
     
  4. Jan 1, 2006 #3
    Is this something you just need to know? It suddenly appears in my Uni notes without any explanation.
     
  5. Jan 1, 2006 #4

    Doc Al

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    Staff: Mentor

    I'm not sure what you're asking. Are you asking "Why is that true?" or "How am I supposed to know that?"
     
  6. Jan 1, 2006 #5
    What I mean is why is this true? Can this be shown from previous statements regarding the interaction of forces. I don't understand where

    [tex]\sum_{j \neq i}F_i_j = \sum_{i < j}(F_j_i + F_i_j)[/tex]

    is coming from.
     
  7. Jan 1, 2006 #6

    Doc Al

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    Staff: Mentor

    The truth of this has nothing to do with physics; it's just a mathematical truism. To see that it's true, make up a small example (where i,j < 5, say) and confirm that both sides are equal.
     
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