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Homework Help: Centre of mass problem

  1. Sep 17, 2006 #1
    Prove that the magnitude of R of the position vector for the centre of mass from an arbitrary origin is given by the equation:
    [tex]M^2R^2 = M\sum m_ir_i^2 - {1\over 2}\sum m_i m_j r_{ij}^2[/tex]

    Well the centre of mass is given by:
    [tex]\vec R = \frac{\sum m_i r_i}{M}[/tex]

    But squaring this doesn't seem to produce the result I require. I need more help.
  2. jcsd
  3. Sep 17, 2006 #2
    I'm not too sure about this but I think it should work if you start working backwards - from the defintion of rij2, then multiply by -0.5mimj and then sum over the indices. My summation rules are a bit rusty. :redface:
    Last edited: Sep 17, 2006
  4. Sep 19, 2006 #3
    Work backwards? I will try that. My summation concepts are pretty outdated too.:yuck: Got to relocate my high school number theory text.
  5. Sep 19, 2006 #4
    This is quite straightforward. Let's start with:

    [itex] M\vec{R} = \sum_i m_i \vec{r}_i [/itex]

    Square it:

    [itex] M^2 R^2 = \sum_{i,j} m_i m_j \vec{r}_i \cdot \vec{r}_j \qquad (1) [/itex]

    Then consider [itex] \vec{r}_{i j} \equiv \vec{r}_i - \vec{r}_j \Longleftrightarrow r^2_{ij} = (\vec{r}_i - \vec{r}_j)^2 = r^2_i - 2 \vec{r}_i \cdot \vec{r}_j + r^2_j \Longleftrightarrow \vec{r}_i \cdot \vec{r}_j = \frac{1}{2}(r^2_i+r^2_j-r^2_{ij}) [/itex]

    Insert the last relationship in (1) and you'll obtain:

    [itex] M^2 R^2 = \frac{1}{2} \sum_{i,j} m_i m_j (r^2_i+r^2_j-r^2_{ij}) = \frac{1}{2} \sum_{i,j} m_i m_j (r^2_i+r^2_j) - \frac{1}{2} \sum_i m_i m_j r^2_{ij} \qquad (2) [/itex]

    Now, consider the first term of (2)'s rhs:

    [itex] \frac{1}{2} \sum_{i,j} m_i m_j (r^2_i+r^2_j) = \frac{1}{2}\sum_i m_i \left[\sum_j m_j (r^2_i + r^2_j) \right] = \frac{1}{2} \sum_i m_i \left[\sum_j m_j r^2_i + \sum_j m_j r^2_j \right] \qquad (3) [/itex]

    But [itex] \sum_j m_j r^2_i = M r^2_i [/itex] so:

    [itex] \frac{1}{2} \sum_i m_i \left[\sum_j m_j r^2_i + \sum_j m_j r^2_j \right] = \frac{1}{2} \sum_i m_i \left[ M r^2_i + \sum_j m_j r^2_j \right] = \frac{1}{2} \left[ M \sum_i m_i r^2_i + \sum_i m_i \sum_j m_j r^2_j \right] = \frac{1}{2} \left[ M \sum_i m_i r^2_i + M \sum_j m_j r^2_j \right] = [/itex]

    [itex] = \frac{1}{2} M \left[ \sum_i m_i r^2_i + \sum_j m_j r^2_j \right] [/itex]

    Because "j" is a summed index, you can call it "i" so (3) becomes:

    [itex] \frac{1}{2} \sum_{i,j} m_i m_j (r^2_i+r^2_j) = \frac{1}{2} M \left[ \sum_i m_i r^2_i + \sum_i m_i r^2_i \right] = M \sum_i m_i r^2_i \qquad (3) [/itex]

    Finally, just insert rhs of (3) in (2) and that's it :smile:
    Last edited: Sep 19, 2006
  6. Sep 19, 2006 #5
    Very nicely done, Emanuel. :approve:
  7. Sep 20, 2006 #6
    Bravo, Emanuel! :smile:
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