Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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 My summation concepts are pretty outdated too.:yuck: Got to relocate my high school number theory text.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:
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook