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Arbitrary choice of origin when linear momentum is zero

  1. Jun 12, 2010 #1
    I read the following in a textbook I'm reading:

    If the total linear momentum of a system of particles is zero, then the angular momentum of the system is independent of the choice of origin.

    It was given without proof and I've been trying to see why this is the case - mainly intuitively speaking, as it doesn't seem entirely obvious to me, if anyone can provide any insight.
  2. jcsd
  3. Jun 13, 2010 #2
    I think you would memorize it best, if you do the calculation by yourself :)
    Try pluging in [itex]\vec{r}'=\vec{r}+\vec{a}[/itex] into the definition of angular momentum and see if [itex]\vec{a}[/itex] cancels given [itex]\sum m_i v_i=0[/itex] (zero total momentum).
  4. Jun 13, 2010 #3
    [tex]\displaystyle \Sigma \Right ( r_i \times m_i v_i \Left ) - \Sigma \Right ( (r_i + \alpha_i) \times m_i v_i \Left )[/tex]
    which gives
    [tex]\Sigma \Right ( \alpha_i \times m_i v_i \Left )[/tex]
    not sure why this should be zero though?
  5. Jun 13, 2010 #4
    "a" ist a constant displacement of origin which is the same for all point. Take it out of the sum.
  6. Jun 13, 2010 #5
    Ah yes, although the mathematical solution still doesn't lend much to the intuitive reason behind it at all tbh.
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