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Angular momentum of a two-particle system

  1. Dec 8, 2005 #1
    angular momentum of a two-particle system
    um heres where i get but idk what to do from here some stuff must be perpendicular always and stuff a lil help would be awesome
    \Sigma_i L_i = \Sigma_i \vec{r_i} \times m_i \vec{v_i} = \Sigma_i (\vec{r_i}+\vec{r_{cm}}) \times m_i (\vec{v_i}+\vec{v_{cm}})[/tex]
    \Sigma_i ((\vec{r_i} \times m_i \vec{v_i}) + (\vec{r_i} \times m_i \vec{v_{cm}}) + (\vec{r_{cm}} \times m_i \vec{v_i}) + (\vec{r_{cm}} \times m_i \vec{v_{cm}}))[/tex]
    Last edited: Dec 8, 2005
  2. jcsd
  3. Dec 8, 2005 #2
    make sure you use the reduced mass [tex]\mu = \frac{m_1+m_2}{m_1 m_2}[/tex]
  4. Dec 8, 2005 #3

    Physics Monkey

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    Recall that the weighted sum of the relative coordinates is zero from the definition of the center of mass. This allows you to kill two terms in your last equation thus recovering the usual statement that "the total angular momentum is the angular momentum of the center of mass plus the angular momentum about the center of mass".
  5. Dec 8, 2005 #4
    can you elaborate further please weighted sum?
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