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Conservation of angular momentum invariance

  1. Sep 15, 2015 #1
    1. The problem statement, all variables and given/known data

    Given a reference frame O' moving at a constant speed $\vec{V}$ in relation to another reference frame O, I want to prove that

    ##\vec{r_{1B}} \times m_1\vec{v_{1B}} + \vec{r_{2B}} \times m_2\vec{v_{2B}} = \vec{r_{1F}} \times m_1\vec{v_{1F}} + \vec{r_{2F}} \times m_2\vec{v_{2F}}##

    in O is equal to

    ##\vec{r'_{1B}} \times m_1\vec{v'_{1B}} + \vec{r'_{2B}} \times m_2\vec{v'_{2B}} = \vec{r'_{1F}} \times m_1\vec{v'_{1F}} + \vec{r'_{2F}} \times m_2\vec{v'_{2F}}##

    in O'. The particles 1 and 2 are colliding (elastic collision). B stands for before the collision and F, after the collision.

    2. Relevant equations

    Galilean transformation gives :

    ##\vec{r} = \vec{r'} + \vec{V}t## and ##\vec{v} = \vec{v'} + \vec{V}##

    3. The attempt at a solution

    Substituting these expressions into the first equation and developing the cross products, I obtain, after having cancelled out 8 terms (because of the conservation of linear momentum),

    ##\vec{r'_{1B}} \times m_1\vec{v'_{1B}} + \vec{r'_{1B}} \times m_1\vec{V} + \vec{r'_{2B}} \times m_2\vec{v'_{2B}} + \vec{r'_{2B}} \times m_2\vec{V} = \vec{r'_{1F}} \times m_1\vec{v'_{1F}} + \vec{r'_{1F}} \times m_1\vec{V} + \vec{r'_{2F}} \times m_2\vec{v'_{2F}} + \vec{r'_{2F}} \times m_2\vec{V}##

    My problem is that I do not know how to cancel out the ##\vec{r'_{1B}} \times m_1\vec{V}## terms... How do I do that? Thanks!
     
    Last edited: Sep 15, 2015
  2. jcsd
  3. Sep 18, 2015 #2
    The sum of these terms across all particles = (total mass * position of the center of mass) X V . This quantity is preserved in collisions, since the center of mass and total mass don't change. [In addition, it is preserved across the Galilean transformation also, since r' - r is a multiple of V so contributed nothing to the cross product]
     
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