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On the moment of inertia of a system of biparticles

  1. Nov 6, 2011 #1
    The moment of inertia of a system of particles, is given by:

    [tex]I_i^{\vphantom{^{\:2}}} = \sum_i \; m_i^{\vphantom{^{\:2}}} \, \vec{r}_i^{\:2}[/tex]
    The moment of inertia of a system of biparticles, is given by:

    [tex]I_{ij}^{\vphantom{^{\:2}}} = \sum_i \; \sum_{j>i} \; \, m_i^{\vphantom{^{\:2}}} m_j^{\vphantom{^{\:2}}} \; \left( \vec{r}_i - \vec{r}_j \right)^2[/tex]
    A system of particles forms a system of biparticles, and from the above equations the following relation can be obtained:

    [tex]I_{ij}^{\vphantom{^{\:cm}}} = M_i^{\vphantom{^{\:cm}}} \: I_i^{\:cm}[/tex]
    where [itex]I_{ij}^{\vphantom{^{\:cm}}}[/itex] is the moment of inertia of the system of biparticles, [itex]M_i^{\vphantom{^{\:cm}}}[/itex] is the mass of the system of particles, and [itex]I_i^{\:cm}[/itex] is the moment of inertia of the system of particles relative to the center of mass.

    For example, the system of particles A, B, C and D forms the system of biparticles AB, AC, AD, BC, BD and CD.

    [itex]{\vphantom{aat}}[/itex]
     
  2. jcsd
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