# On the moment of inertia of a system of biparticles

1. Nov 6, 2011

### motion_ar

The moment of inertia of a system of particles, is given by:

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

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

$$I_{ij}^{\vphantom{^{\:cm}}} = M_i^{\vphantom{^{\:cm}}} \: I_i^{\:cm}$$
where $I_{ij}^{\vphantom{^{\:cm}}}$ is the moment of inertia of the system of biparticles, $M_i^{\vphantom{^{\:cm}}}$ is the mass of the system of particles, and $I_i^{\:cm}$ 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.

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