Editing Moment of inertia (section)

=== Kinetic energy ===
The kinetic energy of a rigid system of particles can be formulated in terms of the [[center of mass]] and a matrix of mass moments of inertia of the system. Let the system of <math>n</math> particles <math>P_i, i = 1, \dots, n</math> be located at the coordinates <math>\mathbf{r}_i</math> with velocities <math>\mathbf{v}_i</math>, then the kinetic energy is<ref name="Marion 1995"/><ref name="Kane"/>
<math display="block">
  E_\text{K} = \frac{1}{2} \sum_{i=1}^n m_i \mathbf{v}_i \cdot \mathbf{v}_i =
    \frac{1}{2} \sum_{i=1}^n
      m_i \left(\boldsymbol{\omega} \times \Delta\mathbf{r}_i + \mathbf{V}_\mathbf{C}\right) \cdot
          \left(\boldsymbol{\omega} \times \Delta\mathbf{r}_i + \mathbf{V}_\mathbf{C}\right),
</math>
where <math>\Delta\mathbf{r}_i = \mathbf{r}_i - \mathbf{C}</math> is the position vector of a particle relative to the center of mass.

This equation expands to yield three terms
<math display="block">
  E_\text{K} =
    \frac{1}{2}\left(\sum_{i=1}^n m_i \left(\boldsymbol{\omega} \times \Delta\mathbf{r}_i\right) \cdot \left(\boldsymbol{\omega} \times \Delta\mathbf{r}_i\right)\right) +
    \left(\sum_{i=1}^n m_i \mathbf{V}_\mathbf{C} \cdot \left(\boldsymbol{\omega} \times \Delta\mathbf{r}_i\right)\right) +
    \frac{1}{2}\left(\sum_{i=1}^n m_i \mathbf{V}_\mathbf{C} \cdot \mathbf{V}_\mathbf{C}\right).
</math>

Since the center of mass is defined by 
<math>
 \sum_{i=1}^n m_i      \Delta\mathbf{r}_i =0</math>
, the second term in this equation is zero. Introduce the skew-symmetric matrix <math>[\Delta\mathbf{r}_i]</math> so the kinetic energy becomes
<math display="block">\begin{align}
  E_\text{K}
    &= \frac{1}{2}\left(\sum_{i=1}^n m_i \left(\left[\Delta\mathbf{r}_i\right] \boldsymbol{\omega}\right) \cdot \left(\left[\Delta\mathbf{r}_i\right] \boldsymbol{\omega}\right)\right) +
         \frac{1}{2}\left(\sum_{i=1}^n m_i\right) \mathbf{V}_\mathbf{C} \cdot \mathbf{V}_\mathbf{C} \\
    &= \frac{1}{2}\left(\sum_{i=1}^n m_i \left(\boldsymbol{\omega}^\mathsf{T}\left[\Delta\mathbf{r}_i\right]^\mathsf{T} \left[\Delta\mathbf{r}_i\right] \boldsymbol{\omega}\right)\right) +
         \frac{1}{2}\left(\sum_{i=1}^n m_i\right) \mathbf{V}_\mathbf{C} \cdot \mathbf{V}_\mathbf{C} \\
    &= \frac{1}{2}\boldsymbol{\omega} \cdot \left(-\sum_{i=1}^n m_i \left[\Delta\mathbf{r}_i\right]^2\right) \boldsymbol{\omega} +
         \frac{1}{2}\left(\sum_{i=1}^n m_i\right) \mathbf{V}_\mathbf{C} \cdot \mathbf{V}_\mathbf{C}.
\end{align}</math>

Thus, the kinetic energy of the rigid system of particles is given by
<math display="block">E_\text{K} = \frac{1}{2} \boldsymbol{\omega} \cdot \mathbf{I}_\mathbf{C} \boldsymbol{\omega} + \frac{1}{2} M\mathbf{V}_\mathbf{C}^2.</math>
where <math>\mathbf{I_C}</math> is the inertia matrix relative to the center of mass and <math>M</math> is the total mass.