Editing Moment of inertia (section)

== Motion in space of a rigid body, and the inertia matrix ==
The scalar moments of inertia appear as elements in a matrix when a system of particles is assembled into a rigid body that moves in three-dimensional space. This inertia matrix appears in the calculation of the angular momentum, kinetic energy and resultant torque of the rigid system of particles.<ref name="Marion 1995"/><ref name="Symon 1971"/><ref name="Tenenbaum 2004"/><ref name="Kane"/><ref name="Tsai">L. W. Tsai, Robot Analysis: The mechanics of serial and parallel manipulators, John-Wiley, NY, 1999.</ref>

{{For|analysis of a spinning top|Precession#Classical (Newtonian)|Euler's equations (rigid body dynamics)}}

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> relative to a fixed reference frame. For a (possibly moving) reference point <math>\mathbf{R}</math>, the relative positions are
<math display="block">\Delta\mathbf{r}_i = \mathbf{r}_i - \mathbf{R}</math>
and the (absolute) velocities are
<math display="block">\mathbf{v}_i = \boldsymbol{\omega} \times \Delta\mathbf{r}_i + \mathbf{V}_\mathbf{R}</math>
where <math>\boldsymbol{\omega}</math> is the angular velocity of the system, and <math>\mathbf{V_R}</math> is the velocity of <math>\mathbf{R}</math>.

=== Angular momentum ===
Note that the [[Cross product#Conversion to matrix multiplication|cross product can be equivalently written as matrix multiplication]] by combining the first operand and the operator into a skew-symmetric matrix, <math>\left[\mathbf{b}\right]</math>, constructed from the components of <math>\mathbf{b} = (b_x, b_y, b_z)</math>:
<math display="block">\begin{align}
  \mathbf{b} \times \mathbf{y}
                           &\equiv \left[\mathbf{b}\right] \mathbf{y} \\
  \left[\mathbf{b}\right] &\equiv \begin{bmatrix}
                 0   & -b_z &  b_y \\
                 b_z &    0 & -b_x \\
                -b_y &  b_x &    0
              \end{bmatrix}.
\end{align}</math>

The inertia matrix is constructed by considering the angular momentum, with the reference point <math>\mathbf{R}</math> of the body chosen to be the center of mass <math>\mathbf{C}</math>:<ref name="Marion 1995"/><ref name="Kane"/>
<math display="block">\begin{align}
  \mathbf{L}
    &= \sum_{i=1}^n m_i\,\Delta\mathbf{r}_i \times \mathbf{v}_i \\
    &= \sum_{i=1}^n m_i\,\Delta\mathbf{r}_i \times \left(\boldsymbol{\omega} \times \Delta\mathbf{r}_i + \mathbf{V}_\mathbf{R}\right) \\
    &= \left(-\sum_{i=1}^n m_i\,\Delta\mathbf{r}_i \times \left(\Delta\mathbf{r}_i \times \boldsymbol{\omega}\right)\right) + \left(\sum_{i=1}^n m_i \,\Delta\mathbf{r}_i \times \mathbf{V}_\mathbf{R}\right),
\end{align}</math>
where the terms containing <math>\mathbf{V_R}</math> (<math>= \mathbf{C}</math>) sum to zero by the definition of [[center of mass]].

Then, the skew-symmetric matrix <math>[\Delta\mathbf{r}_i]</math> obtained from the relative position vector <math>\Delta\mathbf{r}_i = \mathbf{r}_i - \mathbf{C}</math>, can be used to define,
<math display="block">
  \mathbf{L} =
  \left(-\sum_{i=1}^n m_i \left[\Delta\mathbf{r}_i\right]^2\right)\boldsymbol{\omega} =
  \mathbf{I}_\mathbf{C} \boldsymbol{\omega},
</math>
where <math>\mathbf{I_C}</math> defined by
<math display="block">\mathbf{I}_\mathbf{C} = -\sum_{i=1}^n m_i \left[\Delta\mathbf{r}_i\right]^2,</math>
is the symmetric inertia matrix of the rigid system of particles measured relative to the center of mass <math>\mathbf{C}</math>.

=== 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.

=== Resultant torque ===
The inertia matrix appears in the application of Newton's second law to a rigid assembly of particles. The resultant torque on this system is,<ref name="Marion 1995"/><ref name="Kane"/>
<math display="block">\boldsymbol{\tau} = \sum_{i=1}^n \left(\mathbf{r_i} - \mathbf{R}\right) \times m_i\mathbf{a}_i,</math>
where <math>\mathbf{a}_i</math> is the acceleration of the particle <math>P_i</math>. The [[kinematics]] of a rigid body yields the formula for the acceleration of the particle <math>P_i</math> in terms of the position <math>\mathbf{R}</math> and acceleration <math>\mathbf{A}_\mathbf{R}</math> of the reference point, as well as the angular velocity vector <math>\boldsymbol{\omega}</math> and angular acceleration vector <math>\boldsymbol{\alpha}</math> of the rigid system as,
<math display="block">\mathbf{a}_i = \boldsymbol{\alpha} \times \left(\mathbf{r}_i - \mathbf{R}\right) + \boldsymbol{\omega} \times \left( \boldsymbol{\omega} \times \left(\mathbf{r}_i - \mathbf{R}\right) \right) + \mathbf{A}_\mathbf{R}.</math>

Use the center of mass <math>\mathbf{C}</math> as the reference point, and introduce the skew-symmetric matrix <math>\left[\Delta\mathbf{r}_i\right] = \left[\mathbf{r}_i - \mathbf{C}\right]</math> to represent the cross product <math>(\mathbf{r}_i - \mathbf{C}) \times</math>, to obtain
<math display="block">
  \boldsymbol{\tau} = \left(-\sum_{i=1}^n m_i\left[\Delta\mathbf{r}_i\right]^2\right)\boldsymbol{\alpha} +
    \boldsymbol{\omega} \times \left(-\sum_{i=1}^n m_i \left[\Delta\mathbf{r}_i\right]^2\right)\boldsymbol{\omega}
</math>

The calculation uses the identity
<math display="block">
  \Delta\mathbf{r}_i \times \left(\boldsymbol{\omega} \times \left(\boldsymbol{\omega} \times \Delta\mathbf{r}_i\right)\right) + 
    \boldsymbol{\omega} \times \left(\left(\boldsymbol{\omega} \times \Delta\mathbf{r}_i\right) \times \Delta\mathbf{r}_i\right)
 = 0,
</math>
obtained from the [[Jacobi identity]] for the triple [[cross product]] as shown in the proof below:

{{math proof|proof=
<math display="block">\begin{align}
  \boldsymbol{\tau}
    &= \sum_{i=1}^n (\mathbf{r_i} - \mathbf{R})\times (m_i\mathbf{a}_i) \\
    &= \sum_{i=1}^n \Delta\mathbf{r}_i\times (m_i\mathbf{a}_i) \\
    &= \sum_{i=1}^n m_i [\Delta\mathbf{r}_i\times \mathbf{a}_i]\;\ldots\text{ cross-product scalar multiplication} \\
    &= \sum_{i=1}^n m_i [\Delta\mathbf{r}_i\times (\mathbf{a}_{\text{tangential},i} + \mathbf{a}_{\text{centripetal},i} + \mathbf{A}_\mathbf{R})] \\
    &= \sum_{i=1}^n m_i [\Delta\mathbf{r}_i\times (\mathbf{a}_{\text{tangential},i} + \mathbf{a}_{\text{centripetal},i} + 0)] \\
\end{align}</math>
In the last statement, <math>\mathbf{A}_\mathbf{R} = 0</math> because <math>\mathbf{R}</math> is either at rest or moving at a constant velocity but not accelerated, or the origin of the fixed (world) coordinate reference system is placed at the center of mass <math>\mathbf{C}</math>. And distributing the cross product over the sum, we get
<math display="block">\begin{align}
\boldsymbol{\tau} &= \sum_{i=1}^n m_i [\Delta\mathbf{r}_i\times \mathbf{a}_{\text{tangential},i} + \Delta\mathbf{r}_i\times \mathbf{a}_{\text{centripetal},i}] \\
\boldsymbol{\tau}  &= \sum_{i=1}^n m_i [\Delta\mathbf{r}_i\times (\boldsymbol{\alpha} \times \Delta\mathbf{r}_i) + \Delta\mathbf{r}_i\times (\boldsymbol{\omega} \times \mathbf{v}_{\text{tangential},i})] \\
\boldsymbol{\tau} &= \sum_{i=1}^n m_i [\Delta\mathbf{r}_i\times (\boldsymbol{\alpha} \times \Delta\mathbf{r}_i) + \Delta\mathbf{r}_i \times (\boldsymbol{\omega} \times (\boldsymbol{\omega} \times \Delta\mathbf{r}_i))]
\end{align}</math>

Then, the following [[Jacobi identity]] is used on the last term:
<math display="block">\begin{align}
  0 &=
    \Delta\mathbf{r}_i\times (\boldsymbol\omega \times(\boldsymbol\omega\times \Delta\mathbf{r}_i)) + \boldsymbol\omega\times((\boldsymbol\omega\times\Delta\mathbf{r}_i)\times\Delta\mathbf{r}_i) + (\boldsymbol\omega\times\Delta\mathbf{r}_i)\times(\Delta\mathbf{r}_i\times\boldsymbol\omega)\\
    &= \Delta\mathbf{r}_i\times (\boldsymbol\omega\times(\boldsymbol\omega\times\Delta\mathbf{r}_i)) + \boldsymbol\omega\times((\boldsymbol\omega\times\Delta\mathbf{r}_i)\times\Delta\mathbf{r}_i) + (\boldsymbol\omega\times\Delta\mathbf{r}_i)\times -(\boldsymbol\omega\times\Delta\mathbf{r}_i)\;\ldots\text{ cross-product anticommutativity} \\
    &= \Delta\mathbf{r}_i \times (\boldsymbol\omega\times(\boldsymbol\omega\times\Delta\mathbf{r}_i)) + \boldsymbol\omega\times((\boldsymbol\omega\times\Delta\mathbf{r}_i)\times\Delta\mathbf{r}_i) + -[(\boldsymbol\omega\times\Delta\mathbf{r}_i)\times(\boldsymbol\omega\times\Delta\mathbf{r}_i)]\;\ldots\text{ cross-product scalar multiplication} \\
    &= \Delta\mathbf{r}_i\times (\boldsymbol\omega\times(\boldsymbol\omega\times\Delta\mathbf{r}_i)) + \boldsymbol\omega\times((\boldsymbol\omega\times\Delta\mathbf{r}_i)\times\Delta\mathbf{r}_i) + -[0]\;\ldots\text{ self cross-product} \\
  0 &= \Delta\mathbf{r}_i\times (\boldsymbol\omega\times(\boldsymbol\omega\times\Delta\mathbf{r}_i)) + \boldsymbol\omega\times((\boldsymbol\omega\times\Delta\mathbf{r}_i)\times\Delta\mathbf{r}_i)
\end{align}</math>

The result of applying [[Jacobi identity]] can then be continued as follows:
<math display="block">\begin{align}
  \Delta\mathbf{r}_i\times (\boldsymbol\omega\times(\boldsymbol\omega\times\Delta\mathbf{r}_i))
    &= -[\boldsymbol\omega\times((\boldsymbol\omega\times\Delta\mathbf{r}_i)\times\Delta\mathbf{r}_i)] \\
    &= -[(\boldsymbol\omega\times\Delta\mathbf{r}_i)(\boldsymbol\omega\cdot\Delta\mathbf{r}_i) - \Delta\mathbf{r}_i(\boldsymbol\omega\cdot(\boldsymbol\omega\times\Delta\mathbf{r}_i))]\;\ldots\text{ vector triple product} \\
    &= -[(\boldsymbol\omega\times\Delta\mathbf{r}_i)(\boldsymbol\omega\cdot\Delta\mathbf{r}_i) - \Delta\mathbf{r}_i(\Delta\mathbf{r}_i\cdot(\boldsymbol\omega\times\boldsymbol\omega))]\;\ldots\text{ scalar triple product} \\
    &= -[(\boldsymbol\omega\times\Delta\mathbf{r}_i)(\boldsymbol\omega\cdot\Delta\mathbf{r}_i) - \Delta\mathbf{r}_i(\Delta\mathbf{r}_i\cdot(0))]\;\ldots\text{ self cross-product} \\
    &= -[(\boldsymbol\omega\times\Delta\mathbf{r}_i)(\boldsymbol\omega\cdot\Delta\mathbf{r}_i)] \\
    &= -[\boldsymbol\omega\times(\Delta\mathbf{r}_i (\boldsymbol\omega\cdot\Delta\mathbf{r}_i))]\;\ldots\text{ cross-product scalar multiplication} \\
    &= \boldsymbol\omega\times -(\Delta\mathbf{r}_i (\boldsymbol\omega\cdot\Delta\mathbf{r}_i))\;\ldots\text{ cross-product scalar multiplication} \\
  \Delta\mathbf{r}_i\times (\boldsymbol\omega\times(\boldsymbol\omega\times\Delta\mathbf{r}_i))
    &= \boldsymbol\omega\times -(\Delta\mathbf{r}_i (\Delta\mathbf{r}_i \cdot \boldsymbol\omega))\;\ldots\text{ dot-product commutativity} \\
\end{align}
</math>

The final result can then be substituted to the main proof as follows:
<math display="block">\begin{align}
  \boldsymbol\tau
    &= \sum_{i=1}^n m_i [\Delta\mathbf{r}_i\times (\boldsymbol\alpha\times\Delta\mathbf{r}_i) + \Delta\mathbf{r}_i\times (\boldsymbol\omega\times(\boldsymbol\omega\times\Delta\mathbf{r}_i))] \\
    &= \sum_{i=1}^n m_i [\Delta\mathbf{r}_i\times (\boldsymbol\alpha\times\Delta\mathbf{r}_i) + \boldsymbol\omega\times -(\Delta\mathbf{r}_i (\Delta\mathbf{r}_i \cdot \boldsymbol\omega))] \\
    &= \sum_{i=1}^n m_i [\Delta\mathbf{r}_i\times (\boldsymbol\alpha\times\Delta\mathbf{r}_i) + \boldsymbol\omega\times \{0 - \Delta\mathbf{r}_i (\Delta\mathbf{r}_i \cdot \boldsymbol\omega)\}] \\
    &= \sum_{i=1}^n m_i [\Delta\mathbf{r}_i\times (\boldsymbol\alpha\times\Delta\mathbf{r}_i) + \boldsymbol\omega\times \{[\boldsymbol\omega(\Delta\mathbf{r}_i\cdot\Delta\mathbf{r}_i) - \boldsymbol\omega(\Delta\mathbf{r}_i\cdot\Delta\mathbf{r}_i)] - \Delta\mathbf{r}_i (\Delta\mathbf{r}_i \cdot \boldsymbol\omega)\}]\;\ldots\;\boldsymbol\omega(\Delta\mathbf{r}_i\cdot\Delta\mathbf{r}_i) - \boldsymbol\omega(\Delta\mathbf{r}_i\cdot\Delta\mathbf{r}_i) = 0 \\
    &= \sum_{i=1}^n m_i [\Delta\mathbf{r}_i\times (\boldsymbol\alpha\times\Delta\mathbf{r}_i) + \boldsymbol\omega\times \{[\boldsymbol\omega(\Delta\mathbf{r}_i\cdot\Delta\mathbf{r}_i) - \Delta\mathbf{r}_i (\Delta\mathbf{r}_i \cdot \boldsymbol\omega)] - \boldsymbol\omega(\Delta\mathbf{r}_i\cdot\Delta\mathbf{r}_i)\}]\;\ldots\text{ addition associativity} \\
\end{align}</math>
<math display="block">\begin{align}
  \boldsymbol{\tau}
    &= \sum_{i=1}^n m_i [\Delta\mathbf{r}_i\times (\boldsymbol\alpha\times\Delta\mathbf{r}_i) + \boldsymbol\omega\times \{\boldsymbol\omega(\Delta\mathbf{r}_i\cdot\Delta\mathbf{r}_i) - \Delta\mathbf{r}_i (\Delta\mathbf{r}_i \cdot \boldsymbol\omega)\} - \boldsymbol\omega\times\boldsymbol\omega(\Delta\mathbf{r}_i\cdot\Delta\mathbf{r}_i)]\;\ldots\text{ cross-product distributivity over addition} \\
    &= \sum_{i=1}^n m_i [\Delta\mathbf{r}_i\times (\boldsymbol\alpha\times\Delta\mathbf{r}_i) + \boldsymbol\omega\times \{\boldsymbol\omega(\Delta\mathbf{r}_i\cdot\Delta\mathbf{r}_i) - \Delta\mathbf{r}_i (\Delta\mathbf{r}_i \cdot \boldsymbol\omega)\} - (\Delta\mathbf{r}_i\cdot\Delta\mathbf{r}_i)(\boldsymbol\omega\times\boldsymbol\omega)]\;\ldots\text{ cross-product scalar multiplication} \\
    &= \sum_{i=1}^n m_i [\Delta\mathbf{r}_i\times (\boldsymbol\alpha\times\Delta\mathbf{r}_i) + \boldsymbol\omega\times \{\boldsymbol\omega(\Delta\mathbf{r}_i\cdot\Delta\mathbf{r}_i) - \Delta\mathbf{r}_i (\Delta\mathbf{r}_i \cdot \boldsymbol\omega)\} - (\Delta\mathbf{r}_i\cdot\Delta\mathbf{r}_i)(0)]\;\ldots\text{ self cross-product} \\
    &= \sum_{i=1}^n m_i [\Delta\mathbf{r}_i\times (\boldsymbol\alpha\times\Delta\mathbf{r}_i) + \boldsymbol\omega\times \{\boldsymbol\omega(\Delta\mathbf{r}_i\cdot\Delta\mathbf{r}_i) - \Delta\mathbf{r}_i (\Delta\mathbf{r}_i \cdot \boldsymbol\omega)\}] \\
    &= \sum_{i=1}^n m_i [\Delta\mathbf{r}_i\times (\boldsymbol\alpha\times\Delta\mathbf{r}_i) + \boldsymbol\omega\times \{\Delta\mathbf{r}_i \times (\boldsymbol\omega \times \Delta\mathbf{r}_i)\}]\;\ldots\text{ vector triple product} \\
    &= \sum_{i=1}^n m_i [\Delta\mathbf{r}_i\times -(\Delta\mathbf{r}_i \times \boldsymbol\alpha) + \boldsymbol\omega\times \{\Delta\mathbf{r}_i \times -(\Delta\mathbf{r}_i \times \boldsymbol\omega)\}]\;\ldots\text{ cross-product anticommutativity} \\
    &= -\sum_{i=1}^n m_i [\Delta\mathbf{r}_i\times (\Delta\mathbf{r}_i \times \boldsymbol\alpha) + \boldsymbol\omega\times \{\Delta\mathbf{r}_i \times (\Delta\mathbf{r}_i \times \boldsymbol\omega)\}]\;\ldots\text{ cross-product scalar multiplication} \\
    &= -\sum_{i=1}^n m_i [\Delta\mathbf{r}_i\times (\Delta\mathbf{r}_i \times \boldsymbol\alpha)] + -\sum_{i=1}^n m_i [\boldsymbol\omega\times \{\Delta\mathbf{r}_i \times (\Delta\mathbf{r}_i \times \boldsymbol\omega)\}]\;\ldots\text{ summation distributivity} \\
\boldsymbol\tau &= -\sum_{i=1}^n m_i [\Delta\mathbf{r}_i\times (\Delta\mathbf{r}_i \times \boldsymbol\alpha)] + \boldsymbol\omega\times -\sum_{i=1}^n m_i [\Delta\mathbf{r}_i \times (\Delta\mathbf{r}_i \times \boldsymbol\omega)]\;\ldots\;\boldsymbol\omega\text{ is not characteristic of particle } P_i
\end{align}</math>

Notice that for any vector <math>\mathbf{u}</math>, the following holds:
<math display="block">\begin{align}
  -\sum_{i=1}^n m_i [\Delta\mathbf{r}_i\times (\Delta\mathbf{r}_i \times \mathbf{u})]
    &= -\sum_{i=1}^n m_i \left(\begin{bmatrix}
             0             & -\Delta r_{3,i} &  \Delta r_{2,i} \\
            \Delta r_{3,i} &               0 & -\Delta r_{1,i} \\
           -\Delta r_{2,i} &  \Delta r_{1,i} &               0
          \end{bmatrix} \left(\begin{bmatrix}
                          0 & -\Delta r_{3,i} &  \Delta r_{2,i} \\
             \Delta r_{3,i} &               0 & -\Delta r_{1,i} \\
            -\Delta r_{2,i} &  \Delta r_{1,i} & 0
          \end{bmatrix} \begin{bmatrix} u_1 \\ u_2 \\ u_3 \end{bmatrix}
        \right)\right)\;\ldots\text{ cross-product as matrix multiplication} \\[6pt]
    &= -\sum_{i=1}^n m_i \left(\begin{bmatrix}
          0 & -\Delta r_{3,i} & \Delta r_{2,i} \\
          \Delta r_{3,i} & 0 & -\Delta r_{1,i} \\
         -\Delta r_{2,i} & \Delta r_{1,i} & 0
       \end{bmatrix} \begin{bmatrix}
         -\Delta r_{3,i}\,u_2 + \Delta r_{2,i}\,u_3 \\
         +\Delta r_{3,i}\,u_1 - \Delta r_{1,i}\,u_3 \\
         -\Delta r_{2,i}\,u_1 + \Delta r_{1,i}\,u_2
       \end{bmatrix}\right) \\[6pt]
    &= -\sum_{i=1}^n m_i \begin{bmatrix}
         -\Delta r_{3,i}(+\Delta r_{3,i}\,u_1 - \Delta r_{1,i}\,u_3) + \Delta r_{2,i}(-\Delta r_{2,i}\,u_1 + \Delta r_{1,i}\,u_2) \\
         +\Delta r_{3,i}(-\Delta r_{3,i}\,u_2 + \Delta r_{2,i}\,u_3) - \Delta r_{1,i}(-\Delta r_{2,i}\,u_1 + \Delta r_{1,i}\,u_2) \\
         -\Delta r_{2,i}(-\Delta r_{3,i}\,u_2 + \Delta r_{2,i}\,u_3) + \Delta r_{1,i}(+\Delta r_{3,i}\,u_1 - \Delta r_{1,i}\,u_3)
       \end{bmatrix} \\[6pt]
    &= -\sum_{i=1}^n m_i \begin{bmatrix}
         -\Delta r_{3,i}^2\,u_1 + \Delta r_{1,i}\Delta r_{3,i}\,u_3 - \Delta r_{2,i}^2\,u_1 + \Delta r_{1,i}\Delta r_{2,i}\,u_2 \\
         -\Delta r_{3,i}^2\,u_2 + \Delta r_{2,i}\Delta r_{3,i}\,u_3 + \Delta r_{2,i}\Delta r_{1,i}\,u_1 - \Delta r_{1,i}^2\,u_2 \\
         +\Delta r_{3,i}\Delta r_{2,i}\,u_2 - \Delta r_{2,i}^2\,u_3 + \Delta r_{3,i}\Delta r_{1,i}\,u_1 - \Delta r_{1,i}^2\,u_3
       \end{bmatrix} \\[6pt]
    &= -\sum_{i=1}^n m_i \begin{bmatrix}
         -(\Delta r_{2,i}^2 + \Delta r_{3,i}^2)\,u_1 + \Delta r_{1,i}\Delta r_{2,i}\,u_2 + \Delta r_{1,i}\Delta r_{3,i}\,u_3 \\
         +\Delta r_{2,i}\Delta r_{1,i}\,u_1 - (\Delta r_{1,i}^2 + \Delta r_{3,i}^2)\,u_2 + \Delta r_{2,i}\Delta r_{3,i}\,u_3 \\
         +\Delta r_{3,i}\Delta r_{1,i}\,u_1 + \Delta r_{3,i}\Delta r_{2,i}\,u_2 - (\Delta r_{1,i}^2 + \Delta r_{2,i}^2)\,u_3
        \end{bmatrix} \\[6pt]
    &= -\sum_{i=1}^n m_i \begin{bmatrix}
         -(\Delta r_{2,i}^2 + \Delta r_{3,i}^2) &   \Delta r_{1,i}\Delta r_{2,i}         &   \Delta r_{1,i}\Delta r_{3,i} \\
           \Delta r_{2,i}\Delta r_{1,i}         & -(\Delta r_{1,i}^2 + \Delta r_{3,i}^2) &   \Delta r_{2,i}\Delta r_{3,i} \\
           \Delta r_{3,i}\Delta r_{1,i}         &   \Delta r_{3,i}\Delta r_{2,i}         & -(\Delta r_{1,i}^2 + \Delta r_{2,i}^2)
         \end{bmatrix} \begin{bmatrix} u_1 \\ u_2 \\ u_3 \end{bmatrix} \\
    &= -\sum_{i=1}^n m_i [\Delta r_i]^2 \mathbf{u} \\[6pt]
  -\sum_{i=1}^n m_i [\Delta\mathbf{r}_i \times (\Delta\mathbf{r}_i \times \mathbf{u})]
    &= \left(-\sum_{i=1}^n m_i [\Delta r_i]^2\right) \mathbf{u}\;\ldots\;\mathbf{u}\text{ is not characteristic of } P_i
\end{align}</math>

Finally, the result is used to complete the main proof as follows:
<math display="block">\begin{align}
  \boldsymbol{\tau}
    &= -\sum_{i=1}^n m_i [\Delta\mathbf{r}_i \times (\Delta\mathbf{r}_i \times \boldsymbol{\alpha})] + \boldsymbol{\omega} \times -\sum_{i=1}^n m_i \Delta\mathbf{r}_i \times (\Delta\mathbf{r}_i \times \boldsymbol{\omega})] \\
    &= \left(-\sum_{i=1}^n m_i [\Delta r_i]^2\right) \boldsymbol{\alpha} + \boldsymbol{\omega} \times \left(-\sum_{i=1}^n m_i [\Delta r_i]^2\right) \boldsymbol{\omega}
\end{align}</math>
}}

Thus, the resultant torque on the rigid system of particles is given by
<math display="block">\boldsymbol{\tau} = \mathbf{I}_\mathbf{C} \boldsymbol{\alpha} + \boldsymbol{\omega} \times  \mathbf{I}_\mathbf{C} \boldsymbol{\omega},</math>
where <math>\mathbf{I_C}</math> is the inertia matrix relative to the center of mass.

=== Parallel axis theorem ===
{{Main|Parallel axis theorem}}
The inertia matrix of a body depends on the choice of the reference point. There is a useful relationship between the inertia matrix relative to the center of mass <math>\mathbf{C}</math> and the inertia matrix relative to another point <math>\mathbf{R}</math>. This relationship is called the parallel axis theorem.<ref name="Marion 1995"/><ref name="Kane"/>

Consider the inertia matrix <math>\mathbf{I_R}</math> obtained for a rigid system of particles measured relative to a reference point <math>\mathbf{R}</math>, given by
<math display="block">\mathbf{I}_\mathbf{R} = -\sum_{i=1}^n m_i\left[\mathbf{r}_i - \mathbf{R}\right]^2.</math>

Let <math>\mathbf{C}</math> be the center of mass of the rigid system, then
<math display="block">\mathbf{R} = (\mathbf{R} - \mathbf{C}) + \mathbf{C} = \mathbf{d} + \mathbf{C},</math>
where <math>\mathbf{d}</math> is the vector from the center of mass <math>\mathbf{C}</math> to the reference point <math>\mathbf{R}</math>. Use this equation to compute the inertia matrix,
<math display="block">
  \mathbf{I}_\mathbf{R}
 = -\sum_{i=1}^n m_i[\mathbf{r}_i - \left(\mathbf{C} + \mathbf{d}\right)]^2
 = -\sum_{i=1}^n m_i[\left(\mathbf{r}_i - \mathbf{C}\right) - \mathbf{d}]^2.
</math>

Distribute over the cross product to obtain
<math display="block">
  \mathbf{I}_\mathbf{R} =
              - \left(\sum_{i=1}^n m_i [\mathbf{r}_i - \mathbf{C}]^2\right)
              + \left(\sum_{i=1}^n m_i [\mathbf{r}_i - \mathbf{C}]\right)[\mathbf{d}]
  + [\mathbf{d}]\left(\sum_{i=1}^n m_i [\mathbf{r}_i - \mathbf{C}]\right) 
              - \left(\sum_{i=1}^n m_i\right) [\mathbf{d}]^2.
</math>

The first term is the inertia matrix <math>\mathbf{I_C}</math> relative to the center of mass. The second and third terms are zero by definition of the center of mass <math>\mathbf{C}</math>. And the last term is the total mass of the system multiplied by the square of the skew-symmetric matrix <math>[\mathbf{d}]</math> constructed from <math>\mathbf{d}</math>.

The result is the parallel axis theorem,
<math display="block">\mathbf{I}_\mathbf{R} = \mathbf{I}_\mathbf{C} - M[\mathbf{d}]^2,</math>
where <math>\mathbf{d}</math> is the vector from the center of mass <math>\mathbf{C}</math> to the reference point <math>\mathbf{R}</math>.

'''Note on the minus sign''': By using the skew symmetric matrix of position vectors relative to the reference point, the inertia matrix of each particle has the form <math>-m\left[\mathbf{r}\right]^2</math>, which is similar to the <math>mr^2</math> that appears in planar movement. However, to make this to work out correctly a minus sign is needed. This minus sign can be absorbed into the term <math>m\left[\mathbf{r}\right]^\mathsf{T} \left[\mathbf{r}\right]</math>, if desired, by using the skew-symmetry property of <math>[\mathbf{r}]</math>.

=== Scalar moment of inertia in a plane ===
The scalar moment of inertia, <math>I_L</math>, of a body about a specified axis whose direction is specified by the unit vector <math>\mathbf{\hat{k}}</math> and passes through the body at a point <math>\mathbf{R}</math> is as follows:<ref name="Kane"/>
<math display="block">I_L
 = \mathbf{\hat{k}} \cdot \left(-\sum_{i=1}^N m_i \left[\Delta\mathbf{r}_i\right]^2 \right) \mathbf{\hat{k}}
 = \mathbf{\hat{k}} \cdot \mathbf{I}_\mathbf{R} \mathbf{\hat{k}}
 = \mathbf{\hat{k}}^\mathsf{T} \mathbf{I}_\mathbf{R} \mathbf{\hat{k}},</math>
where <math>\mathbf{I_R}</math> is the moment of inertia matrix of the system relative to the reference point <math>\mathbf{R}</math>, and <math>[\Delta\mathbf{r}_i]</math> is the skew symmetric matrix obtained from the vector <math>\Delta\mathbf{r}_i = \mathbf{r}_i - \mathbf{R}</math>.

This is derived as follows. Let a rigid assembly of <math>n</math> particles, <math>P_i, i = 1, \dots, n</math>, have coordinates <math>\mathbf{r}_i</math>. Choose <math>\mathbf{R}</math> as a reference point and compute the moment of inertia around a line L defined by the unit vector <math>\mathbf{\hat{k}}</math> through the reference point <math>\mathbf{R}</math>, <math>\mathbf{L}(t) = \mathbf{R} + t\mathbf{\hat{k}}</math>. The perpendicular vector from this line to the particle <math>P_i</math> is obtained from <math>\Delta\mathbf{r}_i</math> by removing the component that projects onto <math>\mathbf{\hat{k}}</math>.
<math display="block">
  \Delta\mathbf{r}_i^\perp =
  \Delta\mathbf{r}_i - \left(\mathbf{\hat{k}} \cdot \Delta\mathbf{r}_i\right)\mathbf{\hat{k}} =
  \left(\mathbf{E} - \mathbf{\hat{k}}\mathbf{\hat{k}}^\mathsf{T}\right) \Delta\mathbf{r}_i,
</math>
where <math>\mathbf{E}</math> is the identity matrix, so as to avoid confusion with the inertia matrix, and <math>\mathbf{\hat{k}}\mathbf{\hat{k}}^\mathsf{T}</math> is the outer product matrix formed from the unit vector <math>\mathbf{\hat{k}}</math> along the line <math>L</math>.

To relate this scalar moment of inertia to the inertia matrix of the body, introduce the skew-symmetric matrix <math>\left[\mathbf{\hat{k}}\right]</math> such that <math>\left[\mathbf{\hat{k}}\right]\mathbf{y} = \mathbf{\hat{k}} \times \mathbf{y}</math>, then we have the identity
<math display="block">
  -\left[\mathbf{\hat{k}}\right]^2 \equiv
   \left|\mathbf{\hat{k}}\right|^2\left(\mathbf{E} - \mathbf{\hat{k}}\mathbf{\hat{k}}^\mathsf{T}\right) =
   \mathbf{E} - \mathbf{\hat{k}}\mathbf{\hat{k}}^\mathsf{T},
</math>
noting that <math>\mathbf{\hat{k}}</math> is a unit vector.

The magnitude squared of the perpendicular vector is
<math display="block">\begin{align}
  \left|\Delta\mathbf{r}_i^\perp\right|^2
    &=  \left(-\left[\mathbf{\hat{k}}\right]^2 \Delta\mathbf{r}_i\right) \cdot \left(-\left[\mathbf{\hat{k}}\right]^2 \Delta\mathbf{r}_i\right) \\
    &=  \left(\mathbf{\hat{k}} \times \left(\mathbf{\hat{k}} \times \Delta\mathbf{r}_i\right)\right) \cdot
        \left(\mathbf{\hat{k}} \times \left(\mathbf{\hat{k}} \times \Delta\mathbf{r}_i\right)\right)
\end{align}</math>

The simplification of this equation uses the triple scalar product identity
<math display="block">
  \left(\mathbf{\hat{k}} \times \left(\mathbf{\hat{k}} \times \Delta\mathbf{r}_i\right)\right) \cdot
  \left(\mathbf{\hat{k}} \times \left(\mathbf{\hat{k}} \times \Delta\mathbf{r}_i\right)\right) \equiv
  \left(\left(\mathbf{\hat{k}} \times \left(\mathbf{\hat{k}} \times \Delta\mathbf{r}_i\right)\right) \times \mathbf{\hat{k}}\right) \cdot
  \left(\mathbf{\hat{k}} \times \Delta\mathbf{r}_i\right),
</math>
where the dot and the cross products have been interchanged. Exchanging products, and simplifying by noting that <math>\Delta\mathbf{r}_i</math> and <math>\mathbf{\hat{k}}</math> are orthogonal:
<math display="block">\begin{align}
      &\left(\mathbf{\hat{k}} \times \left(\mathbf{\hat{k}} \times \Delta\mathbf{r}_i\right)\right) \cdot
        \left(\mathbf{\hat{k}} \times \left(\mathbf{\hat{k}} \times \Delta\mathbf{r}_i\right)\right) \\
  ={} &\left(\left(\mathbf{\hat{k}} \times \left(\mathbf{\hat{k}} \times \Delta\mathbf{r}_i\right)\right) \times \mathbf{\hat{k}}\right) \cdot
         \left(\mathbf{\hat{k}} \times \Delta\mathbf{r}_i\right) \\
  ={} &\left(\mathbf{\hat{k}} \times \Delta\mathbf{r}_i\right) \cdot \left(-\Delta\mathbf{r}_i \times \mathbf{\hat{k}}\right) \\
  ={} &-\mathbf{\hat{k}} \cdot \left(\Delta\mathbf{r}_i \times \Delta\mathbf{r}_i \times \mathbf{\hat{k}}\right) \\
  ={} &-\mathbf{\hat{k}} \cdot \left[\Delta\mathbf{r}_i\right]^2 \mathbf{\hat{k}}.
\end{align}</math>

Thus, the moment of inertia around the line <math>L</math> through <math>\mathbf{R}</math> in the direction <math>\mathbf{\hat{k}}</math> is obtained from the calculation
<math display="block">\begin{align}
  I_L  &= \sum_{i=1}^N m_i \left|\Delta\mathbf{r}_i^\perp\right|^2 \\
       &= -\sum_{i=1}^N m_i \mathbf{\hat{k}} \cdot \left[\Delta\mathbf{r}_i\right]^2\mathbf{\hat{k}}
        = \mathbf{\hat{k}} \cdot \left(-\sum_{i=1}^N m_i \left[\Delta\mathbf{r}_i\right]^2 \right) \mathbf{\hat{k}} \\
       &= \mathbf{\hat{k}} \cdot \mathbf{I}_\mathbf{R} \mathbf{\hat{k}}
        = \mathbf{\hat{k}}^\mathsf{T} \mathbf{I}_\mathbf{R} \mathbf{\hat{k}},
\end{align}</math>
where <math>\mathbf{I_R}</math> is the moment of inertia matrix of the system relative to the reference point <math>\mathbf{R}</math>.

This shows that the inertia matrix can be used to calculate the moment of inertia of a body around any specified rotation axis in the body.