Editing Moment of inertia (section)

=== 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.