Editing Moment of inertia (section)

=== 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.