Editing Moment of inertia (section)

== Inertia tensor ==

For the same object, different axes of rotation will have different moments of inertia about those axes. In general, the moments of inertia are not equal unless the object is symmetric about all axes. The '''moment of inertia [[tensor]]''' is a convenient way to summarize all moments of inertia of an object with one quantity. It may be calculated with respect to any point in space, although for practical purposes the center of mass is most commonly used.

=== Definition ===

For a rigid object of <math>N</math> point masses <math>m_{k}</math>, the moment of inertia [[tensor]] is given by
<math display="block">
\mathbf{I} = \begin{bmatrix}
I_{11} & I_{12} & I_{13} \\
I_{21} & I_{22} & I_{23} \\
I_{31} & I_{32} & I_{33}
\end{bmatrix}.
</math>

Its components are defined as
<math display="block">I_{ij} \ \stackrel{\mathrm{def}}{=}\  \sum_{k=1}^{N} m_{k}\left(\left\|\mathbf{r}_k\right\|^{2}\delta_{ij} - x_{i}^{(k)}x_{j}^{(k)}\right)</math>
where
* <math>i</math>, <math>j</math> is equal to 1, 2 or 3 for <math>x</math>, <math>y</math>, and <math>z</math>, respectively,
* <math>\mathbf{r}_k = \left(x_1^{(k)}, x_2^{(k)}, x_3^{(k)}\right)</math> is the vector to the point mass <math>m_k</math> from the point about which the tensor is calculated and
* <math>\delta_{ij}</math> is the [[Kronecker delta]].

Note that, by the definition, <math>\mathbf{I}</math> is a [[symmetric tensor]].

The diagonal elements are more succinctly written as
<math display="block">\begin{align}
  I_{xx} \ &\stackrel{\mathrm{def}}{=}\  \sum_{k=1}^{N} m_{k} \left(y_{k}^{2} + z_{k}^{2}\right), \\
  I_{yy} \ &\stackrel{\mathrm{def}}{=}\  \sum_{k=1}^{N} m_{k} \left(x_{k}^{2} + z_{k}^{2}\right), \\
  I_{zz} \ &\stackrel{\mathrm{def}}{=}\  \sum_{k=1}^{N} m_{k} \left(x_{k}^{2} + y_{k}^{2}\right),
\end{align}</math>
while the off-diagonal elements, also called the '''{{Interlanguage link|Produit d'inertie|fr|lt=products of inertia}}''', are
<math display="block">\begin{align}
  I_{xy} = I_{yx} \ &\stackrel{\mathrm{def}}{=}\  -\sum_{k=1}^{N} m_{k} x_{k} y_{k}, \\
  I_{xz} = I_{zx} \ &\stackrel{\mathrm{def}}{=}\  -\sum_{k=1}^{N} m_{k} x_{k} z_{k}, \\
  I_{yz} = I_{zy} \ &\stackrel{\mathrm{def}}{=}\  -\sum_{k=1}^{N} m_{k} y_{k} z_{k}.
\end{align}</math>

Here <math>I_{xx}</math> denotes the moment of inertia around the <math>x</math>-axis when the objects are rotated around the x-axis, <math>I_{xy}</math> denotes the moment of inertia around the <math>y</math>-axis when the objects are rotated around the <math>x</math>-axis, and so on.

These quantities can be generalized to an object with distributed mass, described by a mass density function, in a similar fashion to the scalar moment of inertia. One then has
<math display="block">\mathbf{I} = \iiint_V \rho(x,y,z) \left( \|\mathbf{r}\|^2 \mathbf{E}_{3} - \mathbf{r}\otimes \mathbf{r}\right)\, dx \, dy \, dz,</math>
where <math>\mathbf{r}\otimes \mathbf{r}</math> is their [[outer product]], '''E'''<sub>3</sub> is the 3×3 [[identity matrix]], and ''V'' is a region of space completely containing the object.

Alternatively it can also be written in terms of the [[Cross product#Conversion to matrix multiplication|angular momentum operator]] <math>[\mathbf r]\mathbf x = \mathbf r\times\mathbf x</math>:
<math display="block">\mathbf{I} = \iiint_V \rho(\mathbf{r}) [\mathbf r]^\textsf{T}[\mathbf r] \, dV = -\iiint_{Q} \rho(\mathbf{r}) [\mathbf r]^2 \, dV </math>

The inertia tensor can be used in the same way as the inertia matrix to compute the scalar moment of inertia about an arbitrary axis in the direction <math>\mathbf{n}</math>,
<math display="block">I_n = \mathbf{n}\cdot\mathbf{I}\cdot\mathbf{n},</math>
where the [[dot product]] is taken with the corresponding elements in the component tensors. A product of inertia term such as <math>I_{12}</math> is obtained by the computation
<math display="block">I_{12} = \mathbf{e}_1\cdot\mathbf{I}\cdot\mathbf{e}_2,</math>
and can be interpreted as the moment of inertia around the <math>x</math>-axis when the object rotates around the <math>y</math>-axis.

The components of tensors of degree two can be assembled into a matrix. For the inertia tensor this matrix is given by,
<math display="block">\begin{align}
\mathbf{I} &= \begin{bmatrix}
    I_{11} & I_{12} & I_{13} \\[1.8ex]
    I_{21} & I_{22} & I_{23} \\[1.8ex]
    I_{31} & I_{32} & I_{33}
  \end{bmatrix} = \begin{bmatrix}
    I_{xx} & I_{xy} & I_{xz} \\[1.8ex]
    I_{yx} & I_{yy} & I_{yz} \\[1.8ex]
    I_{zx} & I_{zy} & I_{zz}
  \end{bmatrix} \\[2ex]
  &= \sum_{k=1}^N \begin{bmatrix}
      m_{k} \left(y_{k}^2 + z_{k}^2\right) &
    - m_{k} x_{k} y_{k} &
    - m_{k} x_{k} z_{k} \\[1ex]
    - m_{k} x_{k} y_{k} &
      m_{k} \left(x_{k}^2 + z_{k}^2\right) &
    - m_{k} y_{k} z_{k} \\[1ex]
    - m_{k} x_{k} z_{k} &
    - m_{k} y_{k} z_{k} &
      m_{k} \left(x_{k}^2 + y_{k}^2\right)
  \end{bmatrix}.
\end{align}
</math>

It is common in rigid body mechanics to use notation that explicitly identifies the <math>x</math>, <math>y</math>, and <math>z</math>-axes, such as <math>I_{xx}</math> and <math>I_{xy}</math>, for the components of the inertia tensor.

=== Alternate inertia convention ===
There are some CAD and CAE applications such as SolidWorks, Unigraphics NX/Siemens NX and MSC Adams that use an alternate convention for the products of inertia. According to this convention, the minus sign is removed from the product of inertia formulas and instead inserted in the inertia matrix:
<math display="block">\begin{align}
  I_{xy} = I_{yx} \ &\stackrel{\mathrm{def}}{=}\  \sum_{k=1}^{N} m_{k} x_{k} y_{k}, \\
  I_{xz} = I_{zx} \ &\stackrel{\mathrm{def}}{=}\  \sum_{k=1}^{N} m_{k} x_{k} z_{k}, \\
  I_{yz} = I_{zy} \ &\stackrel{\mathrm{def}}{=}\  \sum_{k=1}^{N} m_{k} y_{k} z_{k}, \\[3pt]
  \mathbf{I} = \begin{bmatrix}
     I_{11} & I_{12} & I_{13} \\[1.8ex]
     I_{21} & I_{22} & I_{23} \\[1.8ex]
     I_{31} & I_{32} & I_{33}
  \end{bmatrix} &= \begin{bmatrix}
     I_{xx} & -I_{xy} & -I_{xz} \\[1.8ex]
     -I_{yx} & I_{yy} & -I_{yz} \\[1.8ex]
      -I_{zx} & -I_{zy} & I_{zz}
  \end{bmatrix} \\[1ex]
  &=
  \sum_{k=1}^{N} \begin{bmatrix}
    m_k \left(y_k^2 + z_k^2\right) & - m_k x_k y_k & - m_k x_k z_k \\[1ex]
    - m_k x_k y_k & m_k \left(x_k^2 + z_k^2\right) & - m_k y_k z_k \\[1ex]
    - m_k x_k z_k & - m_k y_k z_k & m_k \left(x_k^2 + y_k^2\right)
  \end{bmatrix}.
\end{align}</math>

==== Determine inertia convention (principal axes method) ====
If one has the inertia data <math>(I_{xx}, I_{yy}, I_{zz}, I_{xy}, I_{xz}, I_{yz})</math> without knowing which inertia convention that has been used, it can be determined if one also has the [[#Principal axes|principal axes]]. With the principal axes method, one makes inertia matrices from the following two assumptions:

# The standard inertia convention has been used <math>(I_{12} = I_{xy}, I_{13} = I_{xz}, I_{23} = I_{yz})</math>.  
# The alternate inertia convention has been used <math>(I_{12} = -I_{xy}, I_{13} = -I_{xz}, I_{23} = -I_{yz})</math>. 
Next, one calculates the eigenvectors for the two matrices. The matrix whose eigenvectors are parallel to the principal axes corresponds to the inertia convention that has been used.

=== Derivation of the tensor components ===

The distance <math>r</math> of a particle at <math>\mathbf{x}</math> from the axis of rotation passing through the origin in the <math>\mathbf{\hat{n}}</math> direction is <math>\left|\mathbf{x} - \left(\mathbf{x} \cdot \mathbf{\hat{n}}\right) \mathbf{\hat{n}}\right|</math>, where <math>\mathbf{\hat{n}}</math> is unit vector. The moment of inertia on the axis is
<math display="block">I = mr^2 =
  m\left(\mathbf{x} - \left(\mathbf{x}\cdot\mathbf{\hat{n}}\right) \mathbf{\hat{n}}\right)\cdot\left(\mathbf{x} - \left(\mathbf{x}\cdot\mathbf{\hat{n}}\right) \mathbf{\hat{n}}\right) =
  m\left(\mathbf{x}^2 - 2\mathbf{x}\left(\mathbf{x}\cdot\mathbf{\hat{n}}\right)\mathbf{\hat{n}} + \left(\mathbf{x}\cdot\mathbf{\hat{n}}\right)^2\mathbf{\hat{n}}^2\right) =
  m\left(\mathbf{x}^2 - \left(\mathbf{x}\cdot\mathbf{\hat{n}}\right)^2\right).
</math>

Rewrite the equation using [[Transpose|matrix transpose]]:
<math display="block">I =
  m\left(\mathbf{x}^\textsf{T}\mathbf{x} - \mathbf{\hat{n}}^\textsf{T}\mathbf{x}\mathbf{x}^\textsf{T}\mathbf{\hat{n}}\right) =
  m\cdot\mathbf{\hat{n}}^\textsf{T}\left(\mathbf{x}^\textsf{T}\mathbf{x}\cdot\mathbf{E_3} - \mathbf{x}\mathbf{x}^\textsf{T}\right)\mathbf{\hat{n}},
</math>
where '''E'''<sub>3</sub> is the 3×3 [[identity matrix]].

This leads to a tensor formula for the moment of inertia
<math display="block">I =
  m \begin{bmatrix} n_1 & n_2 & n_3 \end{bmatrix} \begin{bmatrix}
    y^2 + z^2 & -xy & -xz \\[0.5ex]
   -yx & x^2 + z^2 & -yz \\[0.5ex]
   -zx & -zy & x^2 + y^2
  \end{bmatrix} \begin{bmatrix}
    n_1 \\[0.7ex]
    n_2 \\[0.7ex]
    n_3
  \end{bmatrix}.
</math>

For multiple particles, we need only recall that the moment of inertia is additive in order to see that this formula is correct.

=== Inertia tensor of translation ===

{{Main|Parallel axis theorem#Tensor generalization}}

Let <math>\mathbf{I}_0</math> be the inertia tensor of a body calculated at its [[center of mass]], and <math>\mathbf{R}</math> be the displacement vector of the body. The inertia tensor of the translated body respect to its original center of mass is given by:
<math display="block">\mathbf{I} = \mathbf{I}_0 + m[(\mathbf{R}\cdot\mathbf{R})\mathbf{E}_3 - \mathbf{R}\otimes\mathbf{R}]</math>
where <math>m</math> is the body's mass, '''E'''<sub>3</sub> is the 3 × 3 identity matrix, and <math>\otimes</math> is the [[outer product]].

=== Inertia tensor of rotation ===

Let <math>\mathbf{R}</math> be the [[rotation matrix#In three dimensions|matrix]] that represents a body's rotation. The inertia tensor of the rotated body is given by:<ref>{{cite web |last1=David |first1=Baraff |title=Physically Based Modeling - Rigid Body Simulation |url=http://graphics.pixar.com/pbm2001/pdf/notesg.pdf |website=Pixar Graphics Technologies}}</ref>
<math display="block">\mathbf{I} = \mathbf{R}\mathbf{I_0}\mathbf{R}^\textsf{T}</math>