moment of inertia tensor

From Physics wiki

1 Angular momentum
2 Kinetic energy
3 Generalization
4 Transformation law
5 Parallel axis theorem
6 Perpendicular axis theorem
7 Stretch rule
8 See also
9 References

Angular momentum

We may imagine that a particle is rigidly connected to some central, fixed point (center of mass), and that $\mathbf{r}\,$ be measured from that point. Then $\hat{\mathbf{r}}\cdot\mathbf{v}= 0\,$ , and

$\mathbf{v} = \boldsymbol{\omega}\times\mathbf{r}\,$ ,

where $\boldsymbol{\omega}\,$ is the angular velocity pseudovector.

The angular momentum of the particle (with respect to the center of mass) is

$\mathbf{L}\,$	$= m \mathbf{r}\times\mathbf{v}\,$ ,
	$= m \mathbf{r}\times( \boldsymbol{\omega}\times\mathbf{r})\,$ ,
	$= m r^2 \boldsymbol{\omega} - m \mathbf{r} (\mathbf{r}\cdot \boldsymbol{\omega})\,$ (using the BAC-CAB identity).

In components

$L_i\,$	$= m \sum_{k} ( r^2 \delta_{ik} \omega_k - r_i r_k \omega_k )\,$ ,
	$= m \sum_{k} ( r^2 \delta_{ik} - r_i r_k )\omega_k\,$ ,
	$= \sum_{k} I_{ik} \omega_k\,$ ,

where $I_{ik}\,$ are the components of the moment of inertia tensor

$\mathbf{I} \equiv m \begin{pmatrix} y^2 + z^2 & - xy & -xz\\ -yx & x^2 + z^2 & -yz \\ -zx & -zy & x^2 + y^2 \\ \end{pmatrix}\,$ .

A rigid body consists of multiple particles rotating with the same angular velocity $\boldsymbol{\omega}\,$ . The angular momentum is then the sum of the angular momentum of each particle, so that

$\mathbf{L} = \mathbf{I}\,\boldsymbol{\omega}\,$ ,

where

$I_{ik} = \sum_{p=1}^n m_p r_p^2 \delta_{ik} - m_p r_{pi} r_{pk}\,$ ,

$\mathbf{I} \equiv \sum_{p=1}^n m_p \begin{pmatrix} y_p^2 + z_p^2 & - x_py_p & -x_pz_p\\ -y_px_p & x_p^2 + z_p^2 & -y_pz_p \\ -z_px_p & -z_py_p & x_p^2 + y_p^2 \\ \end{pmatrix}\,$ .

The diagonal elements of $\mathbf{I}\,$ are called moments of inertia while the off-diagonal elements are called products of inertia.

The angular momentum calculated thusfar represents the angular momentum due to the rotation of the body only, and does not take into account the motion of the center of mass. Thus, the total angular momentum is

$\mathbf{L}_{total}\,$	$= \sum_{p=1}^n m_p(\mathbf{R}_{com} + \mathbf{r}_p) \times (\mathbf{V}_{com} + \mathbf{v}_p)\,$
	$= M \mathbf{R}_{com}\times\mathbf{V}_{com} + \left(\sum_{p=1}^n m_p \mathbf{r}_p\right) \times \mathbf{V}_{com} + \mathbf{R}_{com} \times \frac{d}{dt}\left(\sum_{p=1}^n m_p \mathbf{r}_p\right) + \mathbf{L}\,$
	$= \mathbf{L}_{com} + \mathbf{I} \boldsymbol{\omega}\,$ ,

where we have used the fact that $\sum_{p=1}^n m_p \mathbf{r}_p\,$ is just the center of mass in the center of mass frame, i.e., $0\,$ .

Kinetic energy

Again, considering a single particle rigidly rotating about a fixed point, $\mathbf{v} = \boldsymbol{\omega}\times\mathbf{r}\,$ , the kinetic energy is

$T = \frac{1}{2} m v^2\,$	$= \frac{1}{2} m \sum_i \left( \sum_{jk} \epsilon_{ijk} \omega^j r^k \right) \left( \sum_{lm} \epsilon_{ilm} \omega^l r^m \right)\,$ , (where $\epsilon_{ijk}\,$ is the Levi-Civita symbol)
	$= \frac{1}{2} m \sum_{jklm} \left( \delta_{jl}\delta_{km} - \delta_{jm}\delta_{kl} \right)\left( \omega^j r^k \omega^l r^m \right)\,$ ,
	$= \frac{1}{2} m \sum_{jl} \omega^j \left( \delta_{jl} r^2 - r^j r^l \right) \omega^l\,$ ,

or,

$T = \frac{1}{2} \boldsymbol{\omega}^T \mathbf{I} \boldsymbol{\omega}\,$ ,

or, more generally,

$T = \frac{1}{2} M V_{com}^2 + \frac{1}{2} \boldsymbol{\omega}^T \mathbf{I} \boldsymbol{\omega}\,$ ,

Generalization

Consider calculating the moment of inertia about some other point on the body (which is not fixed) but instantaneously has position $\mathbf{a}\,$ with respect to the fixed point, i.e., $\mathbf{r}' = \mathbf{r} - \mathbf{a}\,$ . Then

$I'_{ik}\,$	$= \sum_{p=1}^n m_p \left(r_p - a\right)^2 \delta_{ik} - m_p (r_p - a)_i (r_p - a)_k\,$ ,
	$= I_{ik} - 2 \mathbf{a} \cdot \sum_{p=1}^n m_p \mathbf{r}_p \delta_{ik} + \sum_{p=1}^n m_p a^2 \delta_{ik} + a_i \sum_{p=1}^n m_p r_{pk} + a_k \sum_{p=1}^n m_p r_{pi} -a_i a_k \sum_{p=1}^n m_p\,$ ,
	$= I_{ik} + M ( a^2 \delta_{ik} - a_i a_k )\,$ .

Transformation law

Under a rotation, $\mathbf{r} \to \mathbf{r}' = \mathbf{O} \mathbf{r}\,$ , where $\mathbf{O}\,$ is some orthogonal matrix,

$\frac{\partial r^{\prime i}}{\partial r^j} = O_{ij}\,$ ,

so that $\boldsymbol{\omega} \to \boldsymbol{\omega}' = \mathbf{O}\boldsymbol{\omega}\,$ . Since the kinetic energy is a scalar, $\mathbf{I}\,$ must transform as

$\mathbf{I} \to \mathbf{I}' = \mathbf{O}\mathbf{I}\mathbf{O}^T\,$ ,

so that

$T' = \boldsymbol{\omega}'^T \mathbf{I}' \boldsymbol{\omega}' = \boldsymbol{\omega} (\mathbf{O}^T \mathbf{O})\mathbf{I}(\mathbf{O}^T \mathbf{O}) \boldsymbol{\omega} = T\,$ .

We may also write

$I'_{ij} = \sum_{kl} \frac{\partial r^{\prime i}}{\partial r^k} \frac{\partial r^{\prime l}}{\partial r^j} I_{kl}\,$ .

Suppose we want to know the moment of inertia about one of the axes of the rotated system. Then recall that the columns of $\mathbf{O}\,$ form the basis vectors in that system:

$\mathbf{O} = \begin{pmatrix} \hat\mathbf{x}' && \hat\mathbf{y}' && \hat\mathbf{z}'\\ \downarrow && \downarrow && \downarrow \\ \\ \end{pmatrix}\,$ .

Now, $I'_{zz} = \sum_{kl} O_{3k} I_{kl} O_{3l} = \hat\mathbf{z}^{\prime T} \mathbf{I}\hat\mathbf{z}'\,$ . Since our rotated system was arbitrary, we obtain the following result: