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Mass Moment of Inertia Tensor

As derived in the previous section, the moment of inertia tensor, in 3D Cartesian coordinates, is a three-by-three matrix $ \mathbf{I}$ that can be multiplied by any angular-velocity vector to produce the corresponding angular momentum vector for either a point mass or a rigid mass distribution. Note that the origin of the angular-velocity vector $ \underline{\omega}$ is always fixed at $ \underline{0}$ in the space (typically located at the center of mass). Therefore, the moment of inertia tensor $ \mathbf{I}$ is defined relative to that origin.


The moment of inertia tensor can similarly be used to compute the mass moment of inertia for any normalized angular velocity vector $ \underline{\tilde{\omega}}=\underline{\omega}/ \vert\vert\,\underline{\omega}\,\vert\vert $ as

$\displaystyle I \eqsp \underline{\tilde{\omega}}^T\mathbf{I}\,\underline{\tilde{\omega}}. \protect$ (B.22)

Since rotational energy is defined as $ (1/2)I\omega^2$ (see Eq.$ \,$(B.7)), multiplying Eq.$ \,$(B.22) by $ \omega^2$ gives the following expression for the rotational kinetic energy in terms of the moment of inertia tensor:

$\displaystyle E_R \eqsp \frac{1}{2}\, \underline{\omega}^T\mathbf{I}\,\underline{\omega} \protect$ (B.23)

We can show Eq.$ \,$(B.22) starting from Eq.$ \,$(B.14). For a point-mass $ m$ located at $ \underline{x}$, we have
\begin{eqnarray*}
I &=& m \left\Vert\,\underline{x}-(\underline{\tilde{\omega}}^...
...nderline{\tilde{\omega}}^T\mathbf{I}\,\underline{\tilde{\omega}}
\end{eqnarray*}
where again $ \mathbf{E}$ denotes the three-by-three identity matrix, and

$\displaystyle \mathbf{I}\eqsp m \left(\left\Vert\,\underline{x}\,\right\Vert^2\mathbf{E}-\underline{x}\underline{x}^T\right), \protect$ (B.24)

which agrees with Eq.$ \,$(B.20). Thus we have derived the moment of inertia $ I$ in terms of the moment of inertia tensor $ \mathbf{I}$ and the normalized angular velocity $ \underline{\tilde{\omega}}$ for a point-mass $ m$ at $ \underline{x}$. For a collection of $ N$ masses $ m_i$ located at $ \underline{x}_i\in{\bf R}^3$, we simply sum over their masses to add up the moments of inertia:

$\displaystyle \mathbf{I}\eqsp \sum_{i=1}^N m_i \left(\left\Vert\,\underline{x}_i\,\right\Vert^2\mathbf{E}
-\underline{x}_i\underline{x}_i^T\right)
$

Finally, for a continuous mass distribution, we integrate as usual:

$\displaystyle \mathbf{I}\eqsp \frac{1}{M}\int_V \rho(\underline{x}) \left(\left...
...underline{x}\,\right\Vert^2\mathbf{E}
-\underline{x}\underline{x}^T\right)\,dV
$

where $ M=\int_V\rho(\underline{x})dV$ is the total mass.

Simple Example

Consider a mass $ m$ at $ \underline{x}=[x,0,0]^T$. Then the mass moment of inertia tensor is

$\displaystyle \mathbf{I}\eqsp m \left(\left\Vert\,\underline{x}\,\right\Vert^2\...
...ay}{ccc}
0 & 0 & 0\\ [2pt]
0 & 1 & 0\\ [2pt]
0 & 0 & 1
\end{array}\right].
$

For the angular-velocity vector $ \underline{\omega}=[\omega,0,0]^T$, we obtain the moment of inertia

\begin{displaymath}
I \eqsp \underline{\tilde{\omega}}^T\mathbf{I}\,\underline{\...
...{array}{c} 1 \\ [2pt] 0 \\ [2pt] 0\end{array}\right]m \eqsp 0.
\end{displaymath}

This makes sense because the axis of rotation passes through the point mass, so the moment of inertia should be zero about that axis. On the other hand, if we look at $ \underline{\omega}=[0,1,0]^T$, we get

$\displaystyle I \eqsp \underline{\tilde{\omega}}^T\mathbf{I}\,\underline{\tilde...
...]
\left[\begin{array}{c} 0 \\ [2pt] 1 \\ [2pt] 0\end{array}\right] \eqsp m x^2
$

which is what we expected.

Example with Coupled Rotations

Now let the mass $ m$ be located at $ \underline{x}=[1,1,0]^T$ so that
\begin{eqnarray*}
\mathbf{I}&=& m \left(\left\Vert\,\underline{x}\,\right\Vert^2...
... & 0\\ [2pt]
-1 & 1 & 0\\ [2pt]
0 & 0 & 2
\end{array}\right].
\end{eqnarray*}
We expect $ \underline{\omega}=[1,1,0]$ to yield zero for the moment of inertia, and sure enough $ \underline{\tilde{\omega}}^T\mathbf{I}\,\underline{\tilde{\omega}}=0$. Similarly, the vector angular momentum is zero, since $ \mathbf{I}\,\underline{\omega}=\underline{0}$. For $ \underline{\omega}=[1,0,0]^T$, the result is

\begin{displaymath}
\mathbf{I}\eqsp
\begin{array}{r}\left[\begin{array}{ccc} 1 ...
...egin{array}{c} 1 \\ [2pt] 0 \\ [2pt] 0\end{array}\right]m = m,
\end{displaymath}

which makes sense because the distance from the axis $ \underline{e}_1$ to $ \underline{x}$ is one. The same result is obtained for rotation about $ \underline{\omega}=\underline{e}_2$. For $ \underline{\omega}=\underline{e}_3$, however, the result is $ I=2m = m \vert\vert\,\underline{x}\,\vert\vert ^2$, as expect.

Off-Diagonal Terms in Moment of Inertia Tensor

This all makes sense, but what about those $ -1$ off-diagonal terms in $ \mathbf{I}$? Consider the vector angular momentumB.4.14):

$\displaystyle \underline{L}\eqsp \mathbf{I}\,\underline{\omega}\eqsp
m\left[\b...
...begin{array}{c} \omega_1 \\ [2pt] \omega_2 \\ [2pt] \omega_3\end{array}\right]
$

We see that the off-diagonal terms $ I_{ij}$ correspond to a coupling of rotation about $ \underline{e}_i$ with rotation about $ \underline{e}_j$. That is, there is a component of moment-of-inertia $ I_{ij}$ that is contributed (or subtracted, as we saw above for $ \underline{\omega}=[1,1,0]^T$) when both $ \omega_i$ and $ \omega_j$ are nonzero. These cross-terms can be eliminated by diagonalizing the matrix [449],B.25as discussed further in the next section.
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Angular Momentum Vector