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
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
is always fixed at
in the space
(typically located at the center of mass). Therefore, the moment of
inertia tensor
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
as
Since rotational energy is defined as




We can show Eq.





where again
denotes the three-by-three identity matrix, and
which agrees with Eq.






For a collection of masses
located at
, we
simply sum over their masses to add up the moments of inertia:



Simple Example
Consider a mass at
. 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].
$](http://www.dsprelated.com/josimages_new/pasp/img2891.png)
![$ \underline{\omega}=[\omega,0,0]^T$](http://www.dsprelated.com/josimages_new/pasp/img2892.png)
![\begin{displaymath}
I \eqsp \underline{\tilde{\omega}}^T\mathbf{I}\,\underline{\...
...{array}{c} 1 \\ [2pt] 0 \\ [2pt] 0\end{array}\right]m \eqsp 0.
\end{displaymath}](http://www.dsprelated.com/josimages_new/pasp/img2893.png)
![$ \underline{\omega}=[0,1,0]^T$](http://www.dsprelated.com/josimages_new/pasp/img2894.png)
![$\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
$](http://www.dsprelated.com/josimages_new/pasp/img2895.png)
Example with Coupled Rotations
Now let the mass be located at
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*}](http://www.dsprelated.com/josimages_new/pasp/img2897.png)
We expect
to yield zero for the moment of inertia, and
sure enough
. Similarly, the vector angular
momentum is zero, since
.
For
, 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}](http://www.dsprelated.com/josimages_new/pasp/img2902.png)





Off-Diagonal Terms in Moment of Inertia Tensor
This all makes sense, but what about those off-diagonal terms in
? Consider the vector angular momentum (§B.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]
$](http://www.dsprelated.com/josimages_new/pasp/img2907.png)




![$ \underline{\omega}=[1,1,0]^T$](http://www.dsprelated.com/josimages_new/pasp/img2910.png)


Next Section:
Principal Axes of Rotation
Previous Section:
Angular Momentum Vector