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Next: Euler's Equations Up: Rigid Body Rotation Previous: Matrix Eigenvalue Theory

Principal Axes of Rotation

We have seen that the moment of inertia tensor, $\tilde{\bf I}$, defined in Section 8.3, takes the form of a real symmetric three-dimensional matrix. It therefore follows, from the matrix theory that we have just reviewed, that the moment of inertia tensor possesses three mutually orthogonal eigenvectors which are associated with three real eigenvalues. Let the $i$th eigenvector (which can be normalized to be a unit vector) be denoted $\hat{\mbox{\boldmath$\omega$}}_i$, and the $i$th eigenvalue $\lambda_i$. It then follows that
\tilde{\bf I}\, \hat{\mbox{\boldmath$\omega$}}_i = \lambda_i\,\hat{\mbox{\boldmath$\omega$}}_i ,
\end{displaymath} (487)

for $i=1, 3$.

The directions of the three mutually orthogonal unit vectors $\hat{\mbox{\boldmath$\omega$}}_i$ define the three so-called principal axes of rotation of the rigid body under investigation. These axes are special because when the body rotates about one of them (i.e., when $\omega$ is parallel to one of them) the angular momentum vector ${\bf L}$ becomes parallel to the angular velocity vector $\omega$. This can be seen from a comparison of Equation (466) and Equation (487).

Suppose that we reorient our Cartesian coordinate axes so the they coincide with the mutually orthogonal principal axes of rotation. In this new reference frame, the eigenvectors of $\tilde{\bf I}$ are the unit vectors, ${\bf e}_x$, ${\bf e}_y$, and ${\bf e}_z$, and the eigenvalues are the moments of inertia about these axes, $I_{xx}$, $I_{yy}$, and $I_{zz}$, respectively. These latter quantities are referred to as the principal moments of inertia. Note that the products of inertia are all zero in the new reference frame. Hence, in this frame, the moment of inertia tensor takes the form of a diagonal matrix: i.e.,

\tilde{\bf I} =
\end{displaymath} (488)

Incidentally, it is easy to verify that ${\bf e}_x$, ${\bf e}_y$, and ${\bf e}_z$ are indeed the eigenvectors of the above matrix, with the eigenvalues $I_{xx}$, $I_{yy}$, and $I_{zz}$, respectively, and that ${\bf L} = \tilde{\bf I}\,\mbox{\boldmath$\omega$}$ is indeed parallel to $\omega$ whenever $\omega$ is directed along ${\bf e}_x$, ${\bf e}_y$, or ${\bf e}_z$.

When expressed in our new coordinate system, Equation (466) yields

{\bf L} = \left(I_{xx}\,\omega_x,\,I_{yy}\,\omega_y,I_{zz}\,\omega_z\right),
\end{displaymath} (489)

whereas Equation (471) reduces to
K = \frac{1}{2}\left(I_{xx}\,\omega_x^{\,2} + I_{yy}\,\omega_y^{\,2}
+ I_{zz}\,\omega_z^{\,2}\right).
\end{displaymath} (490)

In conclusion, there are many great simplifications to be had by choosing a coordinate system whose axes coincide with the principal axes of rotation of the rigid body under investigation. But how do we determine the directions of the principal axes in practice?

Well, in general, we have to solve the eigenvalue equation

\tilde{\bf I}\,\hat{\mbox{\boldmath$\omega$}} = \lambda\,\hat{\mbox{\boldmath$\omega$}},
\end{displaymath} (491)

...y}\right) = \left(\begin{array}{c}0\\ 0\\ 0\end{array}\right),
\end{displaymath} (492)

where $\hat{\mbox{\boldmath$\omega$}} = (\cos\alpha,\,\cos\,\beta,\,\cos\gamma)$, and $\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1$. Here, $\alpha$ is the angle the unit eigenvector subtends with the $x$-axis, $\beta$ the angle it subtends with the $y$-axis, and $\gamma$ the angle it subtends with the $z$-axis. Unfortunately, the analytic solution of the above matrix equation is generally quite difficult.

Fortunately, however, in many instances the rigid body under investigation possesses some kind of symmetry, so that at least one principal axis can be found by inspection. In this case, the other two principal axes can be determined as follows.

Suppose that the $z$-axis is known to be a principal axes (at the origin) in some coordinate system. It follows that the two products of inertia $I_{xz}$ and $I_{yz}$ are zero [otherwise, $(0,\,0,\,1)$ would not be an eigenvector in Equation (492)]. The other two principal axes must lie in the $x$-$y$ plane: i.e., $\cos\gamma=0$. It then follows that $\cos\beta=\sin\alpha$, since $\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1$. The first two rows in the matrix equation (492) thus reduce to

$\displaystyle (I_{xx}-\lambda)\,\cos\alpha + I_{xy}\,\sin\alpha$ $\textstyle =$ $\displaystyle 0,$ (493)
$\displaystyle I_{xy}\,\cos\alpha + (I_{yy}-\lambda)\,\sin\alpha$ $\textstyle =$ $\displaystyle 0.$ (494)

Eliminating $\lambda$ between the above two equations, we obtain
I_{xy} \,(1-\tan^2\alpha) = (I_{xx}-I_{yy})\,\tan\alpha.
\end{displaymath} (495)

But, $\tan( 2\,\alpha) \equiv 2\,\tan\alpha/(1-\tan^2\alpha)$. Hence, Equation (495) yields
\tan (2\,\alpha) = \frac{2\,I_{xy}}{I_{xx} - I_{yy}}.
\end{displaymath} (496)

There are two values of $\alpha$, lying between $-\pi/2$ and $\pi/2$, which satisfy the above equation. These specify the angles, $\alpha$, that the two mutually orthogonal principal axes lying in the $x$-$y$ plane make with the $x$-axis. Hence, we have now determined the directions of all three principal axes. Incidentally, once we have determined the orientation angle, $\alpha$, of a principal axis, we can then substitute back into Equation (493) to obtain the corresponding principal moment of inertia, $\lambda$.

Figure 29: A uniform rectangular laminar.
\epsfysize =2.5in

As an example, consider a uniform rectangular lamina of mass $m$ and sides $a$ and $b$ which lies in the $x$-$y$ plane, as shown in Figure 29. Suppose that the axis of rotation passes through the origin (i.e., through a corner of the lamina). Since $z=0$ throughout the lamina, it follows from Equations (464) and (465) that $I_{xz}= I_{yz} = 0$. Hence, the $z$-axis is a principal axis. After some straightforward integration, Equations (460)-(463) yield

$\displaystyle I_{xx}$ $\textstyle =$ $\displaystyle \frac{1}{3}\,m\,b^2,$ (497)
$\displaystyle I_{yy}$ $\textstyle =$ $\displaystyle \frac{1}{3}\,m\,a^2,$ (498)
$\displaystyle I_{xy}$ $\textstyle =$ $\displaystyle - \frac{1}{4}\,m\,a\,b.$ (499)

Thus, it follows from Equation (496) that
\alpha = \frac{1}{2}\tan^{-1}\left(\frac{3}{2}\frac{a\,b}{a^2-b^2}\right).
\end{displaymath} (500)

The above equation specifies the orientation of the two principal axes that lie in the $x$-$y$ plane. For the special case where $a=b$, we get $\alpha=\pi/4,\,
3\pi/4$: i.e., the two in-plane principal axes of a square lamina (at a corner) are parallel to the two diagonals of the lamina.

next up previous
Next: Euler's Equations Up: Rigid Body Rotation Previous: Matrix Eigenvalue Theory
Richard Fitzpatrick 2011-03-31