next up previous
Next: Euler's equations Up: Rigid body rotation Previous: Rotational kinetic energy


Principal axes of rotation

We have seen that, for a general orientation of the Cartesian coordinate axes, the moment of inertia tensor, $ \textbf{\em I}$ , defined in Section 8.3, takes the form of a real symmetric $ 3\times 3$ matrix. It therefore follows, from the standard matrix theory discussed in Section A.11, 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 $ \skew{3}\hat{\mbox{\boldmath $\omega$}}_i$ , and the $ i$ th eigenvalue $ \lambda_i$ . It then follows that

$\displaystyle \textbf{\em I}\, \skew{3}\hat{\mbox{\boldmath$\omega$}}_i = \lambda_i\,\skew{3}\hat{\mbox{\boldmath$\omega$}}_i ,$ (8.19)

for $ i=1, 3$ .

The directions of the three mutually orthogonal unit vectors $ \skew{3}\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 (8.13) and Equation (8.19).

Suppose that we reorient our Cartesian coordinate axes so they coincide with the mutually orthogonal principal axes of rotation. In this new reference frame, the eigenvectors of $ \textbf{\em 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, $ {\cal I}_{xx}$ , $ {\cal I}_{yy}$ , and $ {\cal I}_{zz}$ , respectively. These latter quantities are referred to as the principal moments of inertia. 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:

$\displaystyle \textbf{\em I} = \left(\begin{array}{ccc} {\cal I}_{xx},&0,&0\\ 0,&{\cal I}_{yy},&0\\ 0,&0,&{\cal I}_{zz} \end{array}\right).$ (8.20)

Incidentally, it is easy to verify that $ {\bf e}_x$ , $ {\bf e}_y$ , and $ {\bf e}_z$ are indeed the eigenvectors of this matrix, with the eigenvalues $ {\cal I}_{xx}$ , $ {\cal I}_{yy}$ , and $ {\cal I}_{zz}$ , respectively, and that $ {\bf L} = \textbf{\em I}\,$$ \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 (8.13) yields

$\displaystyle {\bf L} = \left({\cal I}_{xx}\,\omega_x,\,{\cal I}_{yy}\,\omega_y,{\cal I}_{zz}\,\omega_z\right),$ (8.21)

whereas Equation (8.18) reduces to

$\displaystyle K = \frac{1}{2}\left({\cal I}_{xx}\,\omega_x^{\,2} + {\cal I}_{yy}\,\omega_y^{\,2} + {\cal I}_{zz}\,\omega_z^{\,2}\right).$ (8.22)

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.


next up previous
Next: Euler's equations Up: Rigid body rotation Previous: Rotational kinetic energy
Richard Fitzpatrick 2016-03-31