I am confused about principal axes of inertia.

Consider the compount pendulum in the picture, made of a rectangular plate. I oscillates about a horizontal axis $\hat{a}$ passing through $A$. The center of mass is in $C$.

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On my book it is claimed that, since $\hat{a}$ is parallel to $\hat{c}$ (the axis passing through the cm), which is a principal axis of inertia (it is a symmetry axis), $\hat{a}$ is a principal axis of inertia too. Hence the angular momentum of the compound pendulum, $\vec{L}$ is totally parallel to the axis of rotation $\hat{a}$.

I don't get this: $A$ is not an axis of simmetry, but it is a principal axis of inertia, just because it is parallel to another one? How can that be?


The answer is the parallel axis theorem. It states that the inertia tensor $I_{ij}$ will transform under a translation with vector $\vec{a}$ as $$ I = I^{(\text{cm})} + m \begin{pmatrix} a_2^2 + a_3^2 & - a_1a_2 & -a_1a_3 \\ -a_1a_2 & a_1^2+a_3^2 & -a_2a_3 \\ -a_1a_3 & -a_2a_3 & a_1^2+a_2^2\end{pmatrix}\\ $$ where $I$ is the new intertia tensor and $I^{(\text{cm})}$ is the intertia tensor with center of mass origin. Now assume that we are in the reference frame, where $I^{(\text{cm})}$ is diagonal. Then a translation along one of the principle axis, i.e $\vec{a} = a\; \vec{e}_i$ will conserve the diagonality because all mixing terms $a_i a_j$ with $i\neq j$ will vanish in the transformation matrix.

  • $\begingroup$ Thanks for the answer! Is this true for any traslation along one principle axis? In this post of mine physics.stackexchange.com/questions/246615/… I asked about a barbell with the axis of rotation not passing through the CM but still parallel to a principal axis. Nevertheless in that case the axis of rotation is not a principal axis, how can that be? $\endgroup$ – Sørën Apr 15 '16 at 16:47
  • $\begingroup$ I think your mistake there was that you did not take into account that $\vec{\omega} = \frac{\vec{r}\times\vec{v}}{r^2}$ is not invariant under translation. So you moved the pivot point to calculate $\vec{L}$ but used another point to calculate $\vec{\omega}$ $\endgroup$ – Jannick Apr 15 '16 at 17:14

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