where \(I_1\), \(I_2\), and \(I_3\) are the moments of inertia around the rigid objects principal axes \(\hat{x}_1\), \(\hat{x}_2\), and \(\hat{x}_3\), which are chosen so that the moment of inertia tensor is diagonal. The \(\omega_i\) are the angular velocities and the \(\tau_i\) are the torques about the same axes.
If there are no torques acting on the object then the left hand side of \eqref{eq:eulereqn} is zero. and we can write the equations of motion as
Further more, if we consider the motion of an object that is symmetric about a certain axis then two of the moments of inertia will be equal. Let’s choose \(\hat{x}_1\) to lie in the direction of the axis of symmetry so that \(I_2=I_3\ne I_1\), in which case the first equation in \eqref{eq:eulereqntzero} is zero and \(\dot{\omega}_1=0\) so that the spin about the axis of symmetry of the object is a constant \(\omega_1\).
which is just motion in a circle. We’ve chosen the sign of \(\Omega_\mathrm{obj}\) so that it describes counter-clockwise rotation when viewed against axis \(\hat{x}_1\).Motion in a circle around \(\hat{x}_1\).
Now we have the components, the angular velocity vector is
\[\begin{equation}\vec{\omega} = \begin{pmatrix}
\omega_1 \\
A \cos(\Omega_\mathrm{obj} t+\phi) \\
A \sin(\Omega_\mathrm{obj} t+\phi)
\end{pmatrix}\end{equation}\]
where \(A\), \(\omega_1\), and \(\phi\) are constants set by the initial conditions. The interpretation of these equation is that the vector \(\vec{\omega}\)precesses around the axis of symmetry \(\hat{x}_1\) with and angular frequency \(\Omega_\mathrm{obj}\).
Now that we have the angular velocity vector, finding the angular momentum is trivial since we are using the objects principal axes:
A cylinder spinning about its symmetry axis. On the left, the observer is on the cylinder itself; on the right, the observer is not rotating with the cylinder. Note that \(I_1=1/2 MR^2\) and \(I_2=1/4 MR^2+1/12 M l^2\).
Though we have both \(\vec{\omega}\) and \(\vec{L}\), the equations of motion that we have are in terms of the principal axes of the object. These axes, \(\hat{x}_1(t)\), \(\hat{x}_2(t)\), and \(\hat{x}_3(t)\), are themselves complicated functions of time when viewed from a co-ordinate system not fixed to the object.
3 Motion Relative to “Lab”
So imagine now that we are observing an object which we have thrown spinning into the air. We have worked out what the equations of motion are as seen by a very giddy ant sitting on the object, but what are they relative to our observation point?
Since \(\vec{L}\) is a sum of \(\vec{\omega}\) and \(\hat{x}_1\), it must be the case that all three vectors lie on the same plane. The picture to have in mind for the body-fixed view is a fixed \(\hat{x}_1\) with \(\vec{L}\) and \(\vec{\omega}\) rotating around it at the same rate \(\Omega_\mathrm{obj}\). This can be easily verified with the simulation above.
Now, if there are no torques acting on the body then angular momentum must be conserved. This means that if we pick a co-ordinate system that is not rotating we will find that \(\vec{L}\) remains fixed in orientation and magnitude and \(\vec{\omega}\) and \(\hat{x}_1\) must precess around it. It’s tempting to think that from the point of view of an observer with fixed \(\vec{L}\) the vector \(\hat{x}_1\) would precess with frequency \(\Omega_\mathrm{obj}\) but this is not the case!
In the diagram below we’ve drawn a snapshot of the situation at an instance in time. Labelling the precession of \(\hat{x}_1\) around \(\vec{L}\) by the vector \(\Omega_\mathrm{lab}\) (parrallel to \(\vec{L}\)), the converse, the precession of \(\vec{L}\) around \(\hat{x}_1\) would be the component of \(\Omega_\mathrm{lab}\) in the direction \(\hat{x}_1\), i.e. \(\Omega_\mathrm{lab}\cos\theta\). This must be the same as \(\Omega_\mathrm{obj}\)plus\(\omega_1\). An observer sitting on the object is not directly aware of \(\omega_1\) but as soon as the observer is no longer rotating with the object we have to take it into account. Hence,
Take a moment to convince yourself that this formula makes sense—what happens when \(\theta=\pi/2\)? What happens when \(\omega_1=0\)?Decomposition of the precession angular velocity vector.
Rearranging and expanding \eqref{eq:precessionrelation},
where \(\vec{L} = L \hat{L}\), and we’ve made use of the fact that for any vector \(\vec{a}\), \(\vec{a}\times\vec{a}=0\). We can recognise the final expression as saying the vector \(\hat{x}_1\) rotates with angular velocity \(\Omega_\mathrm{lab}=L/I_2\) around \(\hat{L}\).
Now go back to the simulation and reconcile the left and right perspectives with the theory that has been developed.
