PC1672 Advanced dynamics

4.7 The symmetric top

If a body has an axis of symmetry ${\bf e}_3$ , then this is one of the principal axes of the moment-of-inertia tensor. The other two principal axes ${\bf e}_1$ and ${\bf e}_2$ are can be chosen to be any two orthogonal vectors in the plane perpendicular to ${\bf e}_3$ . The symmetry means that the principal moments about these axes are the same,

$\begin{displaymath}I_1=I_2\end{displaymath}$

This simplifies the equations of motion of a symmetric body. For free rotations, in the absence of any torque, Euler's equations become

$\displaystyle I_1\dot\omega_1+(I_3-I_1)\omega_2\omega_3$	$\textstyle =$	$\displaystyle 0$
$\displaystyle I_2\dot\omega_2+(I_1-I_3)\omega_3\omega_1$	$\textstyle =$	$\displaystyle 0$
$\displaystyle I_3\dot\omega_3$	$\textstyle =$	$\displaystyle 0$

From the third equation we can see that the rate of rotation about the symmetry axis is a constant.

The remaining two equations are thus a pair of coupled linear differential equations for $\omega_1$ and $\omega_2$ . If we define the constant

$\begin{displaymath}\Omega={I_3-I_1\over I_1}\omega_3\end{displaymath}$

then they can be written

$\displaystyle \dot\omega_1+\Omega\omega_2$	$\textstyle =$	$\displaystyle 0$
$\displaystyle \dot\omega_2-\Omega\omega_1$	$\textstyle =$	$\displaystyle 0$

These may be solved by various techniques. One possibility is to decouple the equations, by differentiating the first equation with respect to and using the second to substitute for $\dot\omega_2$ . This leaves us with a familiar-looking equation for $\omega_1$ :

$\begin{displaymath}\ddot\omega_1+\Omega^2\omega_1=0\end{displaymath}$

Alternatively we can introduce the complex ``coordinate'' $\eta=\omega_1+{\rm i}\omega_2$ which satisfies a single first-order equation.

By whatever method, we find that the solutions to these equations are

$\displaystyle \omega_1$	$\textstyle =$	$\displaystyle A\cos(\Omega t+\phi)$
$\displaystyle \omega_2$	$\textstyle =$	$\displaystyle A\sin(\Omega t+\phi)$

The magnitude of the angular velocity $\hbox{\boldmath {$\omega$ }}$ is a constant. For the body-fixed point of view, this vector precesses about the symmetry axis at a rate $\Omega$ .

The angular momentum of the body is

$\begin{displaymath}{\bf L}=(I_1\omega_1,I_1\omega_2,I_3\omega_3)\end{displaymath}$

This lies in the same plane as ${\bf e}_3$ and ${\bf L}$ and so it too precesses about ${\bf e}_3$ .

In contrast, for a space-fixed observer the angular momentum is constant since there is no torque. That observer sees $\hbox{\boldmath {$\omega$ }}$ and ${\bf e}_3$ as precessing about the fixed direction provided by ${\bf L}$ . As discussed in the textbooks, this can be pictured in terms of one cone rolling on another.

In the special case where ${\bf e}_3$ , $\hbox{\boldmath {$\omega$ }}$ and ${\bf L}$ are nearly parallel, we can find the rate of precession seen by the space-fixed observer relatively easily. With respect to the body-fixed axes, $\hbox{\boldmath {$\omega$ }}$ precesses about the symmetry axis at the rate $\Omega$ . For the space-fixed observer, these body-fixed axes are rotating at a rate $\omega\simeq\omega_3$ (about an axis very close to ${\bf e}_3$ ). Hence for that observer the net rate of precession is

$\begin{displaymath}\Omega'\simeq\omega+\Omega={I_3\over I_1}\omega\end{displaymath}$

For an object like a rugby ball with , the rate of precession is slower than the rate of rotation, while for something like a plate with the rate of precession is faster.

Textbook references

K&K 7.7 pages 317-320
F&C 9.5
B&O 7.8, 7.9
M 12.9
M&T 11.9

Mike Birse
17th May 2000