Example: Symmetric Top with no Torque

The primary application of this equation is motion with no torque since its hard to transform a torque into the body frame. We now consider a symmetric top, that is an object with $I^{(1)}=I^{(2)}\equiv I^{(12)}$. The equations are.

$$I^{(k)}\dot{\omega}_k + I^{(j)}\omega_j\omega_i\epsilon_{ijk} = 0$$

$$I^{(3)}\dot{\omega}_3 + I^{(2)}\omega_2\omega_1-I^{(1)}\omega_1\omega_2 = 0$$

$$I^{(3)}\dot{\omega}_3 + (I^{(2)}-I^{(1)})\omega_2\omega_1 = 0$$

$$\dot{\omega}_3 = 0$$

$$I^{(1)}\dot{\omega}_1 + (I^{(3)}-I^{(2)})\omega_3\omega_2 = 0$$

$$\dot{\omega}_1 + \left({I^{(3)}-I^{(2)}\over I^{(1)}}\omega_3\right)\omega_2 =0$$

$$I^{(2)}\dot{\omega}_2 + (I^{(1)}-I^{(3)})\omega_3\omega_1 = 0$$

$$\dot{\omega}_2 - \left({I^{(3)}-I^{(1)}\over I^{(2)}}\omega_3\right)\omega_1 =0$$

$$\Omega\equiv \left({I^{(3)}-I^{(1)}\over I^{(2)}}\omega_3\right)$$

$$\dot{\omega}_1 + \Omega \omega_2 =0$$

$$\dot{\omega}_2 - \Omega \omega_1 =0$$

We have simple coupled oscillators. With the right initial conditions the solution is

$$\omega_1(t)=A\cos(\Omega t) \qquad\qquad\qquad \omega_2(t)=A\sin(\Omega t)$$

$\vec{\omega}$ sweeps out a circle in the $xy$ plane, or a cone in 3D, in the Body system where we have solved the equations. The $z'$ axis is the symmetry axis for the object. $\vec{\omega}$ precesses about the symmetry axis.

Since there is no torque, $\vec{L}$ is constant in the inertial system. The angle between $\vec{\omega}$ and $\vec{L}$ is constant in the inertial system, since $T={1\over 2}\vec{\omega}\cdot\vec{L}$ is constant and the magnitude of $\omega$ is constant. So $\vec{\omega}$ precesses around $\vec{L}$ in the fixed system. We can choose $\hat{z}_I$ along $\vec{L}$ since it is constant. We can show that $\vec{\omega}$, $\vec{L}$, and $\hat{z}_B$ are in a plane. This can be shown by computing $\vec{L}\cdot(\vec{\omega}\times\hat{z}_B)$ and showing it is zero.

The rate of precession is $\Omega={I^{(3)}-I^{(12)}\over I^{(3)}}\omega_3$. This is small if the moments of inertia are nearly the same. Examples of this effect are Frisbees wobbling, a spiral pass in football, and maybe the oblate earth's precession. (The dominant term for the earth is gravity driven so we need to analyze it with an external torque.)