# Gyroscopic Motion

## Alexei Gilchrist

Equations of motion for a gyroscope are derived by considering the torques.

### 1 Orthogonal configuration

Imagine we have a rapidly spinning top in the configuration shown above. The spin $$\vec{\omega}_3$$ is initially directed along the $$x$$ axis, and we are assuming that this is large enough that the angular momentum $$\vec{L}$$ points in the same direction. The top is gently released so it is only held at the pivot. What will happen?

The simplistic intuition, that it will fall down and hit the table is not true as anyone who has played with a gyroscope will tell you. In fact, there is a wonderfully counter-intuitive interplay of forces that results in the top precessing around the $$z$$-axis instead.

When the top is released, the force $$\vec{F}$$ due to gravity will be directed straight down at the centre of mass. Say the centre of mass is at $$\vec{l}$$ from the pivot (and initially directed along $$x$$ axis), then the torque is

$$$\vec{\tau} = \vec{l} \times \vec{F} = \begin{pmatrix} l\\0\\0 \end{pmatrix}\times \begin{pmatrix} 0\\0\\ -F \end{pmatrix} = \begin{pmatrix} 0\\l F\\0 \end{pmatrix} = \begin{pmatrix} 0\\ \tau\\0 \end{pmatrix}.$$$
The torque is directed in the $$y$$ direction. This means that the change in $$L$$ after a small time, $$\Delta L$$, must also be in the $$y$$ direction since we have $$\vec{\tau}=d \vec{L}/dt$$. So in that small time $$\vec{L}$$ will shift to $$\vec{L}'$$, effectively rotating around the $$z$$-axis by a small angle $$\delta \theta$$. This rotation can be represented by the angular velocity $$\vec{\Omega}$$ which points along $$z$$ so that we have
$$$\frac{d\vec{L}}{dt} = \vec{\Omega}\times\vec{L} = \begin{pmatrix} 0\\0\\ \Omega \end{pmatrix}\times\begin{pmatrix} L\\0\\0 \end{pmatrix} = \begin{pmatrix} 0\\ \Omega L\\ 0 \end{pmatrix}.$$$

Equating $$y$$ components we have

$$$\Omega = \frac{\tau}{L},$$$
and $$\Omega$$ is called the precession frequency. So rather than fall down when it’s released, the top instead precesses around the $$z$$-axis!

We’ve set up everything so that the vectors are all orthogonal to one another, to further study the spinning top we need to analyse a more general configuration.

### 2 General configuration

Now consider that the top is at an angle $$\theta$$ from the vertical. Two angular rotations contribute to the motion of the top: the spin $$\vec{\omega}_3$$ and the precession $$\vec{\Omega}$$. Both of these vectors lie in the $$x$$-$$z$$ plane and so the angular momentum will also lie on the $$x$$-$$z$$ plane.

There are two ways to calculate the torque:

$$$\vec{\tau} = \vec{l}\times\vec{F} = \begin{pmatrix} l_x\\ 0\\ l_z \end{pmatrix}\times \begin{pmatrix} 0\\ 0\\ -mg \end{pmatrix} = \begin{pmatrix} 0\\ l_x m g\\ 0 \end{pmatrix} = \begin{pmatrix} 0\\ l m g \sin\theta\\ 0 \end{pmatrix}$$$
(the last step used $$l_x=l\sin\theta$$) and,
$$$\vec{\tau} = \frac{d\vec{L}}{dt} = \vec{\Omega}\times \vec{L} = \begin{pmatrix} 0\\ 0\\ \Omega \end{pmatrix}\times \begin{pmatrix} L_x\\ 0\\ L_z \end{pmatrix} = \begin{pmatrix} 0\\ L_x \Omega\\ 0 \end{pmatrix}$$$
so we have
$$$\label{eq:ytorque} l m g \sin\theta = L_x \Omega$$$
and we need to resolve $$L_x$$.

Referring to the figure above, both angular velocities $$\vec{\omega}_3$$ and $$\vec{\Omega}$$ will contribute to $$L_x$$. The former contributes an amount $$I_3 \omega_3$$ directed along the symmetry axis of the top, where $$I_3$$ is the moment of inertia along that principal axis.

Finding the contribution of $$\vec{\Omega}$$ is a little more involved. The moment of inertia along the two other principal axes is $$I_1$$ but since the top is rotating about the pivot not the centre of mass we need to displace this using the parallel axis theorem to $$I_1' = I_1+ml^2$$. The component of $$\vec{\Omega}$$ along the $$I_1$$ principal axis is $$\Omega \sin\theta$$ so that the angular momentum in that direction is $$I_1' \Omega\sin\theta$$, which contributes $$-I_1' \Omega\sin\theta\cos\theta$$ to $$L_x$$. Substituting these results into \eqref{eq:ytorque} yields

$$$\label{eq:gyroscope} I_1' \cos\theta\;\Omega^2 - I_3\omega_3\;\Omega + l m g = 0$$$

### 3 Solution to motion

Equation \eqref{eq:gyroscope} is a quadratic equation in $$\Omega$$ so we can immediately write down the solution

$$$\Omega = \frac{I_3\omega_3}{2I_1'\cos\theta}\pm\frac{1}{2 I_1'\cos\theta}\sqrt{I_3^2\omega_3^2-4 I_1'm g l\cos\theta}$$$
This equation doesn’t always give real values so there are combinations of parameters that are not allowed.

• If $$\theta\ge\pi/2$$ the solution is always real and any value of $$\omega_3$$ is allowed. This situation corresponds to the top tilted below the horizontal plane of the the pivot.
• If $$\theta<\pi/2$$ the top sits above the horizontal plane, and in order for the equation to yield real solutions we must have
$$$\omega_3 > \frac{2}{I_3}\sqrt{I_1'm g l\cos\theta} = \omega_\mathrm{min}.$$$
Note that there are two solutions for a given spin, a fast ($$+$$’) and slow ($$-$$’) precession.

### 4 General solution

In the treatment above we have tacitly assumed that steady precession was a valid motion of the top and it’s just as well it is. A more general treatment would also allow $$\theta$$ to change in time. The resulting equations of motion are

$\begin{gather*}I_1' \dot{\Omega}\sin\theta -I_3\omega_3\dot{\theta}+2I_1'\Omega\dot{\theta}\cos\theta = 0 \\ I_1'\ddot{\theta} = \left(I_1'\Omega^2\cos\theta - I_3\omega_3\Omega + mgl \right)\sin\theta,\end{gather*}$
and it can be easily verified that the condition $$\dot{\theta}=0$$ leads to \eqref{eq:gyroscope}. See heavy-symmetrical-top for details.

© Copyright 2021 Alexei Gilchrist