Intermediate Axis Theorem
This doesn’t really have anything to do with calculating moments of inertia, but it’s a pretty interesting phenomenon in physics. This also goes by several other names, such as the Dzhanibekov Effect and the Tennis Racket Theorem.
In this post, I will be using the dot notation for time derivatives. In particular,
\[\dot{x} \equiv \tfrac{d}{dt} x \qquad\qquad \ddot{x} \equiv \tfrac{d^2}{dt^2} x\]Explanation of the Phenomenon
Consider a thin rectangular slab where $a > b > c$. For the best results, we want $c \ll b$ and $a \geq 2b$ at least. Otherwise, the dimensions are too similar for the phenomenon to work. Let \(\b{\omega}_{x}\), \(\b{\omega}_{y}\), and \(\b{\omega}_{z}\) denote this object’s perpedicular rotation about the $x$, $y$, and $z$ axes, respectively.
Here’s the interesting and somewhat unintutive part. Rotation around \(\b{\omega}_{x}\) and \(\b{\omega}_{z}\) is stable, but rotation around \(\b{\omega}_{y}\) is unstable. In other words, it is easy to get this object to purely rotate around \(\b{\omega}_{x}\) or \(\b{\omega}_{z}\). However, it is practically impossible to obtain pure rotation around \(\b{\omega}_{y}\). Free rotation about \(\b{\omega}_{y}\) will almost always result in a twist about \(\b{\omega}_{x}\) (we will see why \(\b{\omega}_{x}\) and not \(\b{\omega}_{z}\)).
Real-Life Examples
This is an idealized object that can proxy many everyday objects. Below are a few examples.
Your Smart Phone: If you are daring enough to throw your smartphone in the air, you can see this phenomenon for yourself (maybe do it over your bed or a couch). Pure rotation about \(\b{\omega}_{x}\) and \(\b{\omega}_{z}\) is very easy. However, trying to rotate only around \(\b{\omega}_{y}\) always results in an unwanted twist about \(\b{\omega}_{x}\).
Tennis Rackets: If you don’t want to risk damaging your phone, a tennis racket is a good object to use. In fact, it is one of the objects that motivated the discovery of this phenomenon. This wikipedia article has a good video showing each of the axes. As hard as you may try, it is nearly impossible to freely rotate the racket about \(\b{\omega}_{y}\) without an extra twist about \(\b{\omega}_{x}\).
Skateboarding: In skateboarding, rotation about \(\b{\omega}_{x}\) is called a kickflip and rotation about \(\b{\omega}_{z}\) is called a pop shove it. These are some of the first ground tricks you learn since the board rotation is so stable. However, rotation about \(\b{\omega}_{y}\) is called the impossible since the rotation is inherently unstable and for a long time no one could do the trick. It took the genius of Rodney Mullen to invent a method, which involves guiding the rotation about \(\b{\omega}_{y}\) with your foot to prevent the unwanted rotation. Physics Girl has a really good video about this.
Intuitive Justification
The intermediate axis theorem is unintuitive. I don’t think anyone would infer this phenomenon without directly observing it. An attempt at an intuitive explanation is too long to include here. Veritasium has a really good video with his attempt that I recommend.
Mathematical Proof
Luckily, the mathematical justification is pretty straightforward.
Setup
Let $I_{x}$, $I_{y}$, and $I_{z}$ denote the rotational inertia about the $x$, $y$, and $z$ axes, respectively. In a later post where we derive the rotational inertia of a slab, we know the following.
\[I_{x} = \tfrac{1}{12}M(b^2 + c^2) \qquad I_{x} = \tfrac{1}{12}M(c^2 + a^2) \qquad I_{z} = \tfrac{1}{12}M(a^2 + b^2)\]Additionally, we know that the object is positioned on a principal axis. Therefore the inertia tensor is diagonal.
\[\m{I} = \begin{bmatrix} I_x & 0 & 0 \\ 0 & I_y & 0 \\ 0 & 0 & I_z \end{bmatrix}\]If we assume that $a > b > c$ (as drawn in the diagram), then
\[a > b > c \quad\implies\quad b^2 + c^2 < c^2 + a^2 < a^2 + b^2 \quad\implies\quad I_{x} < I_{y} < I_{z}\]Now, we need to get an expression for the total torque on the system. In classical physics, this is given by the following.
\[\b{\tau} = \frac{d \b{L}}{dt} = \frac{d (\m{I} \b{\omega})}{dt}\]The issue here is that as we rotate the object, the inertia tensor changes. So we actually can’t assume the inertia tensor is a constant. We have to convert to a reference frame where the inertia tensor is constant. Here’s where I’m going to do a bit of handwaving. Euler’s rotation equations do precisely this.
\[\b{\tau} = \m{I} \dot{\b{\omega}} + \b{\omega} \times (\m{I} \b{\omega})\]Unrolling the vector definitions gives the following.
\[\begin{align} &\tau_x = I_{x} \dot{\omega}_{x} + (I_{z} - I_{y}) \omega_{y} \omega_{z} \\[10pt] &\tau_y = I_{y} \dot{\omega}_{y} + (I_{x} - I_{z}) \omega_{z} \omega_{x} \\[10pt] &\tau_z = I_{z} \dot{\omega}_{z} + (I_{y} - I_{x}) \omega_{x} \omega_{y} \end{align}\]Now, we assume that the net torque on the system is $0$ (i.e. we have free rotation). Also, define the following constants. I have set these up so that each constant is positive based on the geometry of the slab.
\[k_{1} \equiv I_{z} - I_{y} > 0 \qquad k_{2} \equiv I_{z} - I_{x} > 0 \qquad k_{3} \equiv I_{y} - I_{x} > 0\]Rearranging gives the following.
\[\begin{align} &I_{x} \dot{\omega}_{x} = -k_{1} \omega_{y} \omega_{z} \\[10pt] &I_{y} \dot{\omega}_{y} = \ \ \ k_{2} \omega_{z} \omega_{x} \\[10pt] &I_{z} \dot{\omega}_{z} = -k_{3} \omega_{x} \omega_{y} \end{align}\]We can already see what makes the “intermediate” axis special. It’s the only term with a positive coefficient.
Stable Rotation Proof
Suppose that we are mostly rotating about \(\b{\omega}_x\) with some perturbation. This means that the contribution due to $\omega_y$ and $\omega_z$ is small (but not necessarily $0$). In particular, $0 \approx \omega_y, \omega_z \ll \omega_x$ and Therefore
\[I_{x} \approx 0 \quad\implies\quad I_{x} \dot{\omega}_{x} \approx 0 \quad\implies\quad \omega_{x} \ \text{is constant}\]Thus, if we neglect the time dependency of $\omega_{x}$.
\[\begin{align} &I_{y} \ddot{\omega}_{y} = \left ( k_{2} \omega_{x} \right ) \; \dot{\omega}_{z} \\[10pt] &I_{z} \ddot{\omega}_{z} = - \left ( k_{3} \omega_{x} \right ) \; \dot{\omega}_{y} \end{align}\]Substituting for the first time derivative and some rearranging.
\[\begin{align} &\ddot{\omega}_{y} = - \left ( \frac{ k_{2} k_{3} \omega_{x}^2 }{I_y I_z} \right ) \; \omega_{y} \\[10pt] &\ddot{\omega}_{z} = - \left ( \frac{ k_{2} k_{3} \omega_{x}^2 }{I_y I_z} \right ) \; \omega_{z} \end{align}\]Notice that every term in the brackets is positive. Even though $\omega_x$ may be either positive or negative, $\omega_x^2$ is always positive. Let $k_4^2 = \frac{ k_{2} k_{3} \omega_{x}^2 }{I_y I_z} > 0$.
\[\ddot{\omega}_y = - k_4^2 \omega_y \quad\implies\quad \omega_y(t) = \omega_y(0) \cos(k_4 t) + \tfrac{1}{k_4}\dot{\omega}_y(0) \sin(k_4 t) \\[10pt] \ddot{\omega}_z = - k_4^2 \omega_z \quad\implies\quad \omega_z(t) = \omega_z(0) \cos(k_4 t) + \tfrac{1}{k_4}\dot{\omega}_z(0) \sin(k_4 t)\]Therefore, since we assumed that $\omega_y$ and $\omega_z$ were initially small and we assumed no external torque (which implies $\dot{\omega}_y(0) = \dot{\omega}_z(0) = 0$), we see that \(\omega_y(t)\) and \(\omega_z(t)\) will remain very small. Thus, a small perturbation in the pure rotation about \(\b{\omega}_x\) will not have a large effect on the motion.
The same analysis can be done to show that rotation around \(\b{\omega}_z\) is also stable. The argument becomes different with rotation around \(\b{\omega}_y\).
Unstable Rotation Proof
The analysis is almost identical. Suppose that we are mostly rotating about \(\b{\omega}_y\) with some perturbation. This means that the contribution due to $\omega_x$ and $\omega_z$ is small (but not necessarily $0$). In particular, $0 \approx \omega_x, \omega_z \ll \omega_y$ and Therefore
\[I_{y} \approx 0 \quad\implies\quad I_{y} \dot{\omega}_{y} \approx 0 \quad\implies\quad \omega_{y} \ \text{is constant}\]Thus, if we neglect the time dependency of $\omega_{y}$.
\[\begin{align} &I_{x} \ddot{\omega}_{x} = - \left ( k_{1} \omega_{y} \right ) \; \dot{\omega}_{z} \\[10pt] &I_{z} \ddot{\omega}_{z} = - \left ( k_{3} \omega_{y} \right ) \; \dot{\omega}_{x} \end{align}\]Substituting for the first time derivative and some rearranging.
\[\begin{align} &\ddot{\omega}_{x} = \left ( \frac{ k_{1} k_{3} \omega_{y}^2 }{I_x I_z} \right ) \; \omega_{x} \\[10pt] &\ddot{\omega}_{z} = \left ( \frac{ k_{1} k_{3} \omega_{y}^2 }{I_x I_z} \right ) \; \omega_{z} \end{align}\]Notice that every term in the brackets is positive. Even though $\omega_y$ may be either positive or negative, $\omega_y^2$ is always positive. Let $k_5^2 = \frac{ k_{1} k_{3} \omega_{y}^2 }{I_x I_z} > 0$.
\[\ddot{\omega}_x = k_5^2 \omega_x \quad\implies\quad \omega_x(t) = \tfrac{1}{2}[\omega_x(0) + \tfrac{1}{k_5}\dot{\omega}_x(0)] e^{k_5 t} + \tfrac{1}{2}[\omega_x(0) - \tfrac{1}{k_5}\dot{\omega}_x(0)] e^{-k_5 t} \\[10pt] \ddot{\omega}_z = k_5 \omega_z \quad\implies\quad \omega_z(t) = \tfrac{1}{2}[\omega_z(0) + \tfrac{1}{k_5}\dot{\omega}_z(0)] e^{k_5 t} + \tfrac{1}{2}[\omega_z(0) - \tfrac{1}{k_5}\dot{\omega}_z(0)] e^{-k_5 t}\]Like before, we assumed that $\omega_x$ and $\omega_z$ were initially small and we assumed no external torque (which implies $\dot{\omega}_x(0) = \dot{\omega}_z(0) = 0$). Therefore, the $e^{-k_5 t}$ will dissipate and the $e^{k_5 t}$ term will dominate. Therefore, the equations become
\[\omega_x(t) \approx \tfrac{1}{2} \omega_x(0) e^{k_5 t} \\[10pt] \omega_z(t) \approx \tfrac{1}{2} \omega_z(0) e^{k_5 t}\]Therefore, even though $\omega_x$ and $\omega_z$ are small, the growing exponential term will eventually dominate, causing a large angular velocity in both the $\omega_x$ and $\omega_z$ directions. In practice, this will cause a twist about \(\b{\omega}_x\), since it has smaller rotational inertia than \(\b{\omega}_z\).
A Short Reflection
It’s interesting to backtrack through the math to find where the asymmetry of \(\b{\omega}_y\) comes from. It first showed up with these equations
\[\begin{align} &I_{x} \dot{\omega}_{x} = - (I_{z} - I_{y}) \omega_{y} \omega_{z} = -k_{1} \omega_{y} \omega_{z} \\[10pt] &I_{y} \dot{\omega}_{y} = \ \ \ (I_{z} - I_{x}) \omega_{z} \omega_{x} = \ \ \ k_{2} \omega_{z} \omega_{x} \\[10pt] &I_{z} \dot{\omega}_{z} = - (I_{y} - I_{x}) \omega_{x} \omega_{y} = -k_{3} \omega_{x} \omega_{y} \end{align}\]where $k_1$, $k_2$, and $k_3$ are positive constants. The fact that there is no negative in front of the second equation is ultimately why the analysis of \(\b{\omega}_y\) is different than that of \(\b{\omega}_x\) and \(\b{\omega}_z\).
The fact that the coefficient in the second equation is $(I_{z} - I_{x})$ rather than $(I_{x} - I_{z})$ ultimately comes from unrolling the cross product of $\b{\omega} \times (\m{I} \b{\omega})$. Written out a bit more.
\[\begin{bmatrix} \omega_x \\ \omega_y \\ \omega_z \end{bmatrix} \times \begin{bmatrix} I_x \omega_x \\ I_y \omega_y \\ I_z \omega_z \end{bmatrix}\]In a way, this grouping order is ultimately just a fundamental property of the cross product. This is why an intuitive justification for the intermediate axis theorem is so difficult to find. It is the result of a quirk of mathematics coupled with the perfect geometry to exploit it.