The Square Cat

Alexei Gilchrist

An extremely simplified model of a cat (just four masses!) is presented, following Putterman and Raz [Am. J. Phys. 76, 1040 (2008)]. This simple model is enough to show how a deformable body is able to rotate itself despite conserving angular momentum.

1 Introduction

One of the classic pieces of lore is that when a cat falls it will always land on its feet. This snippet becomes all the more surprising when you learn about conservation of angular momentum. How is it that a cat can re-orient itself in space in the few seconds it takes it to fall to the ground? The times scale is way too short for the cat to generate significant torque from the air—the cat hasn’t got some large surface like a wing for example. Yet somehow it can twist in a complicated way and, as if by magic, end up feet down.

Now a cat is way too complicated and too quickly becomes bad tempered so we’ll simplify the the problem right down to one we can study in detail and from first principles. We’ll follow the paper by Putterman and Raz [Am. J. Phys. 76, 1040 (2008)]. The “cat” is only four equal masses \(m\) connected by four light rods in a parallelogram as depicted below.

A square cat.

Two of the rods have a fixed length \(b\), and the cat can control the length of the other two rods \(a(t)\) and the angle between the rods \(\theta(t)\). We’ll take the cat’s orientation in space to be measured by the angle of one of its limbs relative to the horizontal: \(\phi\). Of course the cat can trivially change this angle by changing \(\theta\), so we’ll only consider the cat to have rotated if it can change \(\phi\) when measured with the same \(\theta\). The question of how the cat can rotate is now reduced to the following specific question: is there particular functions \(\theta(t)\) and \(a(t)\) that the cat can execute that will result in a change in \(\phi\) when the cat returns to the original shape? We’ll measure everything from the centre of mass since, unlike the cat, we’re not interested in the simple ballistic motion of it’s centre of mass.

2 Angular momentum

Since the angular momentum will be conserved for this system, deriving an expression for it is the first step. Taking \(\vec{r}_j\) to be the vectors from the centre of mass to the masses, and \(\vec{p}_j\) to be the momentum of each mass, then

\[\begin{equation}\label{eq:L} \vec{L} = \sum_{j=1}^{4} \vec{r}_j\times \dot{\vec{p}}_j = m \sum_{j=1}^{4} \vec{r}_j\times \dot{\vec{r}}_j\end{equation}\]

The best way to proceed is to express everything in terms of the vectors \(\vec{u}_1\) and \(\vec{u}_2\) as depicted in the diagram. These vectors are easy to determine, referring to the diagram,

\[\begin{align}\label{eq:ueqns} \vec{u}_1 &= \frac{b}{2}\begin{bmatrix} \cos\phi \\ \sin\phi\end{bmatrix} & \vec{u}_2 &= \frac{a}{2}\begin{bmatrix} \cos(\phi+\theta) \\ \sin(\phi+\theta)\end{bmatrix}.\end{align}\]
Differentiating we get the velocities and consequently the momenta,
\[\begin{align}\label{eq:udoteqns} \dot{\vec{u}}_1 &= \frac{b\dot{\phi}}{2}\begin{bmatrix} -\sin\phi \\ \cos\phi\end{bmatrix} & \dot{\vec{u}}_2 &= \frac{\dot{a}}{2}\begin{bmatrix} \cos(\phi+\theta) \\ \sin(\phi+\theta)\end{bmatrix}+ \frac{a}{2}(\dot{\phi}+\dot{\theta})\begin{bmatrix} -\sin(\phi+\theta) \\ \cos(\phi+\theta)\end{bmatrix}.\end{align}\]
The location of the masses can now be expressed in terms of these vectors
\[\begin{align*}\vec{r}_1 &= \vec{u}_1 + \vec{u}_2 \\ \vec{r}_2 &= \vec{u}_1 - \vec{u}_2 \\ \vec{r}_3 &= -\vec{u}_1 - \vec{u}_2 \\ \vec{r}_4 &= -\vec{u}_1 + \vec{u}_2,\end{align*}\]
and the velocities have the same relationships. Now it’s just a matter of substituting these values into \eqref{eq:L} and calculating the cross products to get
\[\begin{equation}\label{eq:Latheta} L = m \dot{\phi} (a^2+b^2) + m \dot{\theta}a^2.\end{equation}\]

Conservation of angular momentum requires that \(L\) be a constant, for simplicity we’ll take \(L=0\) which corresponds to the cat having no initial angular velocity. We can rearrange the equation as

\[\begin{equation}\label{eq:phidot} \dot{\phi} = - \dot{\theta} \frac{a^2}{a^2+b^2}.\end{equation}\]
This equation looks like it’s impossible to solve as the cat can arbitrarily choose how to modify \(a(t)\) and \(\theta(t)\). However, all we have to do is show that for some function \(a(t)\) and \(\theta(t)\) that returns to the original value, \eqref{eq:phidot} has a non-zero value and maintains a consistent sign.

3 Solution

Separating variables and integrating \eqref{eq:phidot} yields

\[\begin{equation}\label{eq:deltaphi} \Delta \phi = - \int \frac{a^2}{a^2+b^2} d\theta\end{equation}\]
Now there is a wonderful trick described by Putterman and Raz, which is to use Stokes’ theorem:
\[\begin{equation}\iint \limits_S (\vec{\nabla}\times \vec{F})\cdot \vec{dS} = \oint \limits_l \vec{F} \cdot \vec{dl},\end{equation}\]
which relates an integral over the surface \(S\) to an integral around a path \(l\) bounding the surface. To be able to use this result we need to rewrite \eqref{eq:deltaphi} as
\[\begin{equation*}\Delta \phi = - \oint \limits_l \vec{F} \cdot \vec{dl},\end{equation*}\]
where
\[\begin{align*}\vec{F} &= \begin{pmatrix} \frac{a^2}{a^2+b^2} \\ 0 \\ 0\end{pmatrix} & \vec{dl} &= \begin{pmatrix} d\theta \\ da \\ dz \end{pmatrix}.\end{align*}\]
In effect we are adding the term \(\int 0 \;da + \int 0 \;dz\) to \eqref{eq:deltaphi} and considering only motion that returns to the initial shape so it is a closed path in \((\theta,a,z)\) space.

So now applying Stokes’ theorem

\[\begin{equation}\Delta \phi = - \iint \limits_S \left[ \begin{pmatrix} \partial_\theta \\ \partial_a \\ \partial_z \end{pmatrix} \times \begin{pmatrix} \frac{a^2}{a^2+b^2} \\ 0 \\ 0 \end{pmatrix} \right]\cdot \begin{pmatrix} 0 \\ 0 \\ ds \end{pmatrix} = \iint B(\theta,a) \;da\;d\theta\end{equation}\]
where \(\partial_a\equiv \partial/\partial a\) etc, \(ds = da\;d\theta\), and
\[\begin{equation}B(\theta,a) = \frac{2 a b^2}{(a^2+b^2)^2}.\end{equation}\]

The key observation is that \(B(\theta, a) > 0\). This means \(\Delta \phi > 0\), in other words the cat rotated by \(\Delta \phi\) while all the time conserving angular momentum! How large is the rotation? Well, we want to choose a closed path \((\theta(t),a(t))\) so that it surrounds where \(B(\theta, a)\) is largest. Since the function is independent of \(\theta\) the larger its range the better, and for \(a\) we want to capture the peak to have most effect. This will depend on the value of \(b\). The demo below shows \(B(\theta, a)\).

Click figure to download the CDF demo.
The function \(B(\theta, a)\). The amount of rotation the cat can achieve will be the integral of this function over some area. The boundary of that area will define the motion \(\theta(t)\) and \(a(t)\) that the cat has to perform.

As you can see it’s possible to get a significant non-zero rotation. In any case, the cat could just repeat the action, each time acquiring another \(\Delta \phi\) of rotation till it can land on it’s “feet”.

4 Conclusion

Now a real cat is considerably more complicated than this toy model, but we have captured the essence of the trick. The key message is that a deformable body can undergo a series of transformations that will result in an overall rotation. This motion is not in any way in contradiction with the conservation of angular momentum. In fact the rotation arises because angular momentum is conserved.

Watching a cat land on its feet is a great reminder of the richness of phenomena that is possible from the simple laws we have uncovered governing Nature.

© Copyright 2022 Alexei Gilchrist