BETAA version of this post is also available as a YouTube video here.
Recently, while working on my upcoming post about the physics of playground swings, I was analyzing the mechanics of a double pendulum system similar to the following:
The motion of the lower limb can be thought of as a superposition of two rotational motions, ω0 around one axis and ω1 around another axis. It got me thinking about how to write down the kinetic energy of such a superposition.
Now in 2D/3D there is not really any such thing as rotation around two axes1. Two rotations can always be converted into a single axis rotation plus translation. Normally we choose the rotation axis to be around the center of mass of the object, then the kinetic energy is the sum of the rotational kinetic energy (around the center of mass) and translational kinetic energy (of the center of mass, including orbital motion). If we choose a different rotation axis, then this no longer holds and in general there is an extra term.
However, the translational motion of the center of mass can be rather inconvenient to compute in some cases like this one, where the point that is constrained to a circle is not the center of mass. Let us see if we can derive an expression directly in terms of the two rotations. To avoid limiting this analysis to a one dimensional rod, I will consider a more general three-dimensional object here:
We will use the following definitions:
- ω0 is the angular velocity vector for rotation around a first axis (with a direction normal to the plane of rotation per the usual convention),
- ω1 is the angular velocity vector for rotation around a second axis parallel to ω0 (note that this does not include the first rotation, so the object is actually rotating at ω0+ω1 in the rest frame),
- r1 is a vector from the first rotation axis to the second rotation axis, and perpendicular to ω0,
- I0 is the moment of inertia of the object around the first rotation axis,
- I1 is the moment of inertia of the object around the second rotation axis,
- dV is a volumetric element being integrated over,
- r is a vector from the first rotation axis to the volumetric element, and perpendicular to ω0,
- ρ is the mass density at the volumetric element, and
- vector quantities without boldface (e.g. ω0, ω1, r1, r) represent the magnitude of the corresponding vector.
At any point in our object, we can compute velocity as a sum of the two rotational velocities:
Now we can write down the kinetic energy from first principles, and massage the expressions until they are in a convenient form:
![\begin{align*} \text{K.E.} &= \tfrac{1}{2} \int \rho v^2 \mathop{dV} \\ &= \tfrac{1}{2} \int \rho [\vect{r} \times \vect{\omega_0} + (\vect{r}-\vect{r_1}) \times \vect{\omega_1}]^2 \mathop{dV} \\ &= \tfrac{1}{2} \int \rho (\vect{r} \times \vect{\omega_0})^2 \mathop{dV} + \tfrac{1}{2} \int \rho [(\vect{r}-\vect{r_1}) \times \vect{\omega_1}]^2 \mathop{dV} \\[-6pt] & \qquad\qquad + \int \rho (\vect{r} \times \vect{\omega_0}) \cdot [(\vect{r}-\vect{r_1}) \times \vect{\omega_1}] \mathop{dV} \\ & \qquad \text{(using } (\vect{A}+\vect{B})^2 = \vect{A}^2 + \vect{B}^2 + 2\vect{A}\cdot\vect{B} \text{)} \\[6pt] &= \tfrac{1}{2} \left[ \int \rho r^2 \mathop{dV} \right] \omega_0^2 + \tfrac{1}{2} \left[ \int \rho (\vect{r}-\vect{r_1})^2 \mathop{dV} \right] \omega_1^2 \\[-6pt] & \qquad\qquad + \left[ \int \rho \vect{r} \cdot (\vect{r}-\vect{r_1}) \mathop{dV} \right] \omega_0 \omega_1 \\ & \qquad \text{(using } (\vect{A}\times\vect{B})\cdot(\vect{C}\times\vect{D}) = (\vect{A}\cdot\vect{C})(\vect{B}\cdot\vect{D}) - (\vect{A}\cdot\vect{D})(\vect{B}\cdot\vect{C}) \text{,} \\ & \qquad \text{ and all } r \text{ perpendicular to all } \omega \text{)} \\[6pt] &= \tfrac{1}{2} I_0 \omega_0^2 + \tfrac{1}{2} I_1 \omega_1^2 + \left[ \int \rho r^2 \mathop{dV} - \int \rho \vect{r} \cdot \vect{r_1} \mathop{dV} \right] \omega_0 \omega_1 \\ &= \tfrac{1}{2} I_0 \omega_0^2 + \tfrac{1}{2} I_1 \omega_1^2 \\[-3pt] & \qquad + \left[ \int \rho r^2 \mathop{dV} - \tfrac{1}{2} \int \rho r^2 \mathop{dV} + \tfrac{1}{2} \int \rho (\vect{r}-\vect{r_1})^2 \mathop{dV} - \tfrac{1}{2} r_1^2 \int \rho \mathop{dV} \right] \\ & \qquad \text{(using } {-2}\vect{r}\cdot\vect{r_1} = (\vect{r}-\vect{r_1})^2 - \vect{r}^2 - \vect{r_1}^2 \text{)} \\[9pt] \text{K.E} &= \tfrac{1}{2} I_0 \omega_0^2 + \tfrac{1}{2} I_1 \omega_1^2 + \tfrac{1}{2} ( I_0 + I_1 - mr_1^2 ) \omega_0 \omega_1 \end{align*}](https://zmatt.net/wp-content/ql-cache/quicklatex.com-f6cfa8e9c8300ea5d22dbdc45cbd6be2_l3.png "Rendered by QuickLaTeX.com")
If r1 is the center of mass, then I0 = I1 + mr12, and we can show that the above expression reduces to 
which is the expected combination of rotational and translational kinetic energy.
In the more general case where r1 is not the center of mass, I0 is not constant, and this is where the complexity of the center-of-mass oscillation (and kinetic energy oscillation) is hidden. Nonetheless it still may be easier to compute I0 then the general motion of the center of mass. Starting from this nicer form of kinetic energy expression rescued me from a mess of algebra, and perhaps there is something here that will inspire you too.
- I will admit that the title of this post is mathematician clickbait. ↩︎
