What's the Lagrangian for a freely rotating rod? [closed]

Question

This is one of those problems that I thought would be easy, then spent forever on it and realized that I know nothing:

A rigid rod of uniform density has mass $M$ and length $L$ and is free to rotate about its center without a fixed axis. Consider its center to be fixed at the origin of an inertial Cartesian coordinate system, and note that the position of the rod can be specified with the spherical coordinate angles $\theta$ and $\phi$. (Also assume zero gravity.)

1. What is the Lagrangian of this system, in terms of the coordinates $\theta$ and $\phi$?

First, I calculated the moments of inertia for rotation in the pure $\theta$ and pure $\phi$ directions. For $\theta$, this is the usual $\frac{1}{12}ML^2$, but $\phi$ it becomes $\frac{1}{12}ML^2\sin^2\theta$ assuming the rod is held at fixed $\theta$. Then I used $K_{rot}=\frac{1}{2}I\omega^2$ for each type of rotation and added the two:

$$\begin{align} L &= K-U \\ &= K_\theta+K_\phi-0 \\&= \frac{1}{24} M L^2\dot\theta^2+ \frac{1}{24} M L^2\sin^2(\theta) \,\,\dot\phi^2 \end{align}$$

My main concern is that adding the two energies doesn't seem justified, so my next question is:

2. If this is correct, why is adding these two energies justified?

To emphasize the concern, consider this: pure $\theta$-rotation corresponds to $\vec\omega$ in one direction, while pure $\phi$-rotation corresponds to $\vec\omega$ in another direction. These two vectors span only a plane, whereas in general $\vec\omega$ could point anywhere in 3D space, so it isn't at all clear that these angular coordinates simply "add" in a straightforward way.

Trying to find the answer more rigorously, I calculated the inertia tensor (assuming at time $t=0$ the rod is lying in the $z$-$x$ plane) and tried to use $K_{rot}=\frac{1}{2}\vec\omega \cdot (\tilde{I} \vec\omega)$. This leads to my final and most deceptively difficult question:

3. What is $\vec\omega$ in terms of $\theta$, $\dot\theta$, $\phi$, and $\dot\phi$?

Also if there is a cleaner or more straighforward approach for solving this problem, I'd love to know about it!

Answer 1

We can just go back to basics, and integrate the contributions to the kinetic energy along the rod. Let the linear density be $\lambda = M/L$. The square of the velocity in spherical coordinates is

$$v^2 = \dot{r}^2 + r^2(\dot{\theta}^2 + \sin^2 \theta\, \dot{\phi}^2);$$

at all points on the rod we have $\dot{r}=0$ and that the angular part in parentheses is constant. The energy is then

$$E = \frac12 \int_{L/2}^{L/2} \lambda r^2 (\dot{\theta}^2 + \sin^2 \theta\, \dot{\phi}^2).$$

We can just take everything except $r^2$ out of the integral, and we get

$$E = \frac{1}{24} ML^2 (\dot{\theta}^2 + \sin^2 \theta\, \dot{\phi}^2)$$

so your initial guess was right. We can add the two energies because they correspond to orthogonal directions, just like in cartesian coordinates we would say $E = \frac12 m (v_x^2 + v_y^2 + v_z^2)$. You can also see that the thing in parentheses in just $\omega^2$, but hopefully this simpler derivation convinces you that everything is legitimate.