Lagrangian Mechanics Made Simple
Part 2
Rotational Motion

Ron Steinke <rsteinke@w-link.net>

Last month we discussed how to use Lagrangian mechanics to do simple one dimensional motion. This month, we go on to motion in multiple dimensions, and discuss how to handle rotation.

Vectors and Multiple Dimensions

Since we're going to be working in multiple dimensions, it's easiest to write our equations of motion using vectors. The usual notation uses ${\bf q}$ for the position vector,

\begin{displaymath}
{\bf q} = \left(q_1, q_2, q_3\right)  ,
\end{displaymath} (1)

with ${\bf p}$ reserved for the momentum. In this notation, boldface indicates a vector, so ${\bf q}$ is the position vector, and $q$ is its magnitude

Lagrangian mechanics in multiple dimensions is very similar to the one dimensional case. We begin by writing down the kinetic energy $T$ and the potential energy $U$,

$\displaystyle T$ $\textstyle =$ $\displaystyle \frac{1}{2} \dot {\bf q}^2$  
$\displaystyle U$ $\textstyle =$ $\displaystyle 0  .$ (2)

In this case, we have chosen $T$ and $U$ for a particle moving in three dimensions with no external forces. We then write down the Lagrangian,
\begin{displaymath}
L = T - U = \frac{1}{2} \sum_{i=1}^3 \dot q_i^2  .
\end{displaymath} (3)

Next, we find the equations of motion. Since we now have three variables, $q_i$ for $i \in {1,2,3}$, we get three Lagrange equations,

\begin{displaymath}
\frac{\partial L}{\partial q_i} = \frac{d}{dt} \frac{\partial L}{\partial \dot q_i}  .
\end{displaymath} (4)

Inserting the Lagrangian (3) in the Lagrange equations (4) gives the equations of motion,
\begin{displaymath}
\ddot q_i = 0  ,
\end{displaymath} (5)

showing that the free floating particle is not accelerating, as expected.

Rotating Coordinate Systems

The equation of motion (5) is simple enough, since we're working in a fixed rectangular coordinate system. Of course, many other choices of coordinates are possible 1. A particularly interesting choice is a rotating coordinate system. Imagine you are riding inside a large spinning drum, like those found in amusement parks. If you stay at a "fixed" position in the drum, you're really moving, since the drum is rotating. It is useful to describe the physics of object inside the drum in terms of a rotating coordinate ${\bf\bar q}$, which has a fixed value at a given point on the drum. For a drum rotating with frequency $\omega$ about the $z$ axis, we can write your real position ${\bf q}$ in terms of your "drum position" ${\bf\bar q}$,

$\displaystyle q_1$ $\textstyle =$ $\displaystyle \bar q_1 \cos\omega t - \bar q_2 \sin\omega t$  
$\displaystyle q_2$ $\textstyle =$ $\displaystyle \bar q_2 \cos\omega t + \bar q_1 \sin\omega t$  
$\displaystyle q_3$ $\textstyle =$ $\displaystyle \bar q_3  .$ (6)

If you aren't tied to the drum, you can move around inside. Your "drum position" ${\bf\bar q}$ and real position ${\bf q}$ are both functions of the time $t$. By taking the time derivative ${\bf q}$ in (6), we can calculate your velocity in terms of ${\bf\bar q}$,

$\displaystyle \dot q_1$ $\textstyle =$ $\displaystyle \left(\dot {\bar q}_1 - \omega \bar q_2\right) \cos\omega t
- \left(\dot {\bar q}_2 + \omega \bar q_1\right) \sin\omega t$  
$\displaystyle \dot q_2$ $\textstyle =$ $\displaystyle \left(\dot {\bar q}_2 + \omega \bar q_1\right) \cos\omega t
+ \left(\dot {\bar q}_1 - \omega \bar q_2\right) \sin\omega t$  
$\displaystyle \dot q_3$ $\textstyle =$ $\displaystyle \dot {\bar q}_3  .$ (7)

Mechanics of Rotating Systems

We are now in a position to use Lagrange's equation (4) to find the equations of motion in terms of the coordinate ${\bf\bar q}$. Since we know the velocity from (7), we can write down the Lagrangian,

$\displaystyle L$ $\textstyle =$ $\displaystyle \frac{1}{2} \left( \dot q_1^2 + \dot q_2^2 + \dot q_3^2 \right)$  
  $\textstyle =$ $\displaystyle \frac{1}{2} \left[ \left(\dot {\bar q}_1 - \omega \bar q_2\right)...
...left(\dot {\bar q}_2 + \omega \bar q_1\right)^2 + \dot {\bar q}_3^2 \right]  .$ (8)

Notice that the functions $\cos\omega t$ and $\sin\omega t$ don't appear in the Lagrangian. This is a reflection of the time translation invariance of the equations of motion. It doesn't matter how long the drum has been rotating, the equations of motion will be the same. This is a feature of the coordinate system we have chosen, and would not be true for, say, rotation around an ellipse, where the equations of motion would depend on what part of the ellipse you were on.

Looking at Eq. (8), you can see that it's the sum of three squares. This suggests that $L$ is actually the square of some vector, with each of the terms in the sum being the square of one of the vector's components. We use this fact to rewrite $L$ in vector notation,

\begin{displaymath}
L = \frac{1}{2} \left(\dot{\bf\bar q}+ \hbox{\setbox0=\hbox{...
...\kern-.035em\raise.0433em\box0}\times {\bf\bar q}\right)^2  ,
\end{displaymath} (9)

where $\hbox{\setbox0=\hbox{$\omega$}\kern-.035em\copy0\kern-\wd0
\kern.07em\copy0\kern-\wd0
\kern-.035em\raise.0433em\box0}$ is a vector of magnitude $\omega$ pointing along the $z$ axis. We use this simplified form of $L$ to find the equations of motion. The left hand side of Lagrange's equation (4) is
\begin{displaymath}
\frac{\partial L}{\partial {\bf\bar q}} = -\hbox{\setbox0=\h...
...
\kern-.035em\raise.0433em\box0}\times {\bf\bar q}\right)  ,
\end{displaymath} (10)

and the right hand side is
\begin{displaymath}
\frac{d}{dt} \frac{\partial L}{\partial \dot{\bf\bar q}} =
...
...\wd0
\kern-.035em\raise.0433em\box0}\times \dot{\bf\bar q} .
\end{displaymath} (11)

Combining (10) and (11) gives the equation of motion
\begin{displaymath}
\ddot{\bf\bar q}= -2\hbox{\setbox0=\hbox{$\omega$}\kern-.035...
...
\kern-.035em\raise.0433em\box0}\times {\bf\bar q}\right)  .
\end{displaymath} (12)

Pseudo Forces

We have arrived at the result (12), which seems to say that the object at position ${\bf\bar q}$ is accelerating. This may seem mysterious at first, but is actually quite simple. Looking at the two terms on the right hand side of (12), we see that the second, $-\hbox{\setbox0=\hbox{$\omega$}\kern-.035em\copy0\kern-\wd0
\kern.07em\copy0\k...
....07em\copy0\kern-\wd0
\kern-.035em\raise.0433em\box0}\times {\bf\bar q}\right)$, is the familiar centripetal "force." This term is equal to $\omega^2$ multiplied by the part of ${\bf\bar q}$ which is perpendicular to the axis of rotation. It points away from the axis, so the object will tend to get further away with time. The inward force you feel on amusement rides is the wall providing a compensating acceleration to keep you in the "same" place.

The term $-2\hbox{\setbox0=\hbox{$\omega$}\kern-.035em\copy0\kern-\wd0
\kern.07em\copy0\kern-\wd0
\kern-.035em\raise.0433em\box0}\times \dot{\bf\bar q}$ is commonly known as the Coriolis "force." It depends on the velocity $\dot{\bf\bar q}$ instead of the position ${\bf\bar q}$, so it is not as much a part of common experience as the centripetal "force." It causes objects moving towards or away from the rotation axis to be deflected around the axis, and is an important feature of rotational kinematics.