Deriving torque from Euler-Lagrange equation

Question

How could you derive an equation for the torque on a rotating (but not translating) rigid body from the Euler-Lagrange equation? As far as I know from my first class in Classical Mechanics, there is no potential defined for a rotating body, so the only term I see in the Lagrangian is the $\frac{1}{2}I\omega^2$. Since there is no $\theta$, I'm just getting that torque always equals $0$.

Answer 1

The question isn't clear to me, the torque is zero if there's not an applied torque, and it's nonzero if it's not zero, but that's just a tautology. Do you mean that you just want to derive something/anything torque-esque given an applied force? If so, here's one way.

This isn't a job for a potential, it's the job for a generalized force. Consider the 2d case, with a particle at some position $s=(x,y)$, with mass $m$ and a force vector $F$ applied to it.

The first form of the Euler-Lagrange equations I learned was the form:

$$\frac{d}{dt} \frac{\partial T}{\partial \dot q_i}-\frac{\partial T}{\partial q_i}=Q_i$$ where $T$ is the kinetic energy (NOT the lagrangian), $q_i$ are the parameters describing the system, and $Q_i$ is defined as the generalized force $Q_i=\sum_j F_j \frac{\partial s_j}{\partial q_i}$. (Goldstein classical mechanics ch 1)

Then, parameterizing in terms of polar coordinates, we can define $s=(q_1 \cos(q_2),q_1 \sin(q_2))$.

Going through the motions of finding the kinetic energy, generalized forces, and partial derivatives:

$\dot{s}=(\dot{q_1}\cos(q_2)-q_1\sin(q_2) \dot{q_2},\dot{q_1}\sin(q_2)+q_1\cos(q_2) \dot{q_2})$

$\|\dot{s}\|^2=\dot{q_1}^2+q_1^2\dot{q_2}^2$.

$T=\frac{1}{2}m(\dot{q_1}^2+q_1^2\dot{q_2}^2)$

$Q_1=F\cdot(\cos(q_2),\sin(q_2))\equiv F_r$ (define $F_r$ this way)

$Q_2=F\cdot(-q_1\sin(q_2),q_1\cos(q_2))\equiv \tau$ (define $\tau$ this way)

Euler-Lagrange equation for $q_1$:

$\ddot{q_1}m-m\dot{q_2}^2 q_1=F_r$

for $q_2$:

$\frac{d}{dt}(m q_1^2 \dot{q_2})-0=\tau =m q_1^2 \ddot{q_2}+2 m q_1 \dot{q_1} \dot{q_2}$

Denoting $q_1=r$, $q_2=\theta$ for clarity's sake, we wind up with the equations: $$m\ddot{r}-m r \dot{\theta}^2=F_r$$ $$m r^2 \ddot{\theta}+2 m r \dot{r} \dot{\theta}=\tau$$

From which we can identify the torque and whatever we want. If $r$ is constant, we have $F_r=-m v^2/r$, and identifying $m r^2 \dot{\theta}=L$ with the angular momentum, we see $\dot{L}=\tau$.

Answer 2

In my view, you don't derive torque from the Euler-Lagrange equations. It arises naturally from them, as Lagrangian is a formalism of classical mechanics that can explain the behavior of physical systems both in their translations and rotations. For instance, consider a simple pendulum. It's only degree of freedom is $\phi$, the angle denoting its deviation from the vertical.

So you would write the Euler-Lagrange equations as

$$ \frac{\partial \mathcal{L}}{\partial \phi} - \frac{d}{dt}\left( \frac{\partial \mathcal{L}}{\partial \dot \phi}\right) = 0.$$

Let's look at dimensions. $\frac{\partial \mathcal{L}}{\partial \phi}$ has units of energy per radian; that is, $\rm N \cdot m$. The first is canonically the generalized force; in this case, it is a torque.

Then, $\frac{\partial \mathcal{L}}{\partial \dot \phi}$ has units of energy per rotational speed; and thus in this case, the generalized momentum is an angular momentum. So rotational dynamics can arise naturally from minimizing the action of a system. In the case of a simple pendulum, the generalized torque is $-mgl\sin\phi$, as you would expect from the potential $-mgl\cos\phi$.

For a rotating particle with potential $V$, it has Lagrangian

$$ \mathcal{L} = T - V = \frac{1}{2}mR^2\dot\phi^2 - V.$$

Then the generalized force (i.e., torque) is

$$ \tau = \frac{\partial\mathcal{L}}{\partial \phi} = \frac{\partial V}{\partial\phi},$$ which is what we expect from rigid-body dynamics.