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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 $0$.

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It depends how the body is fixed in space. If it is floating, there's no torque and the angular momentum is conserved. But if the body is supported by a pedestal or something like that, the potential energy $mgh$ where $h$ is the height of the center of mass effectively does depend on the angle $\phi$ and the derivative of this potential energy with respect to the angle gives you torque. –  Luboš Motl Mar 27 '13 at 15:26
    
Comment to the question (v1): It is still possible to define the potential energy $V$ of a torsion spring, so that angular system has a Lagrangian formulation. –  Qmechanic Mar 27 '13 at 15:30
    
How does height depend on theta (The angle between the applied force and the radius)? Knowing torque should equal |F||R|sin(theta), the potential should then equal |F||R|cos(theta). If |F||R|cos(theta) = mgh = |F|*h, that would imply that the height is less than the radius. -- What am I missing here? –  JLA Mar 27 '13 at 16:16
    
You know you need to refine the rotation with generalized coordinates (like Euler angles) which will yield generalized forces (torques). –  ja72 Aug 20 '13 at 14:32

1 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 $\dot{L}=\tau$.

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