# How do I treat the Lagrangian in the case of a rigid body?

Here's Exercise 1.11 from Goldstein's Classical Mechanics 3rd edition (the first one after having derived the Lagrangian basically):

Exercise 1.11: Consider a uniform thin disk that rolls without slipping on a horizontal plane. A horizontal force is applied to the center of the disk and in a direction parallel to the center of the disk.

• Derive Lagrange's equations and find the generalized force;

• Discuss the motion if the force is not applied parallel to the plane of the disk.

Now, I realise the approach I'm going to suggest is a bit "primitive" in the sense that doesn't assume much, but I wanted to derive the Lagrangian from "first principles" (since I've just started with Lagrangians).

The disk is made of $N$ particles, so it has $3N$ degrees of freedom. We obviously have several constraints which greatly limit the motion of the disk:

• It has to move in the $xy$ plane;
• It's a rigid body;
• It rolls without slipping.

I want to use the angle $\theta$ with the $x$ axis as my generalised coordinate (first question: can I do that? It looks that from the no slipping condition I can easily derive $X_{CM}$ from $\dot\theta$ and vice versa, but the equations of motion seem kind of strange if I do this). The kinetic energy of the system is is $T=\frac{1}{2}M(\dot X_{CM}^2+\dot\theta^2R^2)=M\dot\theta^2R^2$ because of the no slipping condition. Now, I can use Euler-Lagrange equations in this form:

$$\frac{d}{dt}\bigg(\frac{\partial T}{\partial \dot\theta}\bigg)-\frac{\partial T}{\partial \theta}-F_\theta=0$$

Where $F_\theta$ is the generalised force. Here's my main question: how do I treat the generalised force if I have a rigid body? What do I put in $F_\theta$? I can write the Euler-Lagrange equations for every particle in the body, but that somehow doesn't seem very useful. I thought about integrating the force on a single particle over the body, but something somewhere doesn't feel right. What am I missing?

• Since you are Italian you could have a look to my lecture notes on analytic mechanics for undergraduates (I should translate them in English but I have no spare time)... science.unitn.it/~moretti/dispense.html You could download the first "dispense" you find on that page. I think you should be interested in chapter 8, especially section 8.3. – Valter Moretti Oct 18 '14 at 11:11
• Great Valter, thank you! I'll give a look at it! :) (and it's an awesome resource for the rest of the course too!) – Gennaro Marco Devincenzis Oct 18 '14 at 11:21

Since you would like to derive the lagrangian from first principles, it is worth spending some words on the principle of virtual works and how it yields the lagrangian equations of motion.

The point here is that the rolling without slipping condition guarantees that the constraint forces don't do work, since the point of contact with the ground is always at rest. That is, the D'Alèmbert principle is satisfied. Therefore, Lagrange's generalized equations: $$\frac{\text d}{\text d t}\frac{\partial T}{\partial \dot q ^i}-\frac{\partial T}{\partial q^i}=Q_i$$ hold, where the $Q_i$'s are certain functions which depend only on the external forces. I think that you are well aware of this.

Now, finding the generalized coordinates $q^i$ and the associated generalized forces $Q_i$ is part of the modeling of the problem. In our case, we are allowing only displacements in one dimension, and the non-slipping condition guarantees that the "angle" $-\infty < \theta< +\infty$, or the center of mass position $-\infty <X<\infty$ are both good (global) coordinates.

Now, what about $Q=F_\theta$? If you want to see the system as composed of $N$ massive particles, with positions $\mathbf x ^{(n)}=\mathbf x ^{(n)}(q^1,q^2,\dots,q^n)$ satisfying Newton's equations $$m^{(n)}\ddot {\mathbf{x}} ^{(n)} = \mathbf{f} ^{(n)}+\mathbf R ^{(n)},$$ where the $\mathbf R ^{(n)}$ do no virtual work, I think that Goldstein explains that in that case $$Q_i =\sum _{n=1} ^{3N} \mathbf f ^{(n)} \cdot \frac{\partial \mathbf x ^{(n)}}{\partial q^i}.$$ The RHS of this equation is the so called "virtual work" of the active forces, that is, $Q_i \delta q_i$, is the work done when the system undergoes a displacement parametrized by $q_i \to q_i + \delta q_i$, with all other coordinates (as well as the form of the constraint wrt time, which is constant in the present case), being held fixed. (Also, if $\mathbf f ^{(n)}=-\frac{\partial U}{\partial \mathbf{x} ^{(n)}}$, it is easy to see that $Q_i =-\frac{\partial U}{\partial q^i}$)

In the present case, the virtual work is $F\delta X=-\delta (-FX)$, so $Q_i$ is $-\frac {\partial U}{\partial q^i}$, where $U=-FX$. Writing the lagrangian as $L=T-U$, with the correct form of $T$, you can easily derive the lagrangian.

I want to stress that this procedure, while physically rigorous, is just heuristics, since there's no a priori reason (at least in my discussion) why the intuitive identification of $Q_i$ with the virtual work should reproduce the correct equations of motion. Indeed, if you already have the equations of motion in some form, the easiest route to find out the lagrangian is probably:

1. Write the LHS of Lagrange's generalized equations.
2. Plug-in the (given) equations of motion.
3. Eliminate the constraint forces.
4. Find out $Q_i$ and possibly the lagrangian.

Typically force should be included in your Lagrangian:

$$L=M\dot\theta^2R^2+F \theta r$$

But you could also have included it as $F_{\theta}=F\,r$

They both equate to:

$$2\,M\,R^2\,\frac{d\dot\theta}{dt}-FR=0$$

If you are unsure how forces look after coordinate transforms i suggest you start with a Lagrangian in Cartesian coordinates, and the transform to the generalized coordinates you wish to use.

In cartesian coordinates you then have

$$L = \frac{M}2(\dot x^2 +\dot y^2 + R^2 \omega^2) - Fx$$

As the cylinder is on the ground all the time $\dot y = 0$, and the "no slipping" condition gives $x=R\theta$ and $\dot\theta=\omega$. Changing cordinates to $\theta$ then gives that $\dot x = R\dot\theta$ so

$$L = \frac{M}2(R^2\dot \theta^2 +R^2 \theta^2) - FR\theta = M R^2\dot \theta^2 - FR\theta$$

The gernalized forcce is now $F_\theta = \frac{\partial L}{\partial \theta}=-FR$ and the equations of motion reads

$$2MR^2\ddot\theta = - FR \Leftrightarrow \ddot\theta = - \frac{F}{2MR}$$