# Constraint force on a rod

I really hope someone will take a quick look at the following, I would just love to better understand it...

This exercise is from Arnold's "Mathematical Methods of Classical Mechanics", p. 97 in the chapter on d'Alemberts principle:

A rod of weight P, tilted at an angle of 60° to the plane of a table, begins to fall with initial velocity zero. Find the constraint force of the table at the initial moment, considering the table as

(a) absolutely smooth

(b) absolutely rough

(In the first case, the holonomic constraint holds the end of the rod on the plane of the table, and in the second case, at a given point.)"

I must admit, I am pretty unsure on how to do calculations using this "fancy" mathematical kind of physics.

First off, I'm lost with (a). But I'll have a go at (b): As far as I understood, d'Alemberts principle states, that if $M$ is a constraining manifold, $x(q)$ is a curve in $M$ and $\xi$ is a vector perpendicular to $T_xM$, then $x$ satisfies Lagrange's equations

$\frac{d}{dt} \frac{\partial L}{\partial \dot q} = \frac{\partial L}{\partial q} \qquad L = \frac{{\dot x}^2}{2} - U(x)$

iff for the following inner product, we have

$\left(m \ddot x + \frac{\partial U}{\partial x}, \xi \right) = 0$

I guess that's about right so far? The constraint would in this case essentially be $\mathbb S^1$, since the rod moves on a circe around the point in contact with the table. Can we assume that all the rod's mass is at the center of mass?

Would we then have $U(x) = -gx_2$ in this case (where $x_2$ is the vertical component of $x$)?

If yes, then $\partial U / \partial x = -ge_2$. Where $e_1, e_2$ are the horizontal and vertical unit vectors, respectively.

At the inital moment, we have $x = \cos(60°) e_1 + \sin(60°) e_2$, so by d'Alembert's principle we must have

$\left(m \ddot x + \frac{\partial U}{\partial x}, \cos(60°) e_1 + \sin(60°) e_2 \right) = 0$

or written differently

$m \ddot {x_1} \cos(60°) + m \ddot{x_2} \sin(60°) - \sin(60°)g = 0$

So: Is this correct so far? Or am I way off? Is (a) handled any differently up to this point?

If anyone could recommend a good problem book (with solutions), in which this kind of mathematical approach is used (I don't know if this is how physicists would actually compute stuff??), I would greatly appreciate it.

Kind regards,

Sam

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The first way to consider the problem, as we have a simple rigid body is to consider two points on it, say one end (the one that will touch the table first) and the center of mass (or the other end).

The embedding configuration space is thus $R^2 \times R^2$ (the three coordinates of the end of the rod and the three coordinates of the other) if we consider that the rod can only move in the vertical plane including a line on the table (I hope this is the actual question ...).

The action of the surface lead to the [holonomic] constraint equation that can be written $f(x,y) = 0$ which is the one of a surface (here a curve) in the configuration space $R^2$. In addition to this constraint equation we have the Newton's equation which is $m \frac{d^2}{dt^2} \vec{r} = \vec{F}+\vec{N}$ (where $N$ is the force of constraint. This is for one end. For the other (or for the center of mass) you have two similar equations, with the constraint being that the other end (or the center of mass) is at a given distance to the other end.

In that view, we can consider the points as being free and we have 4 Newtonian equations and 2 constraint equations thus reducing the degrees of freedom to 2.

We can also consider that the constraints do not work and that they are perpendicular to the plane of the table: $\vec{N} = \lambda \vec{\nabla}f$ where $f$ is the one of the constraint equation. Newton's equation can be rewritten $\langle m \frac{d^2}{dt^2} \vec{r} - \vec{F}, \vec{\xi} \rangle = 0$ where $\xi$ is a tangent vector defined as $\langle \vec{\xi} ,\vec{\nabla} f \rangle= 0$ ** (1) **.

This is d'Alembert principle. As the components of $\vec{xi}$ are constrained by (1), this lead to 2N (newton) - 2 (constraints) independant relations. The next step is to choose a system of coordinates where (1) is automatically satisfied: these are the generalized coordinates.

The rod is modeled when in addition to specifying the coordinate of the end that is on the plane you also specify the angle that the rod makes with the surface: we can choose to independent generalized coordinates: $x$ and $\theta$.

The motion of the system (now the system is not free anymore) will take place in the Manifold $M$ where $x$ and $\theta$ are independent coordinates. These $(x, \theta)$ are usually written $q = (q_1, q_2)$. This manifold is the direct product of $R$ and $S^1$ (the angles take values in a 1-torus, a segment whose ends are identified).

It can then be derived that the motion in these coordinates follow Lagrange's equations. The Lagrangian is this case as the natural $L(x, \theta, \dot{x}, \dot{\theta})=T-V$ form.

For your case (b) this problem is simplified: x is fixed and you should only care about $\theta$, you have $L = \frac{m}{2} l^2 \dot{\theta}^2 - l m g sin\theta$ where $l$ is the distance between one end and the center of mass.

From Lagrange's equation you obtain

$m l \ddot{\theta} = -m g cos\theta$

From that you can obtain the complete motion of the rod.

For the case (a) you have to consider a more complete Lagrangian:

$L = \frac{m}{2}( l^2 \dot{\theta}^2 - 2 l \dot{x} sin\theta \dot{\theta} + \dot{x}^2)- l m g sin\theta$

This Lagrangian lead to a more complicated motion; if I did the calculation correctly we have

$\ddot{\theta} = - \frac{g}{l} cos\theta$

$\ddot{x} = l sin \theta \ddot{\theta}$.

To obtain the constraint, you can solve this and then you go back to Newton's equations. In the case (b) you have in addition to force $\dot{x} = \ddot{x} = 0$ to obtain the tangential constraint.

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You have a wrong sign in the equation for (b). $\theta$ must be surely a decreasing function but your equation would imply it's increasing. – Marek Nov 15 '10 at 1:41
@Marek: Right, I corrected. – Cedric H. Nov 15 '10 at 1:45
@Cedric: also, your equations of motion for (a) are wrong. For $\ddot{\theta}$ you forgot to include the term obtained by differentiation of ${1 \over l}\dot{x}\sin(\theta)$ with respect to time and the RHS for $\ddot{x}$ must be a differentation of $l\sin(\theta)\dot{\theta}$. – Marek Nov 15 '10 at 2:09
@Marek: right again, I should really sleep (3 am ...) and I'll look at this later. – Cedric H. Nov 15 '10 at 2:15
Thanks for this detailed answer! I think your answer (as well as Marek's) showed me the approach one should take beautifully. Again, thanks a lot! =) – Sam Nov 15 '10 at 5:16

I will not address your first question as I am not really sure. I've read what you've written and given enough time I'd be probably be able to understand whether the derivation is correct, but from top of my head I can't and, more importantly (and this answers your other question): this is not how physicists think about these problems (at least I don't know anyone who does), because once you learn Lagrange's formalism there's no need to go back to this strange principle that, as far as I know, is only used as a motivation on the road to better formalisms (Lagrange's and Hamilton's).

So, here's what I'd do:

1. In (b), as you correctly state, the endpoint (or any other point for that matter) of the rod will be constrained to $S^1$. Now this gives us a nice way to parametrize the problem: the angle $\varphi$ between the rod and the table (this is all part of the Lagrange formalism. If you haven't yet learned it, I strongly advise you do). Also let us denote by $r$ the distance between the center of mass and the point of contact with the table. We write out the Lagrangian for the center of mass (assuming the rod is homogeneous) $$L(\varphi, \dot{\varphi}) = T(\varphi, \dot{\varphi}) - U(\varphi) = m{r^2\dot{\varphi}^2 \over 2} - mgr\sin(\varphi)$$ Now just apply Lagrange's equations to obtain $$mr^2\ddot{\varphi} = - mgr\cos(\varphi)$$

2. For (a) we should assume smooth table, which means no friction, which means the rod will slide. It is clear that it will be confined to the half-plane perpendicular to the table in which the rod initially lies. We can parametrize this half-plane by $p$: the position of the point of contact of the rod with the table and $\phi$ as before. Again using Lagrange formalism one obtains $$L(p, \varphi, \dot{p}, \dot{\varphi}) = {m \over 2}(\dot{p}^2 -2r\dot{p}\sin(\phi)\dot{\phi} + r^2\dot{\varphi}^2 ) - mgr\sin(\varphi)$$ This is not a particularly nice Lagrangian and one has to wonder whether there are some better coordinates in which the problem simplifies (as in the case (b) above).

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Thanks a lot Marek. So, I understand d'Alemberts principle is less of practical than of theoretical/historical interest, and it should simply be seen as yet another point of view which is mathematically equivalent to Lagrange's and Hamilton's equations / the principle of least action. – Sam Nov 15 '10 at 5:06

I'm just wondering the miss use of "law of cosines",I think using $\cos(\theta)$for substitution of $\sin(\theta)$ that will get a correct solution. : )

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