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?
Thanks for reading (and hopefully thanks for your helpful reply)!
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
Best Answer
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:
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 $$mr^2\ddot{\varphi} = - mgr\cos(\varphi)$$
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).