Are there examples in classical mechanics where D'Alembert's principle fails?

Question

D'Alembert's principle suggests that the work done by the internal forces for a virtual displacement of a mechanical system in harmony with the constraints is zero.

This is obviously true for the constraint of a rigid body where all the particles maintain a constant distance from one another. It's also true for constraining force where the virtual displacement is normal to it.

Can anyone think of a case where the virtual displacements are in harmony with the constraints of a mechanical system, yet the total work done by the internal forces is non-zero, making D'Alembert's principle false?

Answer 1

Given a system of $N$ point-particles with positions ${\bf r}_1, \ldots , {\bf r}_N$; with corresponding virtual displacements $\delta{\bf r}_1$, $\ldots $, $\delta{\bf r}_N$; with momenta ${\bf p}_1, \ldots , {\bf p}_N$; and with applied forces ${\bf F}_1^{(a)}, \ldots , {\bf F}_N^{(a)}$. Then D'Alembert's principle states that

$$\tag{1} \sum_{j=1}^N ( {\bf F}_j^{(a)} - \dot{\bf p}_j ) \cdot \delta {\bf r}_j~=~0. $$

The total force

$${\bf F}_j ~=~ {\bf F}_j^{(a)} +{\bf F}^{(ec)}_j+{\bf F}^{(ic)}_j + {\bf F}^{(i)}_j + {\bf F}_j^{(o)}$$

on the $j$'th particle can be divided into five types:

1. applied forces ${\bf F}_j^{(a)}$ (that we keep track of and that are not constraint forces).

2. an external constraint force ${\bf F}^{(ec)}_j$ from the environment.

3. an internal constraint force ${\bf F}^{(ic)}_j$ from the $N-1$ other particles.

4. an internal force ${\bf F}^{(i)}_j$ (that is not an applied or a constraint force of type 1 or 3, respectively) from the $N-1$ other particles.

5. Other forces ${\bf F}_j^{(o)}$ not already included in type 1, 2, 3 and 4.

Because of Newton's 2nd law ${\bf F}_j= \dot{\bf p}_j$, D'Alembert's principle (1) is equivalent to$^1$

$$\tag{2} \sum_{j=1}^N ( {\bf F}^{(ec)}_j+{\bf F}^{(ic)}_j+{\bf F}^{(i)}_j+{\bf F}_j^{(o)}) \cdot \delta {\bf r}_j~=~0. $$

So OP's question can essentially be rephrased as

Are there examples in classical mechanics where eq. (2) fails?

Eq. (2) could trivially fail, if we have forces ${\bf F}_j^{(o)}$ of type 5, e.g. sliding friction, that we (for some reason) don't count as applied forces of type 1.

However, OP asks specifically about internal forces.

For a rigid body, to exclude pairwise contributions of type 3, one needs the strong Newton's 3rd law, cf. this Phys.SE answer. So if these forces fail to be collinear, this could lead to violation of eq. (2).

For internal forces of type 4, there is in general no reason that they should respect eq. (2).

Example: Consider a system of two point-masses connected by an ideal spring. This system has no constraints, so there are no restrictions to the class of virtual displacements. It is easy to violate eq. (2) if we count the spring force as a type 4 force.

Reference:

H. Goldstein, Classical Mechanics, Chapter 1.

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$^1$It is tempting to call eq. (2) the Principle of virtual work, but strictly speaking, the principle of virtual work is just D'Alembert's principle (1) for a static system.

Answer 2

I have a interesting example:

Consider two blocks moving in a line, and an electric intelligent rod connects them. Everything is frictionless. The rod can make measurements of the coordinates of the two blocks, and change length to always makes sure that $x_2 = 2x_1$. Then we assume the mass of the rod is negligible, so that the forces it gives to the two blocks are exactly opposite. Now, we have an equation of constraint, and whenever the rod changes length and applies a non-zero force, D'Alembert's principle fails.

The way to fix Lagrange equation for this kind of constraint is to add the generalized force $Q_i^{(c)}$ created by the constraint ($0$ if D'Alembert's principle holds) to the right: $$ \frac{\mathrm{d}}{\mathrm{d}{t}} \frac{\partial L}{\partial \dot q_i} - \frac{\partial L}{\partial q_i} = Q_i + Q_i^{(c)} $$

Answer 3

You can have instances where there is no local extremum of the action--for instance, take the lagrangian $L=m\left(\dot x ^{2}+\dot y^{2}\right)$ over the space defined by a crescent embedded in $\mathbb{R}^2$--then, even though the tips of the crescent are both perfectly good starting and ending points in your domain, there is no extremal path connecting them--it would have to be the straight line that leaves the domain of your configuration space.

But this is admittedly a contrived example.