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It seems to me that many different momenta $\dot{\bf p}_j $ can satisfy d'Alembert's principle:

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

in a constrained system.

For example, take two particles connected by a rod, with an applied non-zero force of the same size on each particle, along the line of the rod and in opposite directions (i.e. forces trying to pull the particles apart). In this case, any pair of $\dot{\bf p}_1, \dot{\bf p}_2$ of the same size and in opposite directions along the line of the rod seems to satisfy the d'Alembert's principle (taking into account the dependencies between $\delta {\bf r}_1$ and $\delta {\bf r}_2$ imposed by the rod), even though some of these do not satisfy the constraint.

Am I misunderstanding something, or does d'Alembert's principle not provide any meaningful information in this example?

Qmechanic
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Chen Van Dam
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2 Answers2

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d'Alembert's principle is mostly used with independent generalised coordinates( with the dependent coordinates eliminated using the constraint equations). We can also use dependent coordinates but then we have to also take into account the constraint equations and incorporate them using lagrange multipliers in the zero virtual work principle to get the constraint forces. As for the rod the only possible momentum is p1=p2=0. As nothing else satisfies the constraint(else the rod cannot be rigid and would get deformed). this paper on virtual displacement and Lagrangian dynamics has a very good description of this

Also in this case the only allowed virtual displacement seems to be $\delta r1=\delta r2$. So only choice of values for the $\dot{\bf p}_j $'s satisfying the d'Alembert principle is $\dot{\bf p}_j $=0

pinaki nayak
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  • It was my impression that the constraints are implicitly incorporated as restrictions to the allowed virtual displacements. I thought that as a consequence of this, the equation as it appears in my question should suffice to determine the equations of motion. Hopefully, the paper that you have kindly pointed me to will help me understand why the constraints will nevertheless need to be introduced using Lagrange multipliers. – Chen Van Dam Nov 10 '18 at 18:47
  • The lagrange multipliers are introduced to account for the constraint force. – pinaki nayak Nov 10 '18 at 19:26
  • Please look at the edit – pinaki nayak Nov 10 '18 at 19:36
  • Also if the rod is moving with a constant velocity then p1=p2=const is a solution. Which is what the D'Alembert principle solutions were. – pinaki nayak Nov 10 '18 at 19:45
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  1. There is no ambiguity in the definition of the momentum ${\bf p}_j$ of the $j$'th point particle.

  2. Rather the ambiguity is in the definition of the $j$'th applied force ${\bf F}_j^{(a)}$ among all forces (such as, e.g., gravity force, spring force, constraint force, etc.) that act on the $j$'th point particle, cf. e.g. this Phys.SE post.

Qmechanic
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