Equilibrium configuration in Lagrangian mechanics

In summary, the conversation discusses the necessary and sufficient conditions for equilibrium in a system with scleronomic constraints. It is proven that ∂V/∂qj=0 is a necessary condition for equilibrium, and that it is also a sufficient condition under certain assumptions. However, there may be a mistake in the reasoning, as no one has been able to find the assumption that sneaked in. The person asks for clarification on their question.
  • #1
ralqs
99
1
Suppose we have a system with scleronomic constraints. Is the condition that ∂V/∂qj=0 for generalized coordinates qj a necessary condition for equilibrium? A sufficient condition?

I managed to "prove" that the above condition is necessary and sufficient for any type of holonomic constaint, sclerenomic or rheonomic. This must be a mistake, because I can find an example of a rheonomic system where the equilibrium points don't satisfy ∂V/∂qj=0.

System is in equilibrium iff [itex]\vec{F}_i=0[/itex], where [itex]\vec{F}_i[/itex] is the total force on the ith particle.

Now, [itex]Q_j=\sum_i \vec{F}_i\cdot\frac{\partial \vec{r}_i}{\partial q_j}[/itex] where Qj is the generalized force associated with the jth generalized coordinate. So, if [itex]\vec{F}_i=0[/itex] then Qj = 0. But [itex]Q_j=-\frac{\partial V}{\partial q_j}[/itex], so ∂V/∂qj=0 is a necessary condition for equilibrium.

Now we prove that it is a sufficient condition. To do this, we find the [itex]\vec{F}_i[/itex]'s as a function of the Qjs by making virtual displacements δqj to the generalized coordinates. The the virtual work is
[itex]\delta W = \sum_j Q_j \delta q_j = \sum_i \vec{F}_i \cdot \delta \vec{r}_i[/itex]. Writing [itex]\delta q_j = \sum_i \nabla_i q_j\cdot\delta \vec{r}_i[/itex] (we've tacitly expressed the generalized coordinates as functions of the ri's; [itex]\nabla_i q_j[/itex] stands for [itex]\hat{x}_i\frac{\partial q_j}{\partial x_i}+\hat{y}_i\frac{\partial q_j}{\partial y_i}+\hat{z}_i\frac{\partial q_j}{\partial z_i}[/itex]).

From this, it follows that [itex]\sum_i \vec{F}_i\cdot\delta \vec{r}_i = \sum_i (\sum_j Q_j \nabla_i q_j)\cdot \delta \vec{r}_i[/itex], implying that [itex]\vec{F}_i=\sum_j Q_j \nabla_i q_j[/itex] herefore, if Q_j = 0, system is in equilibrium. QED?

Now, as far as I can tell I haven't used the assumption that the constraints are scleronomic, but maybe the assumption sneaked in there somewhere. However, there *must* be a mistake somewhere. Can anyone spot it?
 
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  • #2
No one has answered my question. I can only assume that I was unclear in formulating it. So please, if there's something in my post that is confusing, let me know so I can clarify what I'm trying to ask.
 

1. What is equilibrium configuration in Lagrangian mechanics?

Equilibrium configuration in Lagrangian mechanics refers to a state in which a system is at rest and there is no net force acting on it. In this state, the system's potential energy is at a minimum, and the forces acting on it are in balance.

2. How is equilibrium configuration determined in Lagrangian mechanics?

Equilibrium configuration is determined by setting the first derivative of the system's Lagrangian with respect to the generalized coordinates to zero. This results in a set of equations that can be solved to find the equilibrium positions of the system.

3. What is the significance of equilibrium configuration in Lagrangian mechanics?

Equilibrium configuration is important because it allows us to analyze the behavior of a system at rest and understand how forces are balanced within the system. It also serves as a starting point for analyzing the system's dynamics and determining its stability.

4. Can a system have multiple equilibrium configurations in Lagrangian mechanics?

Yes, a system can have multiple equilibrium configurations in Lagrangian mechanics. This can occur when the system has more than one stable minimum or when there are multiple sets of generalized coordinates that satisfy the equilibrium conditions.

5. How does the presence of constraints affect the equilibrium configuration in Lagrangian mechanics?

Constraints can restrict the motion of a system and limit the possible equilibrium configurations. In this case, the equilibrium configuration is determined by solving the equations of motion subject to the constraints, rather than setting the first derivative of the Lagrangian to zero.

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