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Equilibrium configuration

  1. May 20, 2012 #1
    1. The problem statement, all variables and given/known data
    Suppose we have a system with scleronomic constraints. Is the condition that [itex]\frac{\partial V}{\partial q_j}=0[/itex] for generalized coordinates qj a necessary condition for equilibrium? A sufficient condition?


    2. Relevant equations
    [tex]-\frac{\partial V}{\partial q_j}=Q_j= \sum_i \vec{F}_i \cdot \frac{\partial \vec{r}_i}{\partial q_j}[/tex]

    where Qj is the generalized for associates with qj.


    3. The attempt at a solution
    I think I managed to prove that the above condition is necessary and sufficient for any type of holonomic constaint, sclerenomic or rheonomic. This makes be believe I made a mistake, so I'd appreciate it if someone could check my work.

    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.
    [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 [itex]\frac{\partial V}{\partial q_j}=0[/itex] 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 [itex]\delta q_j[/itex]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] Therefore, 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.
     
  2. jcsd
  3. May 21, 2012 #2
    To add a little more detail:

    It's necessarily the case that there's a flaw in my proof, because I can find an example of a rheonomic system where the equilibrium points don't satisfy [itex]\frac{\partial V}{\partial q_j}=0[/itex]. Take a point mass on a vertically oriented hoop of radius R rotating w/ angular speed Ω. If [itex]\Omega>\sqrt{\frac{g}{R}}[/itex], it's easy to show that there's an equilibrium position where [itex]\frac{\partial V}{\partial \theta}\neq 0[/itex].
     
  4. May 26, 2012 #3
    I hate bumping my own question, but this question is really bugging me. If you need me to clarify some points, or don't understand what I've written at all, please let me know.
     
  5. Jun 3, 2012 #4
    I noticed this question and bookmarked it to see if anyone would take it up. Since noone has, I will take a stab myself. I'm out of practice so don't know how much help I'll be but maybe I can at least stimulate some discussion.

    One question I have is, why do you introduce F_i and r_i, and what are they intended to represent? I understand that q_j are the generalized coorditates of the system and V is the potential of the system. And, for clarification, scleronomic means the system is not explicitly time-dependent, correct?
     
  6. Jun 3, 2012 #5
    Fi and ri are the total force of the ith particle and the position of the ith particle, respectively. I introduced force because equilibrium, by definition, occurs when all the forces on the particles vanish. I need to bring that in somewhere.

    And yes, you've defined scleronomic correctly.
     
  7. Jun 4, 2012 #6
    Ok, I've worked through your proof. The only point, and it's a minor one, I would bring up with the mathematics, is that i'm not sure you need [itex]\nabla_i[/itex]. I think you just want [itex]\delta q_j = \sum_i \nabla q_j\cdot\delta \vec{r}_i[/itex]. But, this doesn't seem to significantly alter the proof.

    However, I am considering your example of a rheonomic system in 'equilibrium', where you have a point mass on a hoop that's rotating about its diameter. In such a system you do not satisfy [itex]\vec{F}_i = 0[/itex]; it is a dynamic equilibium, not a static one.
     
  8. Jun 4, 2012 #7
    You're absolutely right. I guess the definition for an equilibrium configuration is that [itex]\ddot{q}_j=0[/itex]. I'll have to go back to the drawing board...

    Thanks for your help!
     
  9. Jun 5, 2012 #8
    Well I think you're in good shape as far as the original proof is concerned.

    It is possible the problem specifies scleromic so that static equilibria can in fact exist, and that the problem inherently does not apply to a rheonomous system.

    Extending the consideration to rheonomous systems sounds like fun though. In the case of a rotating hoop for example with generalized coordinates [itex]\theta[/itex] for revolution and [itex]\phi[/itex] for azimuthal, your 'equilibrium' is something like [itex]\theta = \theta_0[/itex] and [itex]\dot{\phi} = C[/itex], right? So in that case your condition [itex]\ddot{q_i}[/itex] is satisfied. I suppose then you'd want to see whether a similar statement can be made about [itex]\frac{\partial{V}}{\partial{q_i}} = 0[/itex] being necessary or sufficient.
     
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