- #1
ralqs
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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 \vec{F}_i=0, where \vec{F}_i is the total force on the ith particle.
Now, Q_j=\sum_i \vec{F}_i\cdot\frac{\partial \vec{r}_i}{\partial q_j} where Qj is the generalized force associated with the jth generalized coordinate. So, if \vec{F}_i=0 then Qj = 0. But Q_j=-\frac{\partial V}{\partial q_j}, 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 \vec{F}_i's as a function of the Qjs by making virtual displacements δqj to the generalized coordinates. The the virtual work is
\delta W = \sum_j Q_j \delta q_j = \sum_i \vec{F}_i \cdot \delta \vec{r}_i. Writing \delta q_j = \sum_i \nabla_i q_j\cdot\delta \vec{r}_i (we've tacitly expressed the generalized coordinates as functions of the ri's; \nabla_i q_j stands for \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}).
From this, it follows that \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, implying that \vec{F}_i=\sum_j Q_j \nabla_i q_j 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?
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 \vec{F}_i=0, where \vec{F}_i is the total force on the ith particle.
Now, Q_j=\sum_i \vec{F}_i\cdot\frac{\partial \vec{r}_i}{\partial q_j} where Qj is the generalized force associated with the jth generalized coordinate. So, if \vec{F}_i=0 then Qj = 0. But Q_j=-\frac{\partial V}{\partial q_j}, 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 \vec{F}_i's as a function of the Qjs by making virtual displacements δqj to the generalized coordinates. The the virtual work is
\delta W = \sum_j Q_j \delta q_j = \sum_i \vec{F}_i \cdot \delta \vec{r}_i. Writing \delta q_j = \sum_i \nabla_i q_j\cdot\delta \vec{r}_i (we've tacitly expressed the generalized coordinates as functions of the ri's; \nabla_i q_j stands for \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}).
From this, it follows that \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, implying that \vec{F}_i=\sum_j Q_j \nabla_i q_j 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?