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To derive Lagrange equation of motion, which are

$$\frac d {dt} \frac {\partial L} {\partial \dot q}-\frac {\partial L} {\partial q} =0$$

we need the virtual work of constraint to be zero , as is said by Goldstein while deriving the equations

We again restrict ourselves to systems for which the virtual work of the forces of constraint vanishes and therefore obtain $$ \sum_{i}\left(\mathbf{F}_{i}^{(a)}-\dot{\mathbf{p}}_{i}\right) \cdot \delta \mathbf{r}_{i}=0 $$ which is often called d'Alembert's principle...

One of the examples using Lagrange equations is of a elastic pendulum as shown

enter image description here

But the virtual displacement isn't perpendicular to the constraint force (spring force) here, so the virtual work of the constraint will not be zero and hence we cannot apply the Lagrange equations.

Why then do we solve this problem using the Lagrange equations?

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If the rigid rod of the pendulum is replaced by a spring, then the constraint force is replaced with a spring force.

The system has 2 position coordinates $(r,\theta)$.

  • On one hand, the rigid pendulum has 1 constraint $r={\rm const}$, so virtual displacements have $\delta r= 0$ and $\delta t=0$. There remains 1 generalized coordinate $\theta$.

  • On the other hand, the elastic pendulum has 0 constraints, so virtual displacements are all displacements frozen in time $\delta t=0$. There remain 2 generalized coordinates $(r,\theta)$.

So the Lagrangian framework still works.

Conversely, the rigid pendulum can in principle be recovered from the elastic pendulum by letting the spring constant $k\to\infty$.

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  • $\begingroup$ Thank you for your response but I don't think it explains 'But the virtual displacement isn't perpendicular to the constraint force (spring force) here, so the virtual work of the constraint will not be zero and hence we cannot apply the Lagrange equations'. Could you please address that. Thank you again :) $\endgroup$
    – Kashmiri
    Commented Aug 26, 2021 at 8:58
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    $\begingroup$ @Kashmiri there is no virtual work since the system is not confined to any submanifold of $\mathbb R^2$. The elastic pendulum is modeled by a free particle moving in a central harmonic potential. There are no additional spatial constraints, the point mass could in principle visit the whole plane! $\endgroup$ Commented Aug 26, 2021 at 9:14
  • $\begingroup$ The string does not constitute a constraint unless $k=\infty$. $\endgroup$
    – Qmechanic
    Commented Aug 26, 2021 at 10:11

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