Lagrangian formalism application on a particle falling system with air resistance
I have this problem, with a first-step resolution:
$$...$$
So, I just don't know why they put the term $\frac{\partial F}{\partial \dot{z}}$ in Euler-Lagrange's equations. Why? I know that the dissipation function isn't a conservative force, but I don't know why the partial derivation. For holonomic constraints we need to partially derivate the function of constraint $f=0$ in order to $q$, the generalized coordinate: $\frac{\partial f}{\partial{q}}$. And we itroduce it in Euler-Lagrange's equation multiplied by Lagrange's multiplier $\lambda$, on RHS.
But we have here some kind of constraint with a velocity $\dot{z}$ dependence. That's why we need to put the term $\frac{\partial F}{\partial \dot{z}}$ in Euler-Lagrange's equations? But the term isn't null and they didn't had the Lagrange multiplier, so is it true that it isn't relationated to the constraints formalism? I'm a little bit condused...
forces lagrangian-formalism drag dissipation
Élio PereiraÉlio Pereira
$\begingroup$ Comment to the post (v2): In a nutshell, OP's post seems to confuse (i) Lagrange's equation $\frac{d}{dt}\frac{\partial T}{\partial \dot{q}^j} - \frac{\partial T}{\partial q^j}=Q_j$ and (ii) Euler-Lagrange's equation $\frac{d}{dt}\frac{\partial L}{\partial \dot{q}^j} - \frac{\partial L}{\partial q^j}=0$. The former does not necessarily arise from an action principle, while the latter always does. The former eq. is relevant for the Rayleigh dissipation function. $\endgroup$ – Qmechanic♦ Apr 5 '15 at 19:47
I would say that this exercise in particular has nothing to do with constraints, because the existence of a dissipation force is not giving you any type of constraint in the coordinates - I mean, it certainly isn't holonomic since it doesn't eliminate any degree of freedom from the problem. It is just more information on how the system behaves along the only coordinate that describes it - z, in this case.
The dissipation function is a bit like a potential. If your particle were only influenced by the gravitational force, you would say that $L=T-U$ where $U$ is the gravitational potential and the Euler-Lagrange equations would be the normal ones. But as you have a friction force, the Euler-Lagrange equations become: \begin{equation} \frac{d}{dt}\frac{\partial L}{\partial \dot{z}} - \frac{\partial L}{\partial z}=Q_j \end{equation}
where $Q_j$ is the generalized force and can be shown by its definition from the D'Alembert's principle that $Q_j=-\frac{ \partial F}{ \partial \dot{z}}$ in this particular case. Indeed, if you have a dissipation function, that's always the form of your equations I think.
blackhole1511blackhole1511
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