How to treat viscous force with Lagrange formalism? I am trying to figure out how to derive the equations of motion of a particle experiencing a viscous force of magnitude $\mathbf{F}=-c\mathbf{v}$, being $c$ constant and $\mathbf{v}$ the velocity of the mass, using Lagrange formalism.
I'd like to have a:

*

*first, a general treatment of the viscous force for the motion of a free particle;


*then, specialize this treatment of the motion of a constrained particle, moving on a known line. As an example, I'm considering the helix, whose point coordinates are described by the parametric equations:
$x=R\cos(\theta)$,
$y=R\sin(\theta)$,
$z=h(\theta)/2\pi$
where $R$ and $h$ are constants.
 A: If you want to use Lagrangian formalism with non-conservative forces you need to evaluate their virtual work
$\delta W = Q_q \, \delta q$,
so that Lagrange equations read
$\dfrac{d}{dt}\dfrac{\partial \mathscr{L}}{\partial \dot{q}} - \dfrac{\partial \mathscr{L}}{\partial {q}} = Q_q$.
Free particle. For the motion of a free particle of mass $m$, experiencing drag $\mathbf{F}^d$, in absence of conservative forces

*

*the Lagrangian function equals the kinetic energy $\mathscr{L} = T = \dfrac{1}{2} m |\mathbf{\dot{r}}|^2 = \dfrac{1}{2} m \mathbf{\dot{r}} \cdot \mathbf{\dot{r}}$, so that its derivatives appearing in Lagrange equations read
$\dfrac{d}{dt}\dfrac{\partial \mathscr{L}}{\partial \mathbf{\dot{r}}} = \dfrac{d}{dt}\left( m \mathbf{\dot{r}} \right) = m \mathbf{\ddot{r}}$$\quad , \qquad$
$\dfrac{\partial \mathscr{L}}{\partial \mathbf{r}} = \mathbf{0}$


*the virtual work reads $\delta W = \mathbf{F}^d \cdot \delta \mathbf{r}$
and the equations of motions read
$ m \mathbf{\ddot{r}} = \mathbf{F}^d$,
as expected.
Your problem: particle on a helix. In your problem, the particle is constrained to move on a helix, so that it has only one degree of freedom, namely $\theta(t)$.

*

*The position of the particle reads
$\mathbf{r}(t) = R \cos \theta(t) \mathbf{\hat{x}} + R \sin \theta(t) \mathbf{\hat{y}} + \dfrac{h \theta(t)}{2\pi} \mathbf{\hat{z}}$


*The velocity of the particle in a Cartesian basis reads
$\mathbf{v} = -R\dot\theta \sin \theta \mathbf{\hat{x}} + R\dot\theta \cos \theta \mathbf{\hat{y}} + \dfrac{h \dot\theta}{2 \pi}\mathbf{\hat{z}}$


*The virtual displacement of the particle reads:
$\delta \mathbf{r} = -R\delta \theta \sin \theta \mathbf{\hat{x}} + R\delta \theta \cos \theta \mathbf{\hat{y}} + \dfrac{h \delta \theta}{2 \pi}\mathbf{\hat{z}}$


*Kinetic energy is
$K = \dfrac{1}{2} m \left( R^2  + \left(\dfrac{h}{2\pi}\right)^2 \right) \dot\theta^2 $


*Virtual work of drag is
$\delta W = \mathbf{F}^d \cdot \delta \mathbf{r} = \underbrace{\mathbf{F}^d \cdot \left[ -R \sin \theta \mathbf{\hat{x}} + R \cos \theta \mathbf{\hat{y}} + \dfrac{h }{2 \pi}\mathbf{\hat{z}} \right]}_{Q_{\theta}} \delta \theta$


*Lagrange equation becomes
$ m \left( R^2  + \left(\dfrac{h}{2\pi}\right)^2 \right) \ddot \theta = \mathbf{F}^d \cdot \left( -R \sin \theta \mathbf{\hat{x}} + R \cos \theta \mathbf{\hat{y}} + \dfrac{h}{2 \pi}\mathbf{\hat{z}} \right)$


*If you model the drag as $\mathbf{F}^d = -c \mathbf{v}$, you can easily find that
$ m \left( R^2  + \left(\dfrac{h}{2\pi}\right)^2 \right) \ddot \theta = - c \left( R^2 + \left(\dfrac{h}{2\pi}\right)^2 \right) \dot \theta$.
And if I made no mistake so far, last equation becomes $m \ddot \theta + c \dot \theta = 0$.
