How to model rotational friction forces in Lagrangian Dynamics? 
A mass $m$ is free to slide on a table and is connected by a string, which passes through a hole in the table, to a mass $M$ which hangs below. Assume that $M$ moves in a vertical line only, and assume that the string always remains taut. 



We can easily compute the lagrangian,
$$ \mathcal{L}=\frac{1}{2}M \dot{r}^2+\frac{1}{2}m(\dot{r}^2+r^2\dot{\theta}^2)+Mg(l-r)$$
If the table surface is frictionless, we can obtain the equation of motion as follows:
$$ \frac{d}{dt} \frac{\partial \mathcal{L}}{\partial \dot{q_i}}=\frac{\partial \mathcal{L}}{\partial q_i} $$
Here degrees of freedom is $2$, so $i=0,1$ and  $q_0=r, q_1=\theta$
But, if there exists a surface friction coefficient $\mu$ (independent of velocity), how can I include the rotational friction in the equations of motion?
Since $\mu$ is independent of velocity, we can use dissipation function, $\mathcal{D_i}=c_i\dot{q_i}$
Then the modified equations of motion become,
$$ \frac{d}{dt} \frac{\partial \mathcal{L}}{\partial \dot{q_i}}=\frac{\partial \mathcal{L}}{\partial q_i}  - \frac{\partial \mathcal{D_i}}{\partial \dot{q_i}},  \hspace{10mm} i=0, 1$$
Therefore, $$ \frac{d}{dt} \frac{\partial \mathcal{L}}{\partial \dot{q_i}}=\frac{\partial \mathcal{L}}{\partial q_i}  - c_i,  \hspace{10mm} i=0, 1$$
Now, my question is how $c_1,c_2$ are related to $\mu$? 
Can we say $c_1=c_2=\mu mg$?
For simplification, we can assume both the mass are point mass, the hole is also like a point and the string has zero radius. My confusion actually derives from how to distribute frictions in tangential and normal direction. Sorry for that.
 A: the position vector of the mass m is:
$$\mathbf R=\left[ \begin {array}{c} r\cos \left( \theta \right) 
\\r\sin \left( \theta \right) \end {array} \right] 
$$
and the friction force components are
$$\mathbf F=-\mu\,m\,g\,\left(\frac{\mathbf v\cdot\mathbf e_r}{v}\,\mathbf e_r+\frac{\mathbf v\cdot\mathbf e_\theta}{v}\,\mathbf e_\theta\right)$$
where
$$\mathbf v=\frac{d\mathbf R}{dt}=\left[ \begin {array}{c} \cos \left( \theta \right) {\dot r}-r\sin
 \left( \theta \right) \dot\theta \\\sin \left( \theta
 \right) {\dot r}+r\cos \left( \theta \right) \dot\theta \end {array}
 \right] ~,v=|\mathbf v|\\
\mathbf e_r=\left[ \begin {array}{c} \cos \left( \theta \right) 
\\ \sin \left( \theta \right) \end {array} \right] 
\\
\mathbf e_\theta= \left[ \begin {array}{c} -\sin \left( \theta \right) 
\\ \cos \left( \theta \right) \end {array} \right]
$$
the generalized forces at RHS of the EL equation is $~\mathbf Q=\mathbf J^T\,\mathbf F~$ where $\mathbf J~$ is the Jacobi matrix
$$\mathbf J=\frac{\partial \mathbf R}{\partial \mathbf q}=
\left[\mathbf e_r~,r\,\mathbf e_\theta\right]
$$
A: Lagrange's equations for a system of particles in the presence of non-conservative forces becomes:
$\frac{d}{dt}\frac{\partial L}{\partial \dot{q}^i} - \frac{\partial L}{\partial q^i} = \sum_{j=1}^2 {\bf F}_{\text{ncon}j} \cdot \frac{\partial {\bf r}_j}{\partial q^i}$,
where ${\bf F}_{\text{ncon}j}$ is the non-conservative force acting on the $j$th particle, and $\frac{\partial {\bf r}_j}{\partial q^i} = \frac{\partial {\bf v}_j}{\partial \dot{q}^i}$.
For the particle on the table, call it particle 2,
${\bf F}_{\text{ncon}2} = \mu m_2 g \left(-\frac{{\bf v}_2}{\|{\bf v}_2\|}\right)$.
This force will appear in both the $r$ and $\theta$ equations of motion. From here, maybe you can find a correspondence to your dissipation function.
A: I think the equations of motion should be something like this:
\begin{align}
(m+M)\,&\frac{d^2 r}{dt^2} \,=\, m\,r\,\left(\frac{d\theta}{dt}\right)^2 \, -\, Mg\, - \, \mu m g\, \frac{\frac{dr}{dt}}{\sqrt{\left(\frac{dr}{dt}\right)^2 \, +\, r^2\left(\frac{d\theta}{dt}\right)^2}}  \\
m\,r^2 \, &\frac{d^2 \theta}{dt^2}  \, = \, - \,2\, m\,r \, \frac{d r}{dt} \left(\frac{d \theta}{dt}\right)^2\,  - \, \mu m g\, \frac{r\,\frac{d\theta}{dt}}{\sqrt{\left(\frac{dr}{dt}\right)^2 \, +\, r^2\left(\frac{d\theta}{dt}\right)^2}}  \\
\end{align}
Basically, you just take the force vector at each point on the surface of the table in inertial cartesian coordinates and decompose it into the basis formed by the vectors tangent to the coordinate lines of the curvilinear polar coordinate system. The reason is because Euler-Lagrange's the equations of motion are decomposed that way.
