= Solution
The <Euler-Lagrange equations> in cylindrical coordinates are
$$
\boxed{\ddot R=R\dot\phi^2-\frac{C\dot z}{R},\qquad
\frac d{dt}(R^2\dot\phi)=0,\qquad
\ddot z=\frac{C\dot R}{R}.}
$$
The last two follow because $\phi$ and $z$ are <cyclic coordinates>. They give the <conserved quantities>
$$
\boxed{\ell=R^2\dot\phi,\qquad k=\dot z-C\log R.}
$$
Since the <Lagrangian> is autonomous, its energy function is conserved as well:
$$
\boxed{\mathcal E=\dot R\frac{\partial L}{\partial\dot R}
+\dot\phi\frac{\partial L}{\partial\dot\phi}
+\dot z\frac{\partial L}{\partial\dot z}-L
=\frac12(\dot R^2+R^2\dot\phi^2+\dot z^2).}
$$
The velocity-linear term cancels from this energy, but it still changes the equations of motion. A dimensionless logarithm means $\log(R/R_0)$ for some fixed reference length; replacing the printed $\log R$ by that expression changes the <Lagrangian> only by a total time derivative and shifts the numerical value of $k$, not the trajectory.
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