Consider a single particle system whose Lagrangian remains the same if the position of the particle is simultaneously (i) rotated by an arbitrary angle s about the z-axis and (ii) shifted by an amount cs in the z-direction (where c is some constant). (The Lagrangian does not have to remain the same if the position is only shifted or only rotated.) Find a conserved quantity of this system. (It is helpful to use cylindrical coordinates.)
so I started by arriving at the following cylindrical coordinates (cscos(s),cssin(s),cs). After that I calculated the kinetic energy and potential energy to find the lagrangian=o.5ms^2c^2Ṡ^2 -mgcs Am I heading in the right direction?
As the comments suggest we can take a general Lagrangian $L$.
Noether's Theorem states that if a change of coordinates in the Lagrangian gives rise in a change of the Lagrangian only up to a total derivative then conserved quantities arise:
\begin{equation}L(q'_i,\dot q'_i,t)=L(q_i,\dot q_i,t)+\epsilon \frac{d}{dt}\alpha(q_i,t) ,\quad\delta q_i=\epsilon\xi^i(q,t) \end{equation} Then we have conserved quantities \begin{equation} J=\frac{\partial L}{\partial \dot q_i}\xi^i \end{equation}
Invariance of Translation in the Z-Direction
Now if we transform our radius vector along the $z$ axis by an infinitesimal amount $\epsilon$ (which IS used as the same $\epsilon$ as above also) we get under this transformation: \begin{equation} z \mapsto z + \epsilon, \quad \delta z=\epsilon , \quad L \mapsto L + 0 \end{equation} Thus: $\epsilon \frac{d}{dt}\alpha(q_i,t)=0$ (as per your question the Lagrangian stays the same).
Thus using Noether's Theorem we can now find the conserved charge:
Now multiplying both sides of J by $\epsilon$ we get:
\begin{equation} J\epsilon=\frac{\partial L}{\partial \dot z}\epsilon\xi \end{equation}
Now by Noether's Theorem $\epsilon\xi=\delta z = \epsilon$, so in fact:
\begin{equation} J\epsilon=\frac{\partial L}{\partial \dot z}\epsilon \implies J=\frac{\partial L}{\partial \dot z} \end{equation}
Is the conserved charge: which is exactly linear momentum in the z direction.
For the Rotation about Z axis
You would need to do a similar computation, except more difficult as we really need to worry about three dimensions. Let i range from 1 to 3, and let summation convention be assumed; with \begin{equation} \delta x_i = \varepsilon_{ijk}\epsilon_jx_k, \quad \delta \vec{r} = \vec{\epsilon} \times \vec{r} \end{equation}
Where $\varepsilon_{ijk}$ is the $3\times 3$ Levi-Civita tensor.