Equations of motion of a particle over a cylinder+conserved quantities

[fluidistic]

Homework Statement

I must determine the conserved quantities+the equations of motion (of the trajectory in fact) of a particle over the surface of a cylinder.

Homework Equations

Lagrangian and Euler-Lagrange's equations.

The Attempt at a Solution

I've found the Lagrangian of the particle to be L=\frac{m}{2}(r^2\dot \phi ^2 + \dot z^2).
Since the Lagrangian doesn't depend explicitly on \phi nor z, the generalized momenta conjugate are conserved (I'm currently having under my eyes Goldstein's book, 1st edition, page 49).
So I can already answer this part of the problem, P_\phi=k_1 and P_z=k_2.
By intuition I know that the angular momentum is conserved and the speed of the particle along the z axis is constant.
I have a problem however with the Lagrange's equations.
For the generalized coordinate q=r I have that \frac{\partial L}{\partial \dot r}=0 and \frac{\partial L}{\partial r}=m r \dot \phi ^2.
This gives me the first equation of motion, namely r\dot \phi ^2=0. Since r\neq 0 (I'm dealing with a cylinder), \dot \phi =0.
Similarly, I get for \phi: \underbrace{2 \dot r \dot \phi}_{=0} + r \ddot \phi =0 \Rightarrow \ddot \phi =0 which isn't a surprise since I already knew that \dot \phi=0.
I also get \ddot z=0.
So... the motion equations are \dot \phi =0 and \ddot z=0?
I think I made a mistake. If I integrate them I get \phi = \text{ constant} which is obviously wrong.
Hmm I'm confused about what I must do.
Edit: It seems that if I hadn't make any mistake, the motion equations should give me the information I already know: \dot \phi = constant and \dot z = constant. By the way I don't see my mistake for the Lagrange equation regarding \phi.

 

[ehild]

The particle moves over the surface of a cylinder, so r is not a coordinate, but a constant, the radius of the cylinder. The only coordinates are phi and z. Do not derive with respect to r.
If r were a variable too, you should include the term 1/2 m r(dot) ^2 into the kinetic energy.


ehild

 

[sgd37]

is there no gravitational force acting on this particle. And since r is constant you cannot consider it to be one of the D.O.F and you can't derive the equation that gives you 0 angular velocity

 

[fluidistic]

Thank you very much guys, I wasn't aware I shouldn't have derived 3 equations of motion because one of the generalized coordinates was a constant.
So I'm left with \ddot \phi = \ddot z =0 which means that both \dot \phi and \dot z are constants.
I realize that \dot \phi =0 means that the angular momentum of the particle with respect to the z-axis is conserved. While the z component of the linear momentum of the particle is also conserved.
Integrating the equations of motion, I reach that of course as you said, r(t)=K. Also \phi(t)=c_2t+c_4 and z(t)=c_1t+c_3.
Thus, \vec r (t)= \begin{bmatrix} K\\ c_2t+c_4 \\ c_1t+c_3 \end{bmatrix}.

Is my answer correct considering they asked for the trajectory?

 

[ehild]

It should be OK if there is no force except the constraints.

ehild

 

[fluidistic]

Thanks for the confirmation. Problem solved.