How to determine the number of constraints in this problem set? I don't understand how to determine the number of constraints in this given problem set  in classical mechanics (picture attached down below).
Now lets take a look at this problem for example where a pendulum consists from a massless rigid bar with a length R. On the endpoint of the bar, a mass m is attached with an electrical charge e > 0 (the other end of the bar is in a constant position z0 on the z-axis). The bar spins with an angular velocity $\dot{Φ}$. (In general, the goal is to determine the degrees of freedom in this given problem set which is a  little bit broader than my problem right now).
What I know is that the degrees of freedom are given by f=3N-k where N is the number of mass-points and k is the number of constraints. I thought that N=2 because the spinning mass is m and the suspension point is another 'mass' located in the point $z_0$.
Now the problem is that N=2 implies k=1 which is weird because I don't know how I can describe all of these different parameters like angles etc. with only one constraint.
My assumption is that there are at least 2 constraints:
k=1: $\vert m-z_0\vert=R$
k=2: m=$\sqrt{x^2+y^2}$
k=3: $cos(ψ)=\frac{z_0}{R}$
I would be infinitely grateful if someone could help me out somehow!

 A: The following may be helpful, if I understand your question.
I believe the only constraint in the problem is provided by the rigid bar attached at $z_{0}$.  This fixes the distance between $z_{0}$ and the position of the mass $m$.  In other words,
$$\lVert \vec{r}\left(x,y,z\right)-\vec{z}_{0} \rVert=R$$
With the set-up as shown, it should be possible to unambiguously locate the mass with $\phi$ and $\psi$ as the two degrees of freedom.
I hope this helps.
A: 
I had this  problem when I solved  the  Foucault pendulum
A point mass in 3D space can move in $x\,y\,z~$ direction.  Because the length of the vector $~||\vec L||=L~$ is constant, you have one constraint equation which is:
$${x}^{2}+{y}^{2}+ \left( z-{z_0} \right) ^{2}-{L}^{2}=0\tag 1$$
your number of the generalized coordinates is now two.
you can solve eq. (1) e.g. for $~z=z(x,y)~$ thus your generalized coordinates are $~x\,,y~$.
but you can choose another generalized   coordinates, those must fulfill the constraint equation (eq. (1)). for example with:
$$\vec{R}= \begin{bmatrix}
  x \\
  y \\
  z \\
\end{bmatrix}=\left[ \begin {array}{c} L\sin \left( \alpha \right) \cos \left( 
\beta \right) \\ L\sin \left( \alpha \right) \sin
 \left( \beta \right) \\ L \left( 1-\cos \left( 
\alpha \right)  \right) \end {array} \right] 
~,z_0=L$$
these fulfill the constraint equation (1)
$\alpha~$ and $\beta~$ are the new generalized coordinates.
