Question: **Edit 1**: May be I should modify my question after getting the answers. I see why $(X\_c, Y\_c, Z\_c, \theta, \phi)$ are legitimate *Dof*'s of the dumb-bell, I never had any problem with that.

Please consider now *any* rigid body in general. It has $6$ *Dof*'s, two examples of which are mentioned below:

* $(X\_c, Y\_c, Z\_c, \alpha, \beta, \gamma)$: $(X\_c, Y\_c, Z\_c)$ are CoM co-ordinates, and $(\alpha, \beta, \gamma)$ are the Eulerian angles (or, the pitch, roll and the yaw) of the rigid body.
* Take a line through any point $(x\_1, y\_1, z\_1)$ inside the rigid body, fix a line through the point using two angles $\theta, \phi$ (or fix a unit vector along the line, using two components). The last $Dof$ is obtained by specifying an angle around this line as axis, $\theta'$ that a point not on this line makes with, say the *horizontal*.

My question is more about the constraint relations. There are $3N$ co-ordinates to begin with, and each constraint relation **eliminates** one redundant co-ordinate. If this is true, because a rigid body has $6$ *Dof*'s, after considering all the constraint relations (minus all the redundant constraints), I will have only $6$ (out of the set of $3N$ co-ordinates to start with). *So can I have the Dof's to be **any*** $6$ *co-ordinates out of the initial* $3N$, *for example* $(x\_1, y\_2, z\_3, x\_4, x\_5, z\_6)$ *- is that sufficient to locate the rigid body*?

If not, the constraint relations only **lowers** the number of *co-ordinates*, doesn't eliminate (remove) though. The *reduced* set of co-ordinates doesn't have to be any of those before reduction, it could be any, just one less in number.

Answer: 
"There is a new suggestion: $(X\_c,Y\_c,Z\_c,θ,ϕ)$ where $(X\_c,Y\_c,Z\_c)$ are the coordinates of the center-of-mass and θ and ϕ are the polar and the azimuthal angles. If this is the effect of the constraint relation, then it doesn't really eliminate one redundant coordinate, but replaces the entire set, with one less coordinate. Am I right?"

Not really. You can consider the set $(X\_c,Y\_c,Z\_c,θ,ϕ,r)$, where $r$ is the distance between the two points. The constraint then simply becomes $r = \sqrt{6}$. This is a much cleaner approach because it addresses the ambiguity you posed in your first question.

If you want to use Cartesians as you suggested, then to address the ambiguity you need to apply a continuity or "no teleportation" condition. That is, if your choice is between $x\_2 = 3, -1$, you pick the value which is closest to the value of $x\_2$ at the preceding time step.
For simplification I will take the 2D space .

The constraint equation is:

$$ \left( x\_{{2}}-x\_{{1}} \right) ^{2}+ \left( y\_{{2}}-y\_{{1}} \right) ^
{2}-{L}^{2}
=0\tag 1$$

thus you obtain three generalized coordinates

solving equation (1) for $x\_2$

$$x\_2=x\_1\pm\sqrt{L^2-(y\_2-y\_1)^2}$$

you get two solution for $x\_2$ but this doesn't affected the equations of motion . to solve the equations of motion the initial conditions must fulfilled the constraint equation. you get this two initial conditions configurations one with $+x\_2~$ and one with $-x\_2$ , but notion is wrong with this two solution.

[![enter image description here](https://i.stack.imgur.com/yQ4aA.png)](https://i.stack.imgur.com/yQ4aA.png)