A.3.3.CC:5.1 - A rigid link, then a variable length

Consider a freely placed link in a plane, with distinguishable endpoints A and B and fixed length 2 metres. Use one Cartesian reference frame. The implicit description is

q = (xA, yA, xB, yB), with (xB - xA)^2 + (yB - yA)^2 = 4.

The tuple (0, 0, 2, 0), in metres, satisfies the relation. The tuple (0, 0, 1, 0) fails it. Each endpoint separately has a valid position in the plane, but the second pair violates the modeled link.

For locating the endpoints, use parameters X, Y and orientation phi:

A = (X, Y); B = (X + 2 cos(phi), Y + 2 sin(phi)).

Every pair of endpoints at distance 2 has such a representation. Orientations differing by a full turn describe the same configuration. Reversing the direction while keeping A fixed changes B, so identifying opposite orientations would lose a distinction of this labeled-endpoint model.

Now allow the link to extend or contract with length 0 < r <= 3. Replace 2 by the variable r. The formerly rejected pair (0, 0, 1, 0) is admitted at r = 1. A later motion calculation needs a law for the changing length and whatever initial data that law requires. If the question concerns force carried by a rigid constraint, retaining its equation can help express the force calculation through a multiplier; eliminating the constraint from position coordinates has not answered that force question.