Monsieur Lagrange pulls a string down through a hole in a horizontal table thereby effecting a rotating (point) mass. A daemon sits on his shoulder and takes careful note of the proceedings. There is no potential function. The Lagrangian is
$$\mathcal{L}= \frac12m \dot{r}^2 +\frac12m r^2\dot{\theta}^2. \tag{1}$$
As Monsieur Lagrange holds himself to a time protocol we have a Lagrangian with constraint
$$\mathcal{L'}= \frac12m \dot{r}^2 +\frac12m r^2\dot{\theta}^2+ \lambda (r-g(t)) .\tag{2}$$
The Euler-Lagrange for $r$ is
\begin{align} \frac{d }{dt }\frac{\partial\mathcal{L'}}{\partial \dot{r}}-\frac{\partial \mathcal{L'}}{\partial {r}}&= m(\ddot{r}-r \dot {\theta} ^2) + \frac{d }{dt }\frac{\partial\lambda(r-g(t))}{\partial \dot{r}}-\frac{\partial \lambda (r-g(t))}{\partial {r}} \\ &\approx m(\ddot{r}-r \dot {\theta} ^2)-\lambda \\ &= 0,\tag{3} \end{align}
where $\approx$ indicates here, and three times subsequently, that possibly some derivatives are multiplied by zero eaux-shell.
The Euler-Lagrange for $\theta$ is
\begin{align}\frac{d }{dt }\frac{\partial\mathcal{L'}}{\partial \dot{\theta}}-\frac{\partial \mathcal{L'}}{\partial {\theta}} &= \frac{d}{dt} (m r^2 \dot{{\theta}}) + \frac{d }{dt }\frac{\partial\lambda(r-g(t))}{\partial \dot{\theta}}-\frac{\partial \lambda (r-g(t))}{\partial {\theta}} \\ &\approx \frac{d}{dt} (m r^2 \dot{{\theta}}) \\ &= 0.\tag{4} \end{align}
We identify $mr^2\dot{\theta}$ as the angular momentum, $L$, a constant of the motion. Without marshalling the full Hamiltonian formalism we can calculate the change in kinetic energy.
$$ T= \frac12m(\dot{r}^2+r^2 \dot{\theta}^2)=\frac12\left(m\dot{r}^2+\frac{L^2}{mr^2}\right).\tag{5} $$
\begin{align} \frac{dT}{dt}&=m\dot{r}\ddot{r} -\frac{L^2}{mr^3}\dot{r}\\ &= \left(m\ddot{r}- \frac{(mr^2\dot{\theta})^2}{mr^3}\right)\dot{r}\\ &=m (\ddot{r} - r \dot{\theta}^2)\dot{r}\\ &= \lambda \dot{r}.\tag{6} \end{align}
No surprises thus far: $\theta$ is "ignorable", angular momentum $L$ is conserved. The Lagrange multiplier $\lambda$ is the total force in the "r" direction equal to the negative of the force exerted by Lagrange who does work as he pulls the mass inward.
Now...the demon has been compiling a table of $\theta$ vs. time and uses the implicit function theory to render r as a function of $\theta$. In the second run Monsieur Lagrange will be so constrained.
THE TRAJECTORY MUST REMAIN UNALTERED.
$$\mathcal{L''}= \frac12m \dot{r}^2 +\frac12m r^2\dot{\theta}^2+ \mu (r-h(\theta)).\tag{7}$$
The Euler-Lagrange for $r$ is
\begin{align} \frac{d }{dt }\frac{\partial\mathcal{L''}}{\partial \dot{r}}-\frac{\partial \mathcal{L''}}{\partial {r}}&= m(\ddot{r}-r \dot {\theta} ^2) + \frac{d }{dt }\frac{\partial\mu (r-h(\theta))}{\partial \dot{r}}-\frac{\partial \mu (r-h(\theta))}{\partial {r}} \\ &\approx m(\ddot{r}-r \dot {\theta} ^2)-\mu \\ &= 0.\tag{8} \end{align}
The Euler-Lagrange for $\theta$ is
\begin{align} \frac{d }{dt }\frac{\partial\mathcal{L''}}{\partial \dot{\theta}}-\frac{\partial \mathcal{L''}}{\partial {\theta}}&= \frac{d}{dt} (m r^2 \dot{{\theta}}) + \frac{d }{dt }\frac{\partial\mu(r-h(\theta))}{\partial \dot{\theta}}-\frac{\partial \mu (r-h(\theta))}{\partial {\theta}} \\ &\approx \frac{d}{dt} (m r^2 \dot{{\theta}}) + \mu \frac{\partial h}{\partial \theta} \\ &= \frac {d L}{d t} + \mu \frac {\dot r}{\dot \theta} \\ &= 0.\tag{9} \end{align}
Now we immediately see that since $\theta$ is no longer "ignorable" angular momentum is no longer conserved. On the other hand the Lagrange multiplier (now $\mu$)is still the total force operating in the $r$ direction. We can again calculate the rate of change of kinetic energy.
\begin{align} \frac{dT}{dt}&=m\dot{r}\ddot{r} -\frac{L^2}{mr^3}\dot{r}+ \frac1{2mr^2} \cdot \frac{d}{dt} L^2\\ &=\left(m\ddot{r}- \frac{(mr^2\dot{\theta})^2}{mr^3}\right)\dot{r}+ \frac1{2mr^2} \cdot 2L \frac{dL}{dt}\\ &= \mu \dot{r}+ \frac1{2mr^2} \cdot 2 mr^2 \dot{\theta}\cdot (-1) \mu \frac {\dot{r}}{\dot{\theta}}\\ &=\mu \dot{r} - \mu \dot{r}\\ &=0.\tag{10} \end{align}
Well I can't see what the mistake is. There don't seem to be any calculational errors. The Demon will only say things like $\theta$ is proceeding from this to that so you must pull in the string from this $r$ to that $r$.
I seem to recall from Electrical Engineering that "a system is 'Observable' if this matrix is of full rank" and "a system is 'Controllable' if this matrix or some other matrix is of full rank or is less than full rank but I don't even see how it applies.
(absorbed into OP's OP from OP's original "answer" in "anticipation" of the answer of @Qmechanic, comment by @ACuriousMind)
Details? The Daemon is in the Details. The construction in the second part is completely general as per Implicit Function Theory. As for one specific example:
We know $L=mr^2\dot{\theta}$ or $\dot{\theta}= \frac{1}{r^2} \frac{L}{m} $. Let us assume the following trial function of time:
$\frac{1}{r^2}=At+B$, (i.e. $r=g(t)=\frac{1}{\sqrt{At+B}}$) ,with A and B positive. Note also that $L$ is (still) a constant of the motion which we take to be positive.
$$\begin{align}\theta =\frac{L}{m}\int\,(At+B)~\mathrm dt &=\frac{L}{m}\left( \frac{1}{2}At^2+Bt\right) \\ \frac{1}{2}At^2+Bt-\frac{m}{L}\theta &=0\\ \therefore ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ t &= \frac{-B+ \sqrt{B^2 + 4 \frac{1}{2} A \frac{m}{L}} \theta}{A}\end{align}$$
Here we have chosen the positive root.
$$r= \frac{1}{\sqrt{At+B}} = \frac{1}{\sqrt{A \frac{-B+ \sqrt{B^2 + 4 \frac{1}{2} A \frac{m}{L}} \theta}{A}+B }} = \frac{1}{\sqrt{\sqrt{B^2+2A \frac{m}{L}\theta}}}$$
(i.e. $r=h(\theta)= \frac{1}{\sqrt{\sqrt{B^2+2A \frac{m}{L}\theta}}}$)
This does not close the issue at hand for there remains the possibility that even as r is being drawn down the Lagrange multiplier, $\lambda$, is null.
$$\begin{align} r &=(At+B)^ {-\frac{1}{2}}\\ \dot {r} &=-~\frac{1}{2}(At+B) ^{-\frac{3}{2}}A\\ \ddot{r} &=+~\frac{3}{4}(At+B)^{-\frac{5}{2}}A^2= \frac{3}{4} A^2 r^5\end{align}$$
But as
$$\lambda=m\left(\ddot{r}-\dot{\theta^2}r\right)=m\left( \ddot{r}-\left(\frac{L}{m}\right)^2 \frac{1}{r^3}\right)=m\left(\frac{3}{4} A^2 r^5-\left(\frac{L}{m}\right)^2\frac{1}{r^3}\right)$$
I'm sure you see that there is but little chance that this Lagrange multiplier will render no work.
I think my original question contained two cons. Neither entailed the "construction" for the second run. The first con was with regard to the constraint itself: $\lambda(r-g(t))$ Now is this "constraint" holonomic?; non-holonomic? As much as you would like it, even if $\lambda$ is allowed to possess explicit time dependence it can not be cast into the form $\lambda$F(all the generalized coordinates, all the generalized velocities).
Alternativelly Qmechanic likes to think in terms of Potential Energy $V := \lambda (g(\theta,t)-r)$. But, by the same token, I remember Transition Probabilities (in Qm 101 (or 102)) and reviling against time-dependandant Potential Energies even just on the right side of Shroedinger's Equation.
On the one hand the more I think of these conundra (the de-Conservation of Angular Momentum and, as Qmechanic emphasized, the re-Conservation of Kinetic Energy) the less botherred I am. On the other hand there is still something that is being missed. The question isn't being asked properly...
