What is the degrees of freedom (Lagrange equation) of two connected spool rolling down two inclines? I'm quite confused as to how to use the Lagrange equation [second type] in a system which features a spool rolling down an incline. I think this particular example is quite representative of what is challenging for me.

the in-extensible rope is wound on two equal homogeneous rollers that can move on steep planes of inclination $α$ and $β$. The mass of a roller is $m$. Determine the acceleration of the unwound part of the rope and the force in it. Ignore friction.


In particular there is a cognitive dissonance I'm unable to resolve with my current understanding: in order to be able to apply the Lagrange equation, we need to presume that at any given time the rope is totally stretched. That being the case, it means that the amount one spool is unwinding is equal to the change in its angular rotation multiplied by the distance from a center of rotation. For the sake of simplicity, we could assume that the instantaneous center of rotation is the same as the center of each spool. [I understand that this might not be the case, but from what I understood in the class this is not an outlandish estimate]. Hence the deflection in the direction of the incline is related to the change in angular coordinate of the center of rotation of the steep; In order words, there is a kinematic restriction which reduces the degree of freedom of the system.
In conclusion, this particular system should be considered having only a single degree of freedom, or otherwise the presumption of the rope being stretched is going to be invalidated, result in impossibility of applying Lagrange's equation for this type of arrangements of spools.
 A: I'll try to give this system a shot, I hope I am right.
First, fix an inertial coordinate system with respect to which the triangle is motionless. In such a coordinate system, denote the distance from the top vertex of the triangle, to the left roller by $q_1$ and the distance from the top vertex of the triangle to the right roller by $q_2$, by $\theta_1$ and $\theta_2$ how much each roller, left and write, has rotated since the beginning of the motion, as well as by $w$ the frictionless sliding displacement of the rope relative to the triangle, the Lagrangian seems to be something like
$$\mathcal{L} \,=\, \frac{I}{2}\, \dot{\theta}_1^2 \,+\, \frac{m}{2}\,\dot{q}_1^2\,+\, mg\cos(\alpha)\,q_1 \,+\,  \frac{I}{2}\, \dot{\theta}_2^2\,+\, \frac{m}{2}\,\dot{q}_2^2\,+\,mg\cos(\beta)\,q_2$$
the first term is the rotational kinetic energy of roller 1, the second term is the translational kinetic energy of roller 1 and the third term is the minus gravitational potential energy of roller 1. The other three terms are the same terms but for roller 2.
However, the rope, which unwinds and slides at the same time, creates a constraint between the rope's displacement, how much unrolling of rope has taken place, and the rotational and translational motion of the two rollers. I am interpreting this system as the rolls having direct contact only with the rope and not with the triangle, while the rope glides on the triangle. So, basically, if at the beginning of the system's evolution, you paint a red dot on the rope that goes exactly over the top vertex of the triangle, after time $t$ the red dot has moved to distance $w$ (this is interpreted as the rope's positional coordinate and let's say it slides without friction from right to left). So, at time $t$, roll 1 on the right has rotated at an angle of $\theta_1$, so it has traversed displacement $r_0\, \theta_1$ to the left along the rope. The rope however, has moved $w$ to the right, so the actual displacement $q_1$ relative to the triangle (relative to the inertial coordinate system) is $$q_1 = r_0\,\theta_1 - w$$
On the other hand, at time $t$, roll 2 on the right has rotated to angle $\theta_2$ and therefore it has traversed $r_0\,\theta_2$ displacement to the right along the rope. The rope itself has moved $w$ to the right, so in total, the displacement $q_2$ relative to the triangle should be
$$q_2 = r_0\,\theta_2 + w$$
As you can see, these are two constraints. Hence
\begin{align}
&\dot{\theta}_1 = \frac{1}{r_0}\,(\dot{q}_1 + \dot{w})\\
&\dot{\theta}_2 = \frac{1}{r_0}\,(\dot{q}_2 - \dot{w})
\end{align}
Hence, the Lagrangian is
$$\mathcal{L} \,=\, \frac{I}{2r_0^2}\,(\dot{q}_1 + \dot{w})^2 \,+\, \frac{m}{2}\,\dot{q}_1^2\,+\, mg\cos(\alpha)\,q_1 \,+\,  \frac{I}{2r_0^2}\,(\dot{q}_2 - \dot{w})^2\,+\, \frac{m}{2}\,\dot{q}_2^2\,+\,mg\cos(\beta)\,q_2$$
so these are three degrees of freedom. However, observe that $w$ is not explicitly present in the Lagrangian, so this fact immidately guarantees a conservation law. For example, if you write the Euler-Lagrange equations, you get
\begin{align}
&\frac{I}{r_0^2}\,(\ddot{q}_1 + \ddot{w}) \,+\, {m}\,\ddot{q}_1\,=\, mg\cos(\alpha)\\
&\frac{I}{r_0^2}\,(\ddot{q}_2 - \ddot{w}) \,+\, {m}\,\ddot{q}_2\,=\, mg\cos(\beta)\\
&\frac{I}{r_0^2}\,(\ddot{q}_1 + \ddot{w}) \,+\,  \frac{I}{r_0^2}\,(\ddot{w} - \ddot{q}_2) \,=\, 0
\end{align}
The last equation allows us to solve for $\ddot{w}$ easily
$$\ddot{w} \,=\, \frac{1}{2}(\ddot{q}_2 - \ddot{q}_1)$$
and reduce the number of equations to
\begin{align}
&\frac{I}{2r_0^2}\,(\ddot{q}_1 + \ddot{q}_2) \,+\, {m}\,\ddot{q}_1\,=\, mg\cos(\alpha)\\
&\frac{I}{2r_0^2}\,(\ddot{q}_1 + \ddot{q}_2)\,+\, {m}\,\ddot{q}_2\,=\, mg\cos(\beta)
\end{align}
and we can even integrate it twice, to obtain another constraint (the conservation law I mentioned earlier)
\begin{align}
&\ddot{q}_1 - \ddot{q}_2 + 2\,\ddot{w} \,=\, 0\\
&\dot{q}_1 - \dot{q}_2 + 2\,\dot{w} \,=\, v_0\\
&{q}_1 - {q}_2 + 2\,{w} \,=\, c_0  + v_0\,t
\end{align}
so
$$w \,=\, \frac{1}{2}\,\big(\,c_0 + v_0 \, t + q_2 - q_1\,\big)$$
Observe the moment of inertia is $I = \frac{1}{2}mr_0^2$, so you can cancel out $m$ and $r_0^2$ from the equations
\begin{align}
&\frac{1}{4}\,(\ddot{q}_1 + \ddot{q}_2) \,+\,\ddot{q}_1\,=\, g\cos(\alpha)\\
&\frac{1}{4}\,(\ddot{q}_1 + \ddot{q}_2)\,+\, \ddot{q}_2\,=\, g\cos(\beta)\\
&w \,=\, \frac{1}{2}\,\big(\,c_0 + v_0 \, t + q_2 - q_1\,\big)
\end{align}
so
\begin{align}
&5\,\ddot{q}_1 + \ddot{q}_2 \,=\, 4g\cos(\alpha)\\
&\ddot{q}_1 + 5\,\ddot{q}_2 \,=\, 4g\cos(\beta)\\
&w \,=\, \frac{1}{2}\,\big(\,c_0 + v_0 \, t + q_2 - q_1\,\big)
\end{align}
From here, solve the linear equations for $\ddot{q}_1$ and $\ddot{q}_2$ and integrate twice and you get the translational motion of each roller as well as the ropes displacement $w$.
