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I know there are many posts on non-holonomic constraints and also many on this exact one but I feel that there is still some confusion on it.

"Consider a disk which rolls without slipping across the horizontal plane, what is the best generalized coordinates that may be used, note that the plane of the disk remains vertical, and free to rotate about the vertical axis. What is the differential equation that describe the rolling constraints, then check if this equation can be integrated or not. Finally check that the constraint holonomic or not."

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Now for this famous question (from Goldstein,chapter 1, section 1.3) there were equations of constraint: \begin{equation} \dot{x} = a\dot{\phi}\sin\theta \\ \dot{y} = -a\dot{\phi}\cos\theta \end{equation} The idea is that maybe if we integrated these 2 equations we can "return" to a function $f_i(x,y,\phi,\theta) = 0$ which would be the holonomic constraint equation.

So in differential form, \begin{equation}\tag{1} dx - a\sin\theta d\phi = 0 \end{equation} and \begin{equation}\tag{2} dy +a\cos\theta d\phi = 0 \end{equation}

They said that these functions were not "integrable".

Question

When they say that the equations are not integrable, what they really mean that these are not exact\perfect differential equations and that we can't find the "potential" function $f_i$ for it, and NOT that we can't integrate them altogether. What I mean is that we can still do this for an example for eq:(1), by integrating both sides, we get \begin{equation}\tag{3} x = a\sin\theta \int d\phi \end{equation} or \begin{equation}\tag{4} x = a\sin\theta \phi + c \end{equation} where c is a constant. (Can I do this?)

conclusion: Am I correct on this assumption:

When they say that the equations are not integrable,what they mean is that we can't find the "potential" function $f_i$ for it, and NOT that we can't integrate them altogether?

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    $\begingroup$ Can I do this? No, because $\theta$ is a function of $t$. $\endgroup$
    – Ghoster
    Commented Jan 9, 2023 at 4:25
  • $\begingroup$ @Ghoster, what do you mean that $\theta$ is a function of $t$? From these equations I see that only $x$ and $\phi$ are functions of $t$. $\endgroup$
    – Reuben
    Commented Jan 9, 2023 at 4:31
  • $\begingroup$ The diagram makes it look like $\theta$ must be changing. I don't have a copy of Goldstein to look at. $\endgroup$
    – Ghoster
    Commented Jan 9, 2023 at 5:43
  • $\begingroup$ The fact that $\dot x$ and $\dot y$ involve $\theta$ but not $\dot\theta$ is not a reason to think that $\theta$ is a constant. Look at the diagram, not at the equation. $\endgroup$
    – Ghoster
    Commented Jan 9, 2023 at 6:36
  • $\begingroup$ the constraint equations are: $~{\frac {d}{dt}}x \left( t \right) -a \left( {\frac {d}{dt}}\phi \left( t \right) \right) \sin \left( \theta \left( t \right) \right) =0~$ $~{\frac {d}{dt}}y \left( t \right) +a \left( {\frac {d}{dt}}\phi \left( t \right) \right) \cos \left( \theta \left( t \right) \right) =0 ~$ $\endgroup$
    – Eli
    Commented Jan 10, 2023 at 17:34

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