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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