Will Y-Direction Motion Cease with X-Direction Acceleration Only?

[haruspex]

I realize now that if $~\dot{\theta}(0)=0$ , then $~\vec{a}(0)~$ must be tangential. So the residual freedom seems to be very limited.

Yes, as I wrote in post #21, you can (only) elect to switch the sign of the tangential acceleration arbitrarily, but if you choose not to you must arrive at $\dot{\omega}=0$.

I don't know what you mean by "cause and effect",

As in, the given differential equation accurately represents the dynamics, so applying it in a simulation should produce the same result.
But here we do not know the basis of the equation, and indeed it cannot represent dynamics because the quadratic in the differential equation has two solutions.

but as in #23
$\dot x^2+x^2=1,\quad x(0)=0,\quad t\ge 0.$
has a solution $x(t)=\sin t$ for all $t\ge 0$, and also
$\dot{x}=\sqrt{1-x^2}$
near the maximum. But obviously $x(t)$ doesn't get stuck at $x=1$ .

The SHM analogy is interesting. In the first order ODE form it has the same issue: two solutions, and simulation would not discover the cyclic nature. But the usual second order ODE has neither of those problems. The algebraic process of deriving the first order from the second order involves multiplying by $\dot x$, thus creating the "stuck" solution.

The resolution comes from higher order derivatives.

Yes, that dawned on me overnight. E.g. if we require $\ddot\omega$ to be continuous then we only get the cyclic solution.