= Solution
{c}
Put $\lambda=\omega^2-g/a>0$. The zero value of the <first integral> in <6/b> gives
$$
\boxed{\left(1+\frac{r^2}{a^2}\right)\dot r^2=\lambda r^2.}
$$
At the endpoint $r=a$, this gives $\dot r^2=\lambda a^2/2$. The absolute <speed> includes radial, azimuthal and vertical components. Since $\dot z=\dot r$ there,
$$
|\mathbf v|^2=\dot r^2+(a\omega)^2+\dot z^2
=2\dot r^2+a^2\omega^2=a^2\left(2\omega^2-\frac ga\right).
$$
Hence the limiting exit <speed> is
$$
\boxed{|\mathbf v|=a\sqrt{2\omega^2-\frac ga}.}
$$
There is a small <initial condition> qualification in the wording. The exact <initial conditions> $r=\dot r=0$ give the stationary solution forever, by <Picard-Lindelöf theorem>. The displayed zero-<first integral> moving branch instead approaches that <unstable equilibrium> as $t\to-\infty$.
For a precise small disturbance, release the bead with zero relative <velocity> at $r=\epsilon>0$. Its <first integral> is $J=-\lambda\epsilon^2/2$, giving
$$
\left(1+\frac{r^2}{a^2}\right)\dot r^2=\lambda(r^2-\epsilon^2),\qquad
|\mathbf v_{\rm exit}|^2=a^2\left(2\omega^2-\frac ga\right)-\lambda\epsilon^2.
$$
The bead initially accelerates outward and reaches the end. Letting $\epsilon\to0^+$ recovers exactly the stated exit <speed>. The conserved $J$ is a <Jacobi integral> associated with the imposed rotation; ordinary <mechanical energy> need not be conserved because the apparatus driving the wire can do <work>.
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