Solution
= Solution
Along $BC$, write $x=(2a,s,0)$. With $\omega=-\alpha\hat z/t$, the tangential equation is
$$
\ddot s=\frac{\alpha^2s-2a\alpha}{t^2}.
$$
A stationary point is $s_0=2a/\alpha$, lying on the side iff $\alpha\ge1$. The general perturbed motion is
$$
s=s_0+C t^{(1+\sqrt{1+4\alpha^2})/2}+D t^{(1-\sqrt{1+4\alpha^2})/2}.
$$
Except on the decaying-mode fine tuning, the growing term carries the bead to $B$ or $C$, according to its sign.