= Solution
Because the contact is frictionless, the horizontal <normal force> is $\mathbf F_h=N\mathbf n$, with $N$ a signed outward component. Any vertical contact component balances <Newtonian gravity> and does not enter the horizontal equations. With $\boldsymbol\omega=\omega\widehat{\mathbf z}$, the <rotating reference frame> equation gives
$$
\mathbf a'=N\mathbf n-2\omega\widehat{\mathbf z}\times(2a\dot\theta\boldsymbol\tau)+\omega^2\mathbf r.
$$
Since $\widehat{\mathbf z}\times\boldsymbol\tau=-\mathbf n$, $\mathbf r\cdot\mathbf n=2a\cos^2\theta$, and $\mathbf r\cdot\boldsymbol\tau=-a\sin2\theta$, its normal and tangential components are
$$
-4a\dot\theta^2=N+4a\omega\dot\theta+2a\omega^2\cos^2\theta,\qquad2a\ddot\theta=-a\omega^2\sin2\theta.
$$
Hence the requested horizontal <force> vector is
$$
\boxed{\mathbf F_h=-2a\bigl(2\dot\theta^2+2\omega\dot\theta+\omega^2\cos^2\theta\bigr)(\cos2\theta,\sin2\theta,0).}
$$
Its magnitude is the absolute value of the scalar coefficient. That coefficient need not always be negative: the constraint can exert either sign of <normal force>, depending on the bead's relative <velocity>. The tangential equation, which does not contain the reaction, is $\boxed{\ddot\theta=-(\omega^2/2)\sin2\theta}$.
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