= Solution
The inertial <acceleration> obtained by differentiating the position used in part (i) is
$$
\ddot r
=R(\ddot\theta\cos\theta-\dot\theta^2\sin\theta-\omega^2\sin\theta)e_r
+2R\omega\dot\theta\cos\theta e_\phi
+R(\ddot\theta\sin\theta+\dot\theta^2\cos\theta)e_z.
$$
Since $N=m\ddot r+mg e_z$, substitution of the equation of motion gives the wire's force on the bead as
$$
\boxed{
N=m\bigl(R\dot\theta^2+R\omega^2\sin^2\theta+g\cos\theta\bigr)
(-\sin\theta\,e_r+\cos\theta\,e_z)
+2mR\omega\dot\theta\cos\theta\,e_\phi }.
$$
The first component is normal to the circle within its plane; the second is perpendicular to the rotating plane.
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