= Solution
Let $e_i'(t)$ be a <Cartesian basis> fixed in the rotating frame. A vector $A=A_i'e_i'$ then satisfies
$$
\left(\frac{dA}{dt}\right)_S
=\dot A_i'e_i'+A_i'\dot e_i'
=\left(\frac{dA}{dt}\right)_{S'}+\omega\times A,
$$
because a <derivative of a body-fixed basis vector> is $\dot e_i'=\omega\times e_i'$. Applying this transport formula twice to $r$ gives, for constant $\omega$,
$$
a_S=a_{S'}+2\omega\times v_{S'}+\omega\times(\omega\times r).
$$
The second and third terms correspond to <Coriolis acceleration> and <centrifugal acceleration> in the rotating description.
For the bead, use cylindrical unit vectors $(e_r,e_\phi,e_z)$ with $e_z$ upward and
$$
r=R(\sin\theta\,e_r-\cos\theta\,e_z),
\qquad \dot e_r=\omega e_\phi.
$$
Projection of $m\ddot r=-mg e_z+N$ along the wire tangent $\cos\theta\,e_r+\sin\theta\,e_z$ eliminates the smooth-wire <normal force> $N$ and gives
$$
\boxed{\ddot\theta=(\omega^2\cos\theta-g/R)\sin\theta}.
$$
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