= Solution
{c}
In <cylindrical coordinates>,
$$
\mathbf e_r=(\cos\theta,\sin\theta,0),\quad
\mathbf e_\theta=(-\sin\theta,\cos\theta,0),\quad
\dot{\mathbf e}_r=\dot\theta\mathbf e_\theta,\quad
\dot{\mathbf e}_\theta=-\dot\theta\mathbf e_r.
$$
Differentiating the <position vector> $\mathbf r=r\mathbf e_r+z\mathbf e_z$ twice gives the <velocity> and <acceleration>
$$
\begin{aligned}
\dot{\mathbf r}&=\dot r\mathbf e_r+r\dot\theta\mathbf e_\theta+\dot z\mathbf e_z,\\
\ddot{\mathbf r}&=(\ddot r-r\dot\theta^2)\mathbf e_r
+(r\ddot\theta+2\dot r\dot\theta)\mathbf e_\theta+\ddot z\mathbf e_z.
\end{aligned}
$$
This derives <acceleration in cylindrical coordinates> rather than assuming it.
At a fixed time, the wire has <parametrization> $r\mapsto r\mathbf e_r+f(r)\mathbf e_z$. Its <derivative> is the nonzero <tangent vector>
$$
\mathbf t=\mathbf e_r+f'(r)\mathbf e_z.
$$
A smooth wire exerts a <constraint force> $\mathbf R$ with $\mathbf R\cdot\mathbf t=0$. Because $g$ is the <gravitational acceleration>, the gravitational <force> is $-mg\mathbf e_z$; the printed $-g\mathbf e_z$ is interpreted per unit mass. Projecting <Newton's second law>, $m\ddot{\mathbf r}=\mathbf R-mg\mathbf e_z$, along $\mathbf t$ eliminates $\mathbf R$:
$$
\ddot r-r\omega^2+f'(r)\ddot z=-gf'(r).
$$
The <chain rule> gives $\ddot z=f'(r)\ddot r+f''(r)\dot r^2$. Rearrangement yields
$$
\boxed{\left(1+f'(r)^2\right)\ddot r
=r\omega^2-f'(r)\left(g+f''(r)\dot r^2\right).}
$$
This is the governing equation for a <bead on a uniformly rotating wire>. Its azimuthal <acceleration> contributes nothing to the projection because $\mathbf e_\theta\cdot\mathbf t=0$.
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