= Solution
{c}
Here <equilibrium> means fixed position relative to the rotating wire: $r=a$ and $z=f(a)$ are constant, while $\theta=\omega t$ continues to change. Substituting $\dot r=\ddot r=0$ into the equation gives
$$
\boxed{f'(a)=\frac{a\omega^2}{g}.}
$$
For the <linear stability> calculation, set $r=a+\epsilon$ and retain terms of first order in $\epsilon$ and $\dot\epsilon$. The term $\dot r^2$ contributes only at second order, and variation of the factor multiplying $\ddot\epsilon$ also contributes only at second order. Thus
$$
\boxed{\left(1+f'(a)^2\right)\ddot\epsilon
=\left(\omega^2-gf''(a)\right)\epsilon.}
$$
The two strict sign cases are treated in the following parts. If $gf''(a)=\omega^2$, this <linearization> has no restoring term and higher orders are needed to classify the <equilibrium>.
For the specified wire,
$$
f'(r)=\frac{\omega^2r}{g}-\sin(r/b),\qquad
f''(r)=\frac{\omega^2}{g}-\frac1b\cos(r/b).
$$
Hence the complete set of positive <equilibrium> radii is
$$
\boxed{a_n=n\pi b,\qquad n=1,2,3,\ldots.}
$$
At $a_n$,
$$
\left[1+\left(\frac{n\pi b\omega^2}{g}\right)^2\right]\ddot\epsilon
=\frac gb(-1)^n\epsilon.
$$
Therefore \b[odd $n$ give a <stable equilibrium>, and even $n$ give an <unstable equilibrium>]. There is no degenerate sign case in this example. The corresponding heights are $f(a_n)=\omega^2n^2\pi^2b^2/(2g)+b(-1)^n$.
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