= Solution
The original diagram shows two equal balls, each of <mass> $m$, attached to the fixed upper pivot and light sliding collar by four equal rods. Each ball describes a horizontal circle of radius $r=L\sin\theta$. The pivot-to-collar distance is $2L\cos\theta$, so the upward collar displacement compresses the <spring> by
$$
\delta=2L(1-\cos\theta).
$$
Thus <Hooke's law> gives a downward spring <force> $F_s=k\delta$ on the collar. Let $T_u,T_l$ be the <tensions> in the upper and lower rods attached to one ball. Vertical <force> balance on the light collar and then on either ball gives
$$
2T_l\cos\theta=F_s,\qquad T_u\cos\theta-T_l\cos\theta=mg.
$$
The radial <force> on one ball supplies its <centripetal acceleration>:
$$
(T_u+T_l)\sin\theta=m\omega^2L\sin\theta.
$$
For $\theta\ne0$, adding the vertical-balance results and eliminating the rod <tensions> gives
$$
\boxed{\omega^2=\frac{mg+F_s}{mL\cos\theta}=\frac{g}{L\cos\theta}+\frac{2k}{m}\frac{1-\cos\theta}{\cos\theta}.}
$$
For the given <centrifugal governor>, $\cos30^\circ=\sqrt3/2$ and
$$
\delta=0.0535898\,\mathrm m,\qquad F_s=26.7949\,\mathrm N,\qquad \omega^2=133.988\,\mathrm{s^{-2}}.
$$
Hence
$$
\boxed{\omega=11.58\,\mathrm{rad\,s^{-1}}\simeq110.5\text{ revolutions per minute}.}
$$
The factor of two in the collar balance matters: both lower rods support the same light collar. The spring force is not an extra downward force of this full magnitude on each ball; its effect on each ball is transmitted through its own lower rod. The formula is the steady <spring-loaded centrifugal governor> balance.
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