= Solution
For this <gyrocompass>, let $\mathbf n,\mathbf w,\mathbf z$ be local north, west and upward unit <vectors>, forming a right-handed basis. <Earth's rotation> has local components
$$
\boldsymbol\Omega_E=\Omega_E\cos\lambda\,\mathbf n+\Omega_E\sin\lambda\,\mathbf z.
$$
Let the spin axis point along $\mathbf u=\cos\theta\,\mathbf n+\sin\theta\,\mathbf w$, and write its dominant <angular momentum> as $J_s\mathbf u$, where $J_s=I_1\omega>0$. In <mechanical equilibrium> relative to <Earth>, this leading <angular momentum> is fixed in the terrestrial <rotating reference frame>. The bearings can exert horizontal <torques> to keep the axle horizontal, but free rotation about the vertical implies $G_z=0$. The vertical component of the <rotating-frame derivative formula> therefore requires
$$
0=(\boldsymbol\Omega_E\times J_s\mathbf u)\cdot\mathbf z
=\Omega_EJ_s\cos\lambda\sin\theta.
$$
Away from either geographical pole, $\cos\lambda\ne0$, hence
$$
\boxed{\theta=0\ \text{or}\ \pi:\quad\text{the axle lies north--south}.}
$$
This is a necessary <equilibrium> condition; the next part distinguishes its stable and unstable spin orientations. The rapid-spin approximation neglects <angular momentum> terms of order $I_2\Omega_E$ compared with $I_1\omega$. At either pole the horizontal component of <Earth's rotation> vanishes, and this north-seeking mechanism gives no preferred horizontal direction. The paper explicitly excludes $90^\circ$; if latitude is signed, $-90^\circ$ must likewise be excluded.
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