= Solution
At $P$, use the local right-handed basis $\mathbf e_E,\mathbf e_N,\mathbf e_U$ pointing east, north and vertically outward. The Earth's rotation vector has components
$$
\boldsymbol\omega=\omega(\sin\theta\,\mathbf e_N+\cos\theta\,\mathbf e_U),\qquad
\mathbf r=R\mathbf e_U.
$$
Initially $\dot{\mathbf r}=0$ in the <rotating reference frame>, so the <Coriolis acceleration> is zero. Evaluating the <centrifugal acceleration> gives
$$
-\boldsymbol\omega\times(\boldsymbol\omega\times\mathbf r)
=\omega^2R(\sin^2\theta\,\mathbf e_U-\sin\theta\cos\theta\,\mathbf e_N).
$$
The <gravitational acceleration> is $-g\mathbf e_U$. Hence
$$
\ddot{\mathbf r}(0)
=-(g-\omega^2R\sin^2\theta)\mathbf e_U
-\omega^2R\sin\theta\cos\theta\,\mathbf e_N.
$$
The small angle $\delta$ from the downward vertical therefore satisfies
$$
\tan\delta=\frac{\omega^2R|\sin\theta\cos\theta|}{g-\omega^2R\sin^2\theta},\qquad
\boxed{\delta\simeq\frac{\omega^2R}{g}|\sin\theta\cos\theta|.}
$$
For the northern hemisphere, $0\leq\theta\leq\pi/2$, this is the printed expression without absolute values. More generally, the printed signed meridional expression specifies the direction of the tilt; an unsigned angle uses its absolute value. The horizontal component is toward the equator and vanishes at both poles and at the equator.
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