= Solution
Let a coordinate frame rotate with <angular velocity> $\boldsymbol\Omega$ relative to an <inertial frame>. For any <vector> $\mathbf u$,
$$
\left(\frac{d\mathbf u}{dt}\right)_{\rm inertial}
=\left(\frac{d\mathbf u}{dt}\right)_{\rm rotating}+\boldsymbol\Omega\times\mathbf u.
$$
Applying this twice to position, with coincident nonaccelerating origins, gives
$$
\mathbf a_{\rm inertial}=\mathbf a_{\rm rotating}
+2\boldsymbol\Omega\times\mathbf v_{\rm rotating}
+\boldsymbol\Omega\times(\boldsymbol\Omega\times\mathbf r)
+\dot{\boldsymbol\Omega}\times\mathbf r.
$$
<Newton's second law> written using rotating coordinates therefore contains the <Coriolis force>
$$
\boxed{\mathbf F_C=-2m\boldsymbol\Omega\times\mathbf v_{\rm rotating}.}
$$
The other inertial corrections are the <centrifugal force> and, for changing <rotation> rate, the <Euler force>. An accelerating origin introduces an additional translational <fictitious force>.
The <Coriolis force> depends on <velocity> and is perpendicular to it, so $\mathbf F_C\cdot\mathbf v_{\rm rotating}=0$: \b[the <Coriolis force> does no <work> in the <rotating frame>]. It bends trajectories without directly changing their rotating-frame speed. It is absent for motion parallel to the <rotation> axis.
For locally horizontal motion on Earth at latitude $\lambda$, take $x$ east, $y$ north and $z$ up. Keeping the vertical component $\Omega\sin\lambda$ gives
$$
\dot v_x=fv_y,\qquad \dot v_y=-fv_x,
\qquad f=2\Omega\sin\lambda.
$$
Horizontal motion is deflected right in the northern hemisphere and left in the southern hemisphere. In the ideal absence of other horizontal <forces> these equations give inertial circles with <angular frequency> $|f|$. The same velocity-dependent correction explains the <precession> of a <Foucault pendulum>, whose precession-rate magnitude is $\Omega|\sin\lambda|$. Terrestrial vertical motions also involve Earth's horizontal <rotation> component, so the reduced horizontal equations are not the full <force> law.
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