= Solution
<Newton's second law> takes its familiar form $m\mathbf a=\mathbf F$ in an <inertial frame>. Axes fixed to a turning apparatus or to the <Earth> form a <rotating reference frame>. Extra <inertial forces> account for the <acceleration> of these coordinates. Their values depend on the chosen <reference frame>; physical interactions such as <gravity>, <tension> and contact <forces> must still be included separately.
Let the rotating origin have inertial <acceleration> $\mathbf A$, and let its axes have <angular velocity> $\boldsymbol\Omega(t)$. For any <vector> $\mathbf q$, the <rotating-frame derivative formula> is
$$
\left(\frac{d\mathbf q}{dt}\right)_{\rm I}=\left(\frac{d\mathbf q}{dt}\right)_{\rm R}+\boldsymbol\Omega\times\mathbf q.
$$
Each rotating <basis vector> changes by $\boldsymbol\Omega\times\mathbf e_i$, which proves this rule by expanding $\mathbf q$ in that basis. Apply it twice to the position relative to the moving origin. With $\mathbf v'$ and $\mathbf a'$ measured in the rotating axes,
$$
\mathbf a_{\rm I}=\mathbf A+\mathbf a'+2\boldsymbol\Omega\times\mathbf v'+\dot{\boldsymbol\Omega}\times\mathbf r+\boldsymbol\Omega\times(\boldsymbol\Omega\times\mathbf r).
$$
Consequently the equation in the <rotating reference frame> is
$$
m\mathbf a'=\mathbf F-m\mathbf A-2m\boldsymbol\Omega\times\mathbf v'-m\dot{\boldsymbol\Omega}\times\mathbf r-m\boldsymbol\Omega\times(\boldsymbol\Omega\times\mathbf r).
$$
The additional terms are the translational <inertial force>, <Coriolis force>, <Euler force> and <centrifugal force>, respectively. The last two simplify when the <angular velocity> is constant: the <Euler force> vanishes, while the other rotation terms remain.
The <centrifugal force> points away from the rotation axis. If $\mathbf r_\perp$ is the perpendicular displacement from that axis, then
$$
\boxed{\mathbf F_{\rm cf}=m\Omega^2\mathbf r_\perp.}
$$
For uniform rotation it has <potential energy>
$$
\Phi_{\rm cf}=-\frac12m|\boldsymbol\Omega\times\mathbf r|^2,\qquad \mathbf F_{\rm cf}=-\nabla\Phi_{\rm cf}.
$$
A mass at rest relative to a turntable can therefore be described as having an inward <tension> or <friction> balancing its outward <centrifugal force>. In an <inertial frame>, the inward physical <force> instead produces the required <centripetal acceleration>. These are two descriptions of the same motion; one must not insert the rotating-frame outward term into the inertial equation as an additional physical interaction.
A rotating liquid gives a useful quantitative example. For a liquid in <hydrostatic equilibrium> relative to a vessel rotating uniformly about the vertical axis,
$$
\frac{\partial p}{\partial s}=\rho\Omega^2s,\qquad \frac{\partial p}{\partial z}=-\rho g.
$$
Along the <free surface> the <pressure> is constant, so $\Omega^2s\,ds-g\,dz=0$ and
$$
z(s)=z(0)+\frac{\Omega^2s^2}{2g}.
$$
The surface is a <elliptic paraboloid>. This is the balance of <gravity> and the <centrifugal force> in the rotating description. On the <Earth>, the same outward <acceleration> reduces effective <gravity>, most strongly at the equator, and contributes to the planet's equatorial bulge.
The <Coriolis force> depends on relative motion:
$$
\boxed{\mathbf F_{\rm cor}=-2m\boldsymbol\Omega\times\mathbf v'.}
$$
It is perpendicular to both $\boldsymbol\Omega$ and $\mathbf v'$, vanishes for motion along the rotation axis, and reverses when the relative <velocity> reverses. The <dot product> gives
$$
\mathbf F_{\rm cor}\cdot\mathbf v'=0.
$$
Thus the <Coriolis force> does no <work> in the <rotating reference frame>: it changes the direction of motion rather than directly changing the rotating-frame <kinetic energy>. For example, on a horizontal turntable rotating anticlockwise as viewed from above, an outward-moving slider has a clockwise <Coriolis acceleration>, whereas an inward-moving slider has an anticlockwise one. The deflection can be observed by comparing a laboratory camera with a camera rotating with the table.
For slow horizontal motion on the <Earth> at latitude $\lambda$, keep the vertical component $\Omega_E\sin\lambda$ of its <angular velocity>. In east-north coordinates $(u,v)$ the horizontal <Coriolis acceleration> is
$$
(a_E,a_N)=(fv,-fu),\qquad f=2\Omega_E\sin\lambda.
$$
It deflects motion to the right in the northern hemisphere and to the left in the southern hemisphere. In large-scale atmospheric and oceanic motion it can balance a horizontal <pressure gradient>, giving <geostrophic balance>. With $\hat{\mathbf z}$ upward, the steady horizontal equation without <friction> is
$$
f\hat{\mathbf z}\times\mathbf v=-\frac1\rho\nabla_h p.
$$
The resulting <velocity> runs along lines of equal <pressure>. This approximation requires rotation to dominate local inertial changes; it is inappropriate close to the equator where $f$ tends to zero.
The <Foucault pendulum> gives another terrestrial example. In the <small-angle approximation>, its horizontal coordinates satisfy
$$
\ddot x+\omega_0^2x=f\dot y,\qquad \ddot y+\omega_0^2y=-f\dot x,\qquad \omega_0^2=\frac g\ell.
$$
To first order in the small ratio $\Omega_E/\omega_0$, the oscillation plane precesses relative to the ground at angular rate $-\Omega_E\sin\lambda$: clockwise in the northern hemisphere, zero at the equator, and one revolution per sidereal day at a pole. The <Coriolis force> explains this changing orientation without requiring <energy> input. Together these examples show why <inertial forces> are useful whenever the apparatus or environment supplies naturally rotating coordinates.
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