= Free particle in rotating coordinates
{title2=$\mathbf r_{\rm rot}=(ut\cos\Omega t,-ut\sin\Omega t)$}
Consider a ball moving horizontally with no net horizontal <force> in the laboratory. Treat its horizontal motion as that of a free particle: $\mathbf r_{\rm lab}=(ut,0)$ with constant $u>0$. Axes rotating counterclockwise at constant <angular velocity> $\boldsymbol\Omega=\Omega\mathbf e_z$ about the launch point assign it the <position>
$$
\mathbf r_{\rm rot}=
\begin{pmatrix}\cos\Omega t&\sin\Omega t\\-\sin\Omega t&\cos\Omega t\end{pmatrix}
\begin{pmatrix}ut\\0\end{pmatrix}
=\begin{pmatrix}ut\cos\Omega t\\-ut\sin\Omega t\end{pmatrix}.
$$
This is an exact <rotation matrix> transformation, with the axes coinciding at $t=0$. Taking $u=0.25\ \mathrm{m\,s^{-1}}$, $\Omega=0.8\ \mathrm{rad\,s^{-1}}$, and $0\leq t\leq3\ \mathrm s$ gives the two views below. At $t=3\ \mathrm s$, the laboratory <position> is $(0.75,0)\ \mathrm m$, but the rotating <position> is approximately $(-0.5530,-0.5066)\ \mathrm m$. Both assign the same distance $0.75\ \mathrm m$ from the origin.
\Image[/coriolis-free-ball-two-frames.png]
{title=One free ball: straight laboratory trajectory and curved rotating-coordinate trajectory}
The entire rotating <acceleration> must include the <centrifugal acceleration> as well as the <Coriolis acceleration>. To see the balance explicitly, write $c=\cos\Omega t$ and $s=\sin\Omega t$. Direct <differentiation> gives
$$
\mathbf v_{\rm rot}=u(c-\Omega ts,-s-\Omega tc),
\qquad
\ddot{\mathbf r}_{\rm rot}=u(-2\Omega s-\Omega^2tc,-2\Omega c+\Omega^2ts).
$$
The two apparent contributions are
$$
\begin{aligned}
\mathbf a_{\rm cor}&=u(-2\Omega s-2\Omega^2tc,-2\Omega c+2\Omega^2ts),\\
\mathbf a_{\rm cf}&=u\Omega^2t(c,-s),\\
\ddot{\mathbf r}_{\rm rot}&=\mathbf a_{\rm cor}+\mathbf a_{\rm cf}.
\end{aligned}
$$
At launch $\mathbf r=0$, so only the <Coriolis acceleration> contributes to the initial bending. Later the <centrifugal acceleration> also contributes; the full curved <trajectory> cannot be explained by keeping only the <Coriolis force>. In particular, $|\mathbf v_{\rm rot}|^2=u^2(1+\Omega^2t^2)$ grows even though the <Coriolis force does no work>: the rotating-frame <kinetic energy> increase is supplied by <work> from the <centrifugal force>. The laboratory <kinetic energy> remains $mu^2/2$.
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