= Solution
The <rotating-frame derivative formula> is $(d\mathbf A/dt)_S=(d\mathbf A/dt)_{S'}+\boldsymbol\omega\times\mathbf A$. Applying it twice to the <position> vector, with constant <angular velocity>, gives
$$
\mathbf a_S=\mathbf a_{S'}+2\boldsymbol\omega\times\mathbf v_{S'}+\boldsymbol\omega\times(\boldsymbol\omega\times\mathbf r).
$$
Solving for $\mathbf a_{S'}$ identifies the displayed terms: its left side is the <acceleration> relative to the <rotating reference frame>, the first right-hand term is the inertial <acceleration>, $-2\boldsymbol\omega\times\mathbf v_{S'}$ is the <Coriolis acceleration>, and $-\boldsymbol\omega\times(\boldsymbol\omega\times\mathbf r)$ is the outward <centrifugal acceleration>.
For a particle of <mass> $m$, replace $\mathbf a_S$ by the actual <force> divided by $m$ using <Newton's second law>. The remaining terms then appear as apparent <forces> in $S'$. There is no <Euler acceleration> because $\boldsymbol\omega$ is constant and no translational acceleration term because the common origin is fixed in the <inertial frame>.
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