= Solution
Use $D=(d/dt)_S$, $D'=(d/dt)_{S'}$, and $\mathbf v'=D'\mathbf r$. By the <rotating-frame derivative formula>, $D\mathbf r=\mathbf v'+\boldsymbol\omega\times\mathbf r$. Apply the <product rule> for the <cross product> and the same <rotating-frame derivative formula> again:
$$
D^2\mathbf r=D\mathbf v'+(D\boldsymbol\omega)\times\mathbf r+\boldsymbol\omega\times D\mathbf r.
$$
We have $D\mathbf v'=D'\mathbf v'+\boldsymbol\omega\times\mathbf v'$ and $D\boldsymbol\omega=D'\boldsymbol\omega$, since $\boldsymbol\omega\times\boldsymbol\omega=0$. Therefore
$$
\boxed{D^2\mathbf r=D'^2\mathbf r+2\boldsymbol\omega\times D'\mathbf r+(D'\boldsymbol\omega)\times\mathbf r+\boldsymbol\omega\times(\boldsymbol\omega\times\mathbf r).}
$$
Rearranging this identity into an <equation of motion in a rotating frame> produces the apparent <Coriolis acceleration>, <Euler acceleration> and <centrifugal acceleration>, with signs opposite to their corresponding terms on the right of this inertial <acceleration> identity. No translational term is needed because the two origins coincide.
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