Let $\{\mathbf e_i(t)\}$ be an orthonormal <basis> fixed in $S'$ and write $\mathbf a=a_i\mathbf e_i$. A basis vector rotating with <angular velocity> $\boldsymbol\omega$ satisfies
$$
\left(\dot{\mathbf e}_i\right)_S=\boldsymbol\omega\times\mathbf e_i.
$$
Differentiating the components and the basis vectors in $S$ therefore gives the <rotating-frame derivative formula>
$$
(\dot{\mathbf a})_S
=\dot a_i\mathbf e_i+a_i\boldsymbol\omega\times\mathbf e_i
=(\dot{\mathbf a})_{S'}+\boldsymbol\omega\times\mathbf a.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / ia / Paper 4 / 12A / a / Solution

If a frame rotates with <angular velocity> $\boldsymbol\Omega$ relative to an inertial frame, the derivatives of any vector $\mathbf a(t)$ satisfy
$$
\left(\frac{d\mathbf a}{dt}\right)_{\rm inertial}
=\left(\frac{d\mathbf a}{dt}\right)_{\rm rotating}
+\boldsymbol\Omega\times\mathbf a.
$$
Applying this formula twice to a <position> gives the <Coriolis acceleration>, <Euler acceleration>, and <centrifugal acceleration> terms.

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 Incoming links: Rotating-frame derivative formula