= Solution
Apply the <rotating-frame derivative formula> first to position and then to its inertial <velocity>:
$$
\left(\frac{d\mathbf r}{dt}\right)_I=\dot{\mathbf r}+\boldsymbol\omega\times\mathbf r,
$$
$$
\left(\frac{d^2\mathbf r}{dt^2}\right)_I
=\ddot{\mathbf r}+2\boldsymbol\omega\times\dot{\mathbf r}
+\boldsymbol\omega\times(\boldsymbol\omega\times\mathbf r).
$$
The angular velocity is constant, so there is no term involving $\dot{\boldsymbol\omega}$. By <Newton's second law> in the <inertial frame>, the inertial acceleration is $\mathbf F/m$. Rearrangement gives the <equation of motion in a rotating frame>:
$$
\boxed{m\ddot{\mathbf r}=\mathbf F-2m\boldsymbol\omega\times\dot{\mathbf r}
-m\boldsymbol\omega\times(\boldsymbol\omega\times\mathbf r).}
$$
The two additional accelerations are the <Coriolis acceleration> and <centrifugal acceleration>.
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