How are quadratic terms of apparent force in a rotating reference frame derived from the usual decomposition into centrifugal and Coriolis terms?

Question

Usually the total acceleration a' seen in a rotating reference frame (like earth) is written down as

$$\vec a' = -\vec \omega \times \left(\vec \omega \times \vec r' \right) - 2 \cdot \vec \omega \times \vec v' \tag{1}$$

https://en.wikipedia.org/wiki/Rotating_reference_frame#Relation_between_accelerations_in_the_two_frames

First term is called usually "centrifugal force" while the right term is the Coriolis force.

But when I consider motion with (zonal) velocity u (along the east) on the equator, the total centrifugal force in radial direction z would be

$$a_z = \frac{(u+\omega R)^2}{R} = \omega^2R+u^2/R+2 \omega \cdot u \tag{2}$$

because the total tangential speed is the sum of earth's rotation and zonal speed.

Now I wonder, where the quadratic term would be derived from equation (1). I see only the first and third term arising from (1) but not the middle...

The answer must be simple, because it is just kinematics, not even physics, but I cannot find it.

Answer 1

Your equation $(1)$ is valid only for a stationary point in the inertial frame. In general, you have: $$a’+\omega’\times (\omega’\times r’)+2\omega’\times v’=a$$ for uniform rotation ($\dot\omega=0$) and no translation.

The more general formula is: $$a’+\omega’\times (\omega’\times r’)+\dot \omega’\times r’+a_t’+2\omega’\times v’=a$$ with $a_t’$ the translation acceleration. In fact you can identify the terms: $$ \omega’\times (\omega’\times r’)+\dot \omega’\times r’+a_t’ $$ as the acceleration in the inertial frame of a point fixed in the non inertial frame coinciding at the given time with the point of interest. This usually makes the calculations more intuitive.

In your example, you have $a’=\frac{u^2}{R}$ since even in the Earth’s reference frame, the point is accelerating from following the curved path of the Equator. The formulas are therefore consistent.

Hope this helps.