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As I tried to say before in this thread, in the mathematical derivation of the inertial forces in rotating reference frames (rotating against the class of inertial frames to be clear), the inertial forces belong to the left-hand side of Newton's equation ##m \vec{a}=\vec{F}##, i.e., from expressing the components of ##\vec{a}## with respect to the rotating basis in terms of the corresponding position-vector components and its time derivatives:Dale said:As others have said, you cannot feel or measure an inertial force like the centrifugal force. All that you can do is to infer it through the motion of the object wrt some frame given all of the directly measurable real forces.
For example, in a rotating reference frame you do not measure the centrifugal force on a co-rotating object. You measure the real centripetal force, and then because the object is not accelerating you infer that there must exist a centrifugal “inertial” force which balances the real centripetal force.
$$\vec{a}' = \mathrm{D}_t^2 \vec{x}' = \ddot{\vec{x}}' + 2 \vec{\omega}' \times \dot{\vec{x}} + \vec{\omega}' \times (\vec{\omega}' \times \vec{x}') + \dot{\vec{\omega}}' \times \dot{\vec{x}}'.$$
Here ##\vec{x}'## etc. are the components of vectors wrt. the rotating basis, and dots denots usual time derivatives, while ##\mathrm{D}_t## is the "covariant time derivative",
$$\mathrm{D}_t \vec{V}'=\dot{\vec{V}}' + \vec{\omega}' \times \vec{V}',$$
and ##\vec{\omega}'## are the components with respect to the rotating basis of the momentary angular velocity of the rotating basis wrt. an arbitrary inertial basis.