May the Centrifugal Force be with you

The centrifugal force enters into the picture in a frame that rotates relative to an inertial frame. If the angular speed of the rotating frame is $\omega_0$ , then the centrifugal force on a particle at distance $a$ from the axis of rotation is $F'_{cf} = \boxed{m\omega_0^2 r}$ in the outward direction, never mind what the motion of the particle is, or what other forces are acting.

To complete the analysis: From a non-rotating frame, the particle must be subject to a centripetal net force, resulting from an applied force $F_{applied} = F_{net} = -m\omega^2 r,$ where the negative sign indicates inward direction. When this particle is viewed from the rotating frame, it appears to be traveling with relative angular velocity $\omega' = \omega - \omega_0$ . In this frame, we conclude that $F'_{net} = -m(\omega')^2r = -m(\omega - \omega_0)^2 r,$ but we also know that there is a centrifugal force $F'_{cf} = m\omega_0^2 r$ , and well as a Coriolis force $F'_{cor} = 2m\omega_0\omega' r$ . Therefore the particle must be subject to an applied force $F'_{applied} = F'_{net} - F'_{cf} - F'_{cor} = -m(\omega')^2 r - m\omega_0^2 r - 2m\omega_0\omega' r \\ = -m((\omega')^2 + 2\omega_0\omega' + \omega_0^2)r = -m(\omega' + \omega_0)^2 r = -m\omega^2 r.$ Therefore in both systems we draw the same conclusion about the applied force that is driving the rotating particle.