Background:
Let’s say we have a trajectory $\vec{y}(t)$ expressed in coordinates of an accelerated reference frame. In an inertial reference frame the trajectory is given by: $\vec{x}(t)=R(t)\vec{y}(t)+ \vec{b}(t)$, $R \in SO(3), \vec{b} \in \mathbb{R}^3$. By differentiation we find that:
$ m\ddot{\vec{y}}=R^T\vec{F}_x -2mR^T\dot{R}\dot{\vec{y}}-mR^T\ddot{R}\vec{y}-mR^T\ddot{\vec{b}} \tag{1} $
Where $\vec{F}_x$ are the forces acting on the object in $x$ coordinates. Noticing that $R^T\dot{R}$ is anti-symmetric we can write: $R^T\dot{R}\vec{y}=: \vec{\omega} \times \vec{y}$, thus we have our equation of motion in y-coordinates:
$ m\ddot{\vec{y}}=\vec{F}_y -2m\vec{\omega} \times {\dot{\vec{y}}}-m \dot{\vec{\omega}}\times\vec{y}-m \vec{\omega} \times (\vec{\omega} \times \vec{y})-m\vec{a} \tag{2} $
Where $\vec{a}$ is the acceleration of the origin of the non inertial reference frame in y-coordinates.
We call $-m \vec{\omega} \times (\vec{\omega} \times \vec{y}) \tag{3} $
the centrifugal force (in y-coordinates). I hope I‘m correct until now.
Question:
I don’t understant, however, how this term makes sense. Let’s say I‘m sitting on a carousel facing inwards. I therefore am the accelerated reference frame and describe the world in y-coordinates (Let’s say my head is at the origin of the y-coordinates, I am looking along the $y_1$-axis to the carousel‘s center and the $y_3$ axis is normal to the ground). I would expect to observe a force pointing towards the negative $y_1$ direction, wouldn’t I? But if I plug everything into $-m \vec{\omega} \times (\vec{\omega} \times \vec{y}) =-m\omega^2 \hat{y_3} \times(\hat{y_3} \times0)$ I get 0! (I chose $\vec{y}=0$ because my head is at the origin). Why doesn’t my head (and everything else on the $y_3$ Axis) experience a centrifugal force?
(Follow up question: Am I correct to express $\vec{\omega}$ in the y basis?)
Thank you:)
