user079622 said:
Normal force dont have component to the right when door opens constantly or accelerate(increase door rpm).
It has component to the right only when door decelerate(rpm slows down)Because of inerita, glass resist to change position in space, centripetal force(friciton between plastic and glass bottom/back) is not enough,so glass slide.
I don't know, why we still discuss this. There are at least two postings, solving the (idealized) problem. Just once again:
We assume a mass that can glide without friction along a rod, which is rotating with constant angular velocity in a horizontal plane. There's no net external force, i.e., the Lagrangian is
$$L=T=\frac{m}{2} \dot{\vec{r}}^2.$$
The position vector can be described as
$$\vec{r}(t)=r(t) \begin{pmatrix} \cos (\omega t) \\ \sin(\omega t) \\ 0 \end{pmatrix}.$$
The derivative is
$$\dot{\vec{r}} = \dot{r} \begin{pmatrix} \cos (\omega t) \\ \sin(\omega t) \\ 0 \end{pmatrix} + r \omega \begin{pmatrix}-\sin(\omega t) \\ cos(\omega t) \\0 \end{pmatrix}.$$
Thus
$$L=\frac{m}{2}(\dot{r}^2+r^2 \omega^2).$$
The equation of motion is
$$\mathrm{d}_t \frac{\partial L}{\partial \dot{r}}=m \ddot{r}=\frac{\partial L}{\partial r}=m r \omega^2.$$
The solution is
$$r(t)=r_0 \cosh(\omega t) + \frac{v_0}{\omega} \sinh(\omega t).$$
The force on the mass is
$$\vec{F}=m \ddot{\vec{r}}=m (\ddot{r}-\omega^2 r) \begin{pmatrix} \cos (\omega t) \\ \sin(\omega t) \\ 0 \end{pmatrix} + 2m \omega \dot{r} \begin{pmatrix}-\sin(\omega t) \\ cos(\omega t) \\0 \end{pmatrix} = 2 m \omega^2 [A \sinh(\omega t) + B \cosh(\omega t)]\begin{pmatrix}-\sin(\omega t) \\ \cos(\omega t) \\0 \end{pmatrix}.$$
This force is the force imposed by the rod to keep the point on the rod.
