Lagrangian formalism for non-inertial reference frames I was solving the exercise where the massless ring with radius $R$ is rotating around axis (shown in the picture) with angular velocity $\omega$. On the ring is a point-object with mass $m$ which moves freely without friction. I want to write equation for the system with Lagrangian formalism and I have a couple of questions about it.

I will use $J_{0} = mR^2$ for moment of inertia of the point-object.
If I write the kinetic energy for system with intuition I get
$$T = \frac{(J_0 + mR^2)}{2}\omega^2 = \frac{1}{2}m[r^2 + r^2]\omega^2,$$
but if I try to do it in non-inertial frame of reference I get
$$T = \frac{1}{2}m[\vec{v_{rel}} + \vec{\omega}\times(\vec{r_{rel}} + \vec{r_0})]$$
where $r_{rel}$ is position of point-object in non-inercial reference frame (with origin in center of ring) and $r_0$ is vector from origin of inertial reference frame to the non-inertial reference frame.
$$T = \frac{1}{2}m[\dot r \vec{e_r}+ r\dot\phi \vec{e_\phi} + \omega\vec{e_z}\times ((r-r\cos\phi)\vec{e_r} + r\sin\phi \;\vec{e_\phi})]^2$$
$$T = \frac{1}{2}m[r^2 + r^2 -2rr\cos\phi]\omega^2$$ this expression has one more term that the initial one. What did I do wrong?
Did I make mistake when going from inertial to non-inertial reference frame?
How could I solve this exercise with use of the moved circle equation $R^2 = (x+R)^2 + y^2$?
 A: 
How could I solve this exercise with use of the moved circle equation

just rotate the position vector to the mass.
$$\vec \rho= \left[ \begin {array}{ccc} \cos \left( \omega\,t \right) &-\sin
 \left( \omega\,t \right) &0\\ \sin \left( \omega\,t
 \right) &\cos \left( \omega\,t \right) &0\\  0&0&1
\end {array} \right]
\,\begin{bmatrix}
x -R\\
  y \\
  0 \\
\end{bmatrix}$$
I assumed that the rotation is about the z axes .
the kinetic energy is now
$$T=\frac m2 \vec{\dot{\rho}}\cdot\vec{\dot\rho}\\\\
\begin{align*}
&\vec{\dot{\rho}}=
 \left[ \begin {array}{c} -\cos \left( \omega\,t \right) {\it \dot{x}}+
 \left(  \left( R+x \right) \omega-{\it \dot{y}} \right) \sin \left( \omega
\,t \right) -\cos \left( \omega\,t \right) \omega\,y
\\  -\sin \left( \omega\,t \right) {\it \dot{x}}-\sin
 \left( \omega\,t \right) \omega\,y+ \left(  \left( -R-x \right)
\omega+{\it \dot{y}} \right) \cos \left( \omega\,t \right)
\\  0\end {array} \right]
\end{align*}  $$
and you have one holonomic constraint equation
$$R^2=x^2+y^2$$
