Oscillations of a cylinder inside a cylinder Please read the whole thing I'm asking for a concept not the problem itself, but I have to show the problem to explain myself

Find the period of the small oscillations of a cylinder of radius r that rolls without
slide inside a cylindrical surface of radius R.

I tried going with the conservation of mechanic energy $$E_m$$ which is equal to $$E_m=E_k+E_p+E_r$$ being $k, p, r$ kinetic, potential and rotational (energy). We know that $E_k=\frac{1}{2}mv^2=\frac{1}{2}m(\frac{d\theta}{dt}(R-r))^2$, being $\theta$ the angle of the small cilinder from the center of the big cilinder with respect to the vertical, and $E_p=-mg(R-r)\cos(\theta)$ but for $E_k=\frac{1}{2}I\omega_{cm}^2$ I had no idea on how to continue, since $\omega_{cm}$ here refers to the rotation of the small cylinder with respect to it's center of mass, not the rotation of the small cylinder with respect to the big cylinder so we can't do $\omega_{cm}=\frac{d\theta}{dt}$
Doubt. So I looked the solution and it said that the velocity of the small cylinder with respect to the big cylinder was equal to the velocity of the points inside the small cylinder with respect to the center of masses of the small cylinder, why is that? That makes no sense to me. The procedure they make afterwards is $v_{cm}=v_{O}\Rightarrow \omega_{cm}=\frac{R-r}{r}\omega_0=\frac{R-r}{r}\frac{d\theta}{dt}$ being $\omega_0$ the rotation of the small cylinder with respect to the center of the big cylinder.
 A: Well honestly I couldn't undertsand the paragraph part of what are you trying to say but mathematically equation is correct, let me explain my way.
There are two types of rotations of small cylinder, one on its own axis, and one about the big cylinder's axis. Let us take the case of a sliding cylinder, not a rolling, one. We can write:
$$v_{cm}=\omega (R-r)$$
where $\omega$ is rotation about big cylinder's axis, and $v_{cm}$ is velocity of small cylinder's center of mass. Now here due to pure rolling of small cylinder $v_{cm}=\omega_{cm}r$. Here $\omega_{cm}$ is about the small cylinder's axis. This will yield the same result that you got.
Hope it helps, ask any doubts.
Edit:

The pure rolling means that bottom-most point of sphere or its point of contact with ground remains at rest with repect to ground. Two motions are going on simultaneosuly: Translational and rotational, which is summed up in the above image and leads us to: $$v_{cm}=\omega_{cm}r$$
A: 
starting with the rolling condition
$$r\,\varphi=(R-r)\theta\tag 1$$
where $~\varphi~$ is the rotation of the small cylinder (generalized coordinate)
the position vector to the center of the small cylinder is
$$\vec R_c=\begin{bmatrix}
    x \\
    y \\
  \end{bmatrix}=(R-r)\begin{bmatrix}
    \sin(\theta(\varphi)) \\
    \cos(\theta(\varphi)) \\
  \end{bmatrix}\quad\Rightarrow\\
T_k=\frac m2 (\dot x^2+\dot y^2)
$$
the potential energy is $~U=-m\,g\,y~$ and the kinetic energy of the small cylinder $T_r=\frac 12\,I_c\,\dot\varphi^2=\frac 14\,m\,r^2\,\dot\varphi^2$
The Langragian
$$L=T_k+T_r-U=L(\varphi~,\dot\varphi)$$
with EL and $~\sin(a\,\varphi)=a\,\varphi$
$$\omega^2=\frac 23\frac{g}{R-r}$$
