Angular velocity of a cylinder rolling inside of another cylinder Problem

Find the kinetic energy of a homogeneous cylinder of radius $a$ rolling inside a cylindrical surface of radius $R$.


My attempt
First, I known that the kinetic energy of the rolling cylinder "measured" in the center of the external cylinder is the sum of kinetic energy of the center of mass plus the rotational kinetic energy of the body about the center of mass.
$$T = \frac{1}{2}M|\dot{\vec{r}_{cm}}|^{2} + \frac{1}{2}\int_{V}\rho|\vec{\omega} \times \left(\vec{r}-\vec{r}_{cm}\right)|^2dV$$
This leads to
$$T = \frac{1}{2}M|\dot{\vec{r}_{cm}}|^{2} + \frac{1}{2}\omega^{T}I_{cm}\omega$$
where $I_{cm}$ is the inertia tensor with the origin in the center of mass. In this particular case, the vector $\vec{\omega}$ is perpendicular to the screen, if we  asume that this direction is the z direction, and the inertia tensor $I_{cm}$ is diagonal because that direction is a principal axe, we obtain:
$$\vec{\omega} = \omega \hat{z} \\ T = \frac{1}{2}M|\dot{\vec{r}_{cm}}|^{2} + \frac{1}{2}I_{z,cm}\omega^2$$
The first term is easy to obtain:
$$\frac{1}{2}M|\dot{\vec{r}_{cm}}|^{2} = \frac{1}{2}M\left(R-a\right)^2\dot{\phi}^2$$
In order to obtain an expression for the second term we need to know $\omega$. The angular velocity can be obtained using the rolling condition (translational velocity and rotational velocity cancels out in the contact point):
$$\vec{v_{rot}} + \vec{v_{tr}} = \vec{0}$$
In the contact point, these two velocity vectors are in the same line; but in opposite directions, for that reason, we can write:
$$|\vec{v_{rot}}| = |\vec{v_{tr}}|$$
Before this point I have some questions.
Questions
In the book: (Course of theoretical physics) L.D Landau, E. M. Lifshitz the solution says:
$$a\omega = (R - a)\dot{\phi}$$
Therefore
$$\omega = \frac{(R - a)}{a}\dot{\phi}$$
I think that this is the "rolling condition". If this is true, this implies that the rotational velocity in the contact point is:
$$|\vec{v_{rot}}| = a\omega$$
For me this is correct, but the translational velocity in the contact point would be:
$$|\vec{v_{tr}}| = (R-a)\dot{\phi}$$
But this is the translational velocity of the center of mass, not the translational velocity of the contact point. The translational velocity of the contact point should be:
$$|\vec{v_{tr}}| = R\dot{\phi}$$
And $\omega$ would be $\omega = \frac{R}{a}\dot{\phi}$.
Note: In other problems where translational movement is in a constant direction, the translational velocity of all points of the rigid body is the same, but in this case, the translational movement is circular, for that reason the translational velocity in the contact point and the translational velocity in the center of mass are different.

*

*I would like to know what I am doing wrong?

*Why the translational velocity of the contact point is the translational velocity of the center of mass
 A: When in doubt, it’s best to revert to the most general definition. The no slip condition is simply $v=0$ for the contact point that moves with the cylinder. Its velocity has two contributions: the translation of the small cylinder’s axis, I’ll name it $O$, and the rotation about $O$. You get:
$$
v=(R-a)\dot \phi-\omega a
$$
and setting $v=0$ gives you LL’s solution.
The mistake in your reasoning is to replace the translation motion of $O$ by a rotational motion of the small cylinder about $O’$ the axis of the big cylinder. For $O$ this dos not change anything, but for the contact point the two are not equivalent. This is because by definition $\omega$ captures all the rotation motion of the small cylinder from your kinetic energy argument. Therefore you cannot add additional rotation motion, and can only add translation motion.
It’s a matter of precisely defining in what reference frame you measure angular velocity. Your reasoning would be correct if $\omega$ represented the angular velocity of the cylinder in the reference frame rotating about $O’$ at velocity $\dot \phi$ (in this frame, $\phi$ would be constant). Your rolling condition would be correct, but your calculation of the kinetic energy wouldn’t be as you would miss a contribution due to the $\dot \phi$ rotation. If you amend the formula for kinetic energy, then yes your approach would be correct.
Hope this helps.
A: Kinetic energy of a rigid body can be written referred to the center of mass as
$K = \dfrac{1}{2} m |\mathbf{v}_G|^2 + \dfrac{1}{2} \boldsymbol{\omega} \cdot \mathbb{I}_G \cdot \boldsymbol{\omega} $.
If we assume pure rolling, the system has one degree of freedom: let's use $\phi$ as the only degree of freedom.
The angular velocity of the body reads $\boldsymbol{\omega} = \dot{\theta} \mathbf{\hat{k}}$, being $\theta$ the angle of rotation of the disk, $\mathbf{\hat{k}}$ the unit vector pointing outwards the plane of the screen.
The velocity of the center of mass reads
$\mathbf{v}_G = a \dot{\theta} \mathbf{\hat{t}} = (R-a)\dot{\phi} \mathbf{\hat{t}}$, so that the kinematic constraints reads
$a\dot{\theta} = (R-a)\dot\phi$,
and the kinetic energy reads
$K = \dfrac{1}{2} m (R-a)^2 \dot{\phi}^2 + \dfrac{1}{2} I_G \dfrac{(R-a)^2}{a^2}\dot\phi^2$,
being $I_G = \frac{1}{2}m a^2$ the moment of the inertia of a disk with uniform mass distribution w.r.t. its center of mass. Thus, we can rearrange the formula as
$K = \dfrac{3}{4} m (R-a)^2 \dot{\phi}^2 = \\
\quad = \dfrac{3}{4} m a^2 \dot{\theta}^2$
