Provided info:
A hollow cylinder (mass $m1$, radius $R$, $\delta$ << $R$) is pivoted on two mass-less bearings. In the hollow cylinder a plate (mass $m2$, radius $r$) rolls without sliding.
The moments of inertia are denoted by $I_i$.
The movement happens in a 2D plane (Fig. 2).
The task at hand is to "Find a connection between $\dot{\theta}$, $\dot{\psi}$ and $\dot{\phi}$."
The provided solution states that:
$(R-r)\dot{\theta} = R\dot{\psi} - r\dot{\phi}$, meaning that $\delta$ can be disregarded.
My own approach was to say that the distance $r\dot{\phi}$ covered by the rotation $\dot{\phi}$ of the plate would equal the distance $-R\dot{\theta}$ when the cylinder is standing still. Adding the potential shift in the surface caused by the rolling of the cylinder leads me to believe that the relationship should be:
$$r\dot{\phi} = R(\dot{\psi} - \dot{\theta})$$
Would appreciate any input on the error in my solution.
There is one tiny aspect that you did not use "rolls without sliding". When the surface of the cylinder rotates, this will cause an additional rotation for the $m_2$ object. So while you consider the motion of the center of mass, you need to account for the fact that the instantaneous velocities of $m_2$ and $m_1$ at the point of contact are the same.
For a circular object of radius $r$, rolling on a straight surface with angular velocity $\dot{\phi}$, the velocity at the point of contact is $0$, but the center of object (at height $r$) moves with a velocity $r \dot{\phi}$. In this case, we need to add the velocity of the surface. Accounting for the directions of increasing $\dot{\phi}$ and $\dot{\psi}$, you get the velocity of the center of $m_2$ is $R\dot{\psi}-r\dot{\phi}$. Now if you write the same velocity in terms of $\dot{\theta}$ you get $(R-r)\dot{\theta}$. By equating the two, the answer is then $$(R-r)\dot{\theta}=R\dot{\psi}-r\dot{\phi}$$