Cylinder inside a cylinder - moment of inertia 
A homogeneous cylinder with radius a and mass m rolls in a hollow cylinder with radius R. Determine the kinetic energy of the cylinder as function of $\dot{\theta}$. 


I'm sorry for reposting this, but I found this question this morning while browsing and tried myself to it, but didn't get even an approach. I would have commented on the original post but I don't have enough reputation yet (no idea how to get reputation). 
Anyway, my thoughts so far: 
Since the inner cylinder does not slide I can assume $v=r\dot{\theta}$, right? As the original poster I was thinking that the kinetic energy in this case would be the rotational energy, so $E=I\dot{\theta}^2=\frac{1}{2}MR^2\dot{\theta}^2$. But then again, the problem says to determine the kinetic energy as a function of $\dot{\theta}$, which means that the rotational energy is not constant. 
I don't know how to approach this. Anyone got any tips? 
 A: You correctly identified there are two angles of interest, labeled $\theta$ and $\phi$. I actually want to pick two different angles for my analysis - see this diagram:

First thing to note is that if the cylinder rolls without sliding, the length of the green arc and the red arc must be the same. 
Length of green arc: $(\phi + \theta) a$
Length of red arc: $\theta R$
$$(\phi + \theta) a = \theta R\\
\theta = \frac{a}{R-a}\phi$$
(You could put a minus sign in there - the rotation for one is clockwise when the other is anticlockwise). The total energy of the cylinder is the rotational energy ($\frac12 I \dot\phi^2$) plus the linear kinetic energy $\frac12 m v^2 = \frac12 m ((R-a)\dot\theta)^2$
You can simplify this (express everything in terms of $\theta$) and set the sum of kinetic and potential energy constant. You end up with an equation for simple harmonic motion (for small displacements) with an apparently greater inertia (longer period) than would be expected from a mass at the end of a string of length $R-a$ ;
See how far you get with that nudge.
