Cylinder inside a cylinder - rotation. 
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 $\dot{\theta}$. 

Alright, I found this problem inside an old textbook of mine. Strangely enough there was no drawing to this, so I was kind of baffled about where $\theta$ actually is. 
I tried drawing this and while drawing I noticed two angles that might fit the description. 

Although while imagining the smalle cylinder rolling inside I thought it should be the angle I coincidentally also denoted as $\theta$, right? 
So I tried approaching this problem with this angle, but I just don't know how to start there. A couple of weeks ago we talked about rotational energy and I was thinking that maybe it only has rotational energy since there is no translational energy if I'm not mistaken?  So, $E=I\omega ^2$? But I don't know how to get $\omega=\dot{\theta}$. 
 A: If you can draw a vector diagram using the centre of the outer cylinder as an origin, you should be able to write i,j vector equations for the position of the centre of mass of the inner cylinder and the point of contact of the inner cyliner with the outer cylinder.  By differentiating the equations with respect to time you should get a relationship between PHI and THETA. 
The kinetic energy would be the total of the angular momentum around the axis through the centre of the inner cylinder and the linear momentum (instaneously) of the centre of mass of the inner cylinder.   You should be able to express the linear momentum as an expression in theta/phi by the rolling condition that the inner cylinder point of contact, so then you should have the expression for the kinetic energy as a function of theta (PHI?).
A: So the hard part is working out the moment of inertia of the small cyclinder. We are told it's uniform so it has inertia around it's centre of $\frac{1}{4}ma^2$ So now we use the parallel axes theorem and it's inertia around the centre of the hollow cylinder is $(\frac{1}{4}a^2+(R-a)^2)m$. So now we use the formula for inertia and get $ E=\frac{1}{2}I\theta^2=\frac{1}{2}(\frac{1}{4}a^2+(R-a)^2)^\theta^2 $
