Q:

Coin rolling, but not sliding. Kinetic energy

A homogeneous coin of mass M rolls, without sliding, along the x-axis with the axis of rotation being parallel to the z-axis. Let $\Theta$ be moment of inertia regarding that axis and $\vec{V}$ be the velocity of the center of gravity.
a) Show that the kinetic energy is
$T=\frac{1}{2}MV^2+\frac{1}{2}\Theta \dot{\phi}^2$.
(king's theorem)
b) Use the theorem in a) to determine the kinetic energy of the coin.
c) Now compute the kinetic energy of the coin without using the theorem.

Okay, before I start into getting to my struggles with this problem I wanted to show my sketch of this exercise to see if I even understoof it correctly.

At first I thought that the axis of rotation was going through the center of mass but while drawing it I realized that it had to be contact point of the of the coin with the x-z-plane, right? Since it has no z-value I just denoted it with $x'$.
I got some issues understanding the difference between a) and b). Isn't the kinetic energy I need to show in a) also the kinetic energy of the coin?
And about that king's theorem. I never heard of it before so I looked it up on Wiki. But it's some kind of graph theory. How is that of any use in this exercise?
So I tried myself on c) since I don't need to use the theorem there. But I don't know how to approach this.
I mean, isn't the velocity of the coin constant? Meaning it's just $T=\frac{1}{2}MV^2$?
Although, truth be told, it doesn't say that there isn't any friction.
That's why I'm with this type of problems.
Edit
a)$\sum \frac{1}{2}m_iv_i^2=\sum\frac{1}{2}(v_{cm}+v_{i,cm})^2=\frac{1}{2}M(V+R\dot{\phi})^2=\frac{1}{2}MV^2+\frac{1}{2}MR\dot{\phi}V+\frac{1}{2}MR^2\dot{\phi}^2=\frac{1}{2}MV^2+\frac{1}{2}MR^2\dot{\phi}^2+\frac{1}{2}MR^2\dot{\phi}^2=\frac{1}{2}MV^2+MR^2\dot{\phi}^2\neq\frac{1}{2}MV^2+\frac{1}{2}\Theta \dot{\phi}^2$
$I_{coin}=\frac{1}{4}MR^2+MR^2$, right? So did I make some arithmetic mistake or did I use König's theorem wrongly?
b) I'm not so sure about this one. Isn't it just the same as in a)? So basically, $T=\frac{1}{2}MV^2+\frac{1}{2}\Theta\dot{\phi}^2$?
c) The approach here is just to determine the rotational energy, meaning $E_{rot}=\frac{1}{2}I\dot{\phi}^2$, right? Which then just is $\frac{1}{2}(\frac{1}{4}MR^2+MR^2)\dot{\phi}^2$?

A:

You are right that the actual axis of rotation is goes through the contact point with the ground. To see this, roll your bike back and forth. You'll see that the contact point is always standing still: the rest of the wheel turns around it and as it turns, the axis of rotation shift to the next point that makes contact with the ground (it's the static friction that makes the wheel and thus the bike go forward).
However, it is easier to look it in a different way: the coin makes a translation, while it is rotating around the center of mass. This is what you use to derive king's theorem: for a collection of particles (of which a coin is clearly an example) you can say
$$K = \sum \frac{1}{2}m_i \vec{v}_i^2=\sum \frac{1}{2}m_i (\vec{v}_{cm}+\vec{v}_{i,cm})^2$$ 
with $\vec{v}_{i,cm}$ the velocity of particle $i$ relative to the center of mass. If you work this out and use that for a rotation relative to the center of mass $v_{i,cm}=r_{i,cm}\omega$ (I prefer $\omega$ over $\dot{\phi}$, but it  is the same), you'll get there.
Now you know the formula is ok, you can use it to calculate K (question b), but if you're not quite sure (question c), you can still depart from your first view: the rotation of the coin around the axis throug $x'$, in which case the kinetic energy is just the rotational energy.
Important: for b) you need the moment of inertia relative to the center of mass and for c) you need it with respect to $x'$.

