How can I solve the moment of inertia? 
The source tells me to use the formula for a ring, but it is not possible, as the portions are nearer to the axis than a normal ring. How can I find the moment of inertia?
 A: You know that to solve this problem you will have to use the integral form for MoI.
$\int_0^Mr^2dm$ where $dm$ is the mass element (geometry of the problem) and $r$ is the distance from the axis of rotation.
You express the mass element in terms of $r$ so you get linear density.
Ex. Rod of mass $M$ has linear density $M/L$ so you get $dm=\frac MLdr$
Ex.2 A solid cylinder has volume density of $dm = \rho L2\pi rdr$ (density * length * circumference or $\rho dV$)
etc.
So, you find the distribution of mass (geometry) and plug it into the integral.    
Another way you could solve it is by using the Parallel Axis Theorem ($PAT$)
$I_{parallel} = I_{cm} + Md^2$.
You know that the $I=\frac{MR^2}2$ for the whole ring.
To sum up, imagine that you have a whole ring, which you cut on both sides and you get a sort "brackety" system () (removing the "middle" part from  a nice round circle) which you can then solve using the integral method of $PAT$ method ($\frac{MR^2}2 + Md^2$ where $d$ is the distance from the axis and the same point where you cut to get one portion of the ring $->$ substituting)
A: Use the parallel axis theorem for calculating the moment of inertia relative to a point which is not the center of mas (COM).  See for example here in Wikipedia.
$I=I_{\rm COM} - m  d^2 ,$
where $d$ is the distance between COM and the point of rotation.
