Moment of inertia
The definition of (mass) moment of inertia of a point mass is
$$
I=r^2m
$$
However in the real world you don't encounter point masses, but objects with non-zero volume (finite density). And leads to an integral to determine moment of inertia
$$
I=\int_m{r^2dm}=\int_V{\rho(r)r^2dV}=\int_x{\int_y{\int_z{\rho(x,y,z)(x^2+y^2+z^2)dz}dy}dx}
$$
The solutions of this integral of a few bodies, with constant non-zero density within geometric volume and zero density outside of it, can be found here. For example the moment of inertia of thin rod rotating around its center of mass is equal to $I=\frac{mL^2}{12}$ and for a solid cylinder $I=\frac{mL^2}{2}$.
Experimental setup
In your experimental a string is on one end connected to the hanging mass, lead over the pulley and then wounded around a drum (the other end is also connected to the drum). This drum is of the object from which you would like to determine its moment of inertia and it is assumed that it can rotate freely (without slip) around its axis.
According to your documentation you measure how far the pulley has rotated, I will call this angle $\theta$, and its first and second time derivative $\omega=\dot{\theta}$ and $\alpha=\ddot{\theta}$.
The displacement of the hanging mass is related to the angular displacement of the pulley and its radius, $r_p$, assuming that the string does not slip, so
$$
s=r_p\theta
$$
where $s$ is the vertical downward displacement of the hanging mass.
This displacement is equal to the amount of string unrolled from the drum (assuming that the string is not elastic), which means that the angular displacement of the object from which you would like to determine the moment of inertia, I will call this $\theta_I$, can be calculated from this the other way around using the radius of the drum $r_d$
$$
s=r_p\theta=r_d\theta_I\rightarrow\theta_I=\frac{s}{r_d}=\frac{r_p}{r_d}\theta
$$
This linear correlation also applies to the $\omega$ and $\alpha$.
The only force applied on this system (which can perform work) is gravity on the hanging mass. Using all this you can derive the equation of motion (possibly using free body diagrams and tension in the string).
The force $m_1g\sin\theta$ provides the moment $Rm_1g\sin\theta$ ($R$ is the radius of the flywheel), so the equation of motion becomes:
$I\ddot\theta+Rm_1g\sin\theta=0$.
For small $\theta$, then $\sin\theta \approx \theta$, so we get:
$\large{\ddot\theta+\frac{Rm_1g}{I}\theta=0}$.
The solution of this classic DE is:
$\theta(t)=\theta_0 \cos(\sqrt{\frac{Rm_1g}{I}}t)$, with period $T$:
$T=2\pi\sqrt{\frac{I}{Rm_1g}}$, so that:
$\Large{I=\frac{Rm_1gT^2}{4\pi^2}}$.
I've seen a method for experimentally determining the moment of inertia of a flywheel and I'm not sure whats the reasoning behind it. You attach a small weight m1 to the flywheel's edge, it's important that m1≪M, M being the mass of the flywheel, so we can approximate that the small mass does not change the MOI of the flywheel.
The derivation above is for that case described in your question. But it doesn't correspond to the formula you gave.
For the formula you gave, the inertial moment of $m_1$ is actually taken into account as follows. The actual inertial moment, including that of $m_1$ is $I'$:
$I'=I+m_1R^2$.
All we have to do now is substitute $I$ with $I'$, so that:
$T=2\pi\sqrt{\frac{I+m_1R^2}{Rm_1g}}$, so that:
$\large{I+m_1R^2=\frac{Rm_1gT^2}{4\pi^2}}$
$\Large{I=\left( \frac{T^2 g}{4 \pi ^2R} - 1\right) m_1 R^2}$.
As in your formula, but only if you do allow for the effect of $m_1$ on the total moment of inertia.
Best Answer
I(ring) = Idisk(R2) - Idisk(R1).
The trick is figuring out the mass.
Mass of R2-sized disk would be MR2 = M*(pi*R2*R2)/((pi*R2*R2)-(pi*R1*R1))
Mass of R1-sized disk would be MR1 = MR2*(pi*R1*R1)/(pi*R2*R2)
So I(ring) = 1/2MR2*(R2*R2) - 1/2MR1*(R1*R1)
I guess the pis can come out:
Ugh, this means:
So, yep.