Moment of Inertia of a Ring and Cylindrical Shell
Formulas
The moment of inertia of a ring with radius $a$ and mass $m$, when the axis of rotation passes through the center of the ring, is:
Perpendicular to the plane formed by the ring, it is $I=ma^{2}$.
Parallel to the plane formed by the ring, it is $I=\dfrac{1}{2}ma^{2}$.
Derivation
Consider a thin, uniform circular ring (or cylindrical shell) with radius $a$ and mass $m$. There are cases where the axis of rotation is perpendicular to the plane formed by the ring or parallel to it.
When the axis of rotation passes through the center of the ring and is perpendicular to the plane formed by the ring
The formula for calculating the moment of inertia is $\displaystyle I=\int r^{2}dm$, and since the distance from the axis of rotation to the mass element is always the constant radius $a$, it follows that:
$$ I_{z}=\int a^{2}dm=a^{2}\int dm=ma^{2} $$
■
When the axis of rotation passes through the center of the ring and is parallel to the plane formed by the ring
According to the perpendicular axis theorem $I_{z}=I_{x}+I_{y}$, and whether the axis of rotation is the $x$-axis or the $y$-axis, the configuration is the same, resulting in $I_{x}=I_{y}$. Therefore, it follows that:
$$ \begin{align*} && 2I_{x} &= I_{z}=ma^{2} \\ \implies && I_{x} &= \dfrac{1}{2}ma^{2} \end{align*} $$
■