Moment of Inertia of a Uniform Thin Circular Ring about any Diameter of the Ring

Let $I_d$ be the moment of inertia of the circular ring about the diameter $XX’$. Let $YY’$ be another diameter of the ring perpendicular to the diameter $XX’$.

Since the ring is symmetrical about any of its diameters, the moment of inertia of the ring about the diameter $YY’$ is also equal to $I_d$. According to the theorem of perpendicular axes:

M.I. of the ring about ZZ’ = M.I. about XX’ + M.I. about YY’

i.e.

$\displaystyle I={{I}_{d}}+{{I}_{d}}=2{{I}_{d}}$

$\displaystyle \Rightarrow {{I}_{d}}=\frac{1}{2}I$

or

$\displaystyle {{I}_{d}}=\frac{1}{2}M{{R}^{2}}$

Example: The moment of inertia of a circular ring about an axis passing through its centre and perpendicular to its plane is $200\, \text{g cm}^2$. If the radius of the ring is $5 \, \text{cm}$, find the mass of the ring.

Solution:

Here, $I = 200 \, \text{g cm}^2$, $R = 5 \, \text{cm}$

Now, moment of inertia of the ring:

\[
I = M R^2
\]

\[
M = \frac{I}{R^2} = \frac{200}{(5)^2} = 8 \, \text{g}
\]