TITLE: Calculate moment of inertia of Koch snowflake
QUESTION [23 upvotes]: That's just a fun question. Please, be creative.
Suppose having a Koch snowflake.
The area inside this curve is having the total mass $M$ and the length of the first iteration is $L$ (a simple equilateral triangle of side $L$). Assume the density being uniform.
If the curve is on $xy$ plane, the rotation is done on the $z$ axis, around the center of the curve.
Calculate its moment of inertia.

REPLY [9 votes]: An answer based off Mark McClure's but involving less theory and invoking the parallel axis theorem.
If we assume that the moment of inertia is always proportional to the mass, and proportional to the square of a characteristic length scale (in this case the length of the diameter as in Mark's answer), then 
$$
I = \gamma Ml^2,
$$
where $\gamma$ is a constant. Using the principle of superposition,
$$
I = I_{\text{centre}} + 6I_{\text{edge}},
$$
where the centre snowflake is in blue and the edge snowflake's are red/brown. Now $$\displaystyle I_{\text{centre}} = \gamma \frac{M}{3}\left(\frac{l}{\sqrt{3}}\right)^2 = \frac{\gamma}{9}Ml^2 = \frac{I}{9}.$$
Using the parallel axis theorem, $\displaystyle I_{\text{edge}} = I_{\text{COM}} + Md^2$, where $I_{\text{COM}}$ is the moment of inertia through the centre of mass, and
$$\displaystyle I_{\text{COM}} = \gamma \frac{M}{9}\left(\frac{l}{3}\right)^2 = \frac{\gamma}{81}Ml^2 = \frac{I}{81},$$
and $d = l/3$, so
$$\displaystyle I_{\text{edge}} = \frac{I}{81}+ \frac{Ml^2}{9},$$
So
\begin{align*}
I &= \frac{I}{9} + 6\left(\frac{I}{81}+ \frac{Ml^2}{9}\right), \\
I & = \frac{15I}{81} + \frac{2Ml^2}{3}, \\
\frac{22I}{27} & = \frac{2Ml^2}{3}, \\
I & = \frac{Ml^2}{11}
\end{align*}
If we let $l=3$ we recover Mark's answer.