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Some neat tricks and manipulation can show that the moment of inertia of the Cantor Set with respect to an axis perpendicular to it and which passes through its center is $\frac{1}{8}ml^2$ where m is the total mass of the set and l it's longitude (understanding that you can consider an homothecy of this set). Of course this doesn't itself use the integral definition of the moment of inertia, but of course trying to just integrate the following (keeping in mind that x is the distance from a point to the axis of rotation and dm is the mass differential: $dm=\rho dx = \frac{m}{l}dx$) :

\begin{equation} I = \int_C x^2 dm = \frac{m}{l} \int_C x^2 dx \end{equation}

Gives of course 0, as the cantor set has Lebesgue measure cero. For physical "reasons" a finite number is expected, that's why I thought that maybe I need to consider some kind of "fractional integral" over the Hausdorff dimension of the Cantor's set. This dimension is $d=\displaystyle \frac{log 2}{log 3}$. So maybe should I try to compute this?

\begin{equation} \int x^2 \chi_C dm_d \end{equation}

With $dm_d$ the d-dimensional Hausdorff measure, and $\chi_C$ the characteristic function over the Cantor's set.

If so, or if you think I should do something totally different, please feel free to tell me, I'm no real expert in this.

Rafa
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1 Answers1

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I'll take $m=1$ and $l=1$. The natural "uniform" mass distribution on the standard Cantor set is the distribution of the random variable $$ X:=\sum_{n=1}^\infty {\xi_n\over 3^n}, $$ where $\xi_1,\xi_2,\ldots$ are i.i.d. with $\Bbb P[\xi_n=0]=\Bbb P[\xi_n=2]=1/2$. (This distribution is also the normalized Hausdorff measure of dimension $d=\log 2/\log 3$.) Clearly $\Bbb E[X] =1/2$, while the variance of $X$ (a.k.a. the moment of inertia about the vertical axis throough the mean of $X$) is $$ \eqalign{ \Bbb E[(X-1/2)^2] &=\Bbb E[(\sum_n{\xi_n-1\over 3^n})^2]\cr &=\Bbb E\sum_{m,n}{\xi_m-1\over 3^m}{\xi_n-1\over 3^n}\cr &=\sum_{m,n} 3^{-m-n}\Bbb E[(\xi_m-1)(\xi_n-1)]}. $$ The expectations in this double sum are $0$ or $1$ according as $m\not=1$ or $m=n$; therefore $$ \eqalign{ \Bbb E[(X-1/2)^2] &=\sum_{n=1}^\infty 3^{-2n} = {1\over 8}.\cr } $$

John Dawkins
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