About the Moment of Inertia of the Cantor Set

Question

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.

Answer 1

I am a highschool student from India and I got this question in one of my weekend test.

For this problem lets have a look at this picture: (don't mind the quality of image)

let's say MOI about axis 1 is $$ I = Km\ell^2 $$

Where,

• ( K ) is a proportionality constant
• ( m ) is the mass of final body
• ( $\ell$ ) is the length of the body

Since, the right half of the body is just the scaled down version of original body, the mass of this right half is $\frac{m}{2}$ and the length is $ \frac{\ell}{3} $.

Then, MOI of The Right half About Axis 2

is:
$$ I = K.\left( \frac{m}{2} \right).\left( \frac{\ell}{3} \right)^2 $$

$$ =\frac{I}{18} $$

$$ \quad (\text{since } Km\ell^2 = I) $$

Now, for MOI of the Right Segment About Axis 1 (passing through the center of the original structure), we use the parallel axis theorem:
$$ I_{\text{axis 1}} = I_{\text{axis 2}} + M d^2 $$
where:

• $ I_{\text{axis 2}} = \frac{I}{18} $ (MOI about its center),
• $ M = \frac{m}{2} $ (mass of the right segment),
• $ d = \frac{\ell}{3} $ (distance from the center of the segment to Axis 1).

Substitute:
$$ I_{\text{axis 1}} = \frac{I}{18} + \frac{m}{2} \left( \frac{\ell}{3} \right)^2 $$
$$ I_{\text{axis 1}} = \frac{I}{18} + \frac{m}{2} \cdot \frac{\ell^2}{9} $$
$$ I_{\text{axis 1}} = \frac{I}{18} + \frac{m\ell^2}{18} $$

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Total MOI for Both Segments : Since the left and right segments are symmetric, the MOI for the left segment about Axis 1 will be the same as that of the right segment.

Thus, the total MOI is:
$$ I = 2 \left( \frac{I}{18} + \frac{m\ell^2}{18} \right) $$
Simplify:
$$ I = \frac{2I}{18} + \frac{2m\ell^2}{18} $$
$$ I = \frac{I}{9} + \frac{m\ell^2}{9} $$

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Final Result
The total moment of inertia is:
$$ I = \frac{m\ell^2}{8} $$

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When I was solving the question I had no knowledge of what Cantor Set is? or what is Hausdorff measure? In fact, I had no idea that this is a fractal geometry related question. So I just tried to use the concepts at hand.

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Answer 2

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 through the mean of $X$) is $$ \eqalign{ \Bbb E[(X-1/2)^2] &=\Bbb E\left[\left(\sum_n{\xi_n-1\over 3^n}\right)^2\right] \\ &=\Bbb E\sum_{m,n}{\xi_m-1\over 3^m}{\xi_n-1\over 3^n} \\ &=\sum_{m,n} 3^{-m-n}\Bbb E[(\xi_m-1)(\xi_n-1)].} $$ The expectations in this double sum are $0$ or $1$ according as $m\not=1$ or $m=n$; therefore $$ \Bbb E[(X-1/2)^2] =\sum_{n=1}^\infty 3^{-2n} = {1\over 8}. $$