Moment of inertia of a part of a fractal

Question

A physics problem that I was solving recently went as follows:

Take a square plate of side l , and remove the “middle” square (1/9 of the area). Then remove the “middle” square from each of the remaining eight squares, and so on, forever. Let the final object have mass m, Moment of inertia of square plate about an axis passing through center and perpendicular to plane is $\frac {xml^{2}}{16}$Find the value of x?

So my point of interest is the MOI of one of the 8 remaining squares. Before commenting on it's MOI let's comment on it's mass If $\sigma$ is the mass per unit area ( at whatever points that are not removed) Therefore $ m =\sigma \int dx dy $ and $A =\sigma \int d\frac x3 d\frac y3 = \frac m9$ But via conservation of mass the mass should obviously be m/8 which of these is correct. Similarly is the moment of inertia of the small portion $\frac{I}{72}$ or $\frac{I}{81} $

Also is this the right forum for thus qn.

Answer 1

The process resembles the construction of the Sierpinski carpet, a fractal formed by iteratively subdividing a square and removing central portions:

1. Start with a square of side $\text{L}$. Its area is $\text{L}^2$;
2. Remove the "middle" square with $\frac{\displaystyle1}{\displaystyle9}$ of the area, i.e., area $\frac{\displaystyle\text{L}^2}{\displaystyle9}$. The side length of this square is $\sqrt{\frac{\displaystyle\text{L}^2}{\displaystyle9}}=\frac{\displaystyle\text{L}}{\displaystyle3}$. This can be visualized by dividing the square into a $3×3$ grid of nine smaller squares, each with side $\frac{\displaystyle\text{L}}{\displaystyle3}$ and area $\frac{\displaystyle\text{L}^2}{\displaystyle9}$, and removing the central one. The remaining area is $\text{L}^2-\frac{\displaystyle\text{L}^2}{\displaystyle9}=\frac{\displaystyle8\text{L}^2}{\displaystyle9}$, consisting of eight smaller squares;
3. For each of the eight remaining squares (side $\frac{\displaystyle\text{L}}{\displaystyle3}$), remove their central square with area $\frac{\displaystyle1}{\displaystyle9}\left(\frac{\displaystyle\text{L}}{\displaystyle3}\right)^2=\frac{\displaystyle\text{L}^2}{\displaystyle81}$. Total area removed is $8\cdot\frac{\displaystyle\text{L}^2}{\displaystyle81}=\frac{\displaystyle8\text{L}^2}{\displaystyle81}$. Remaining area is $\frac{\displaystyle8\text{L}^2}{\displaystyle9}-\frac{\displaystyle8\text{L}^2}{\displaystyle81}=\left(\frac{\displaystyle8\text{L}}{\displaystyle9}\right)^2$.

Generally, after $\text{n}$-stages, the remaining area is $\text{L}^2\left(\frac{\displaystyle8}{\displaystyle9}\right)^\text{n}$, and as $\text{n}\space\to\space\infty$, this approaches zero. However, the problem states the final object has mass $\text{m}$, suggesting that despite the area vanishing, the fractal retains a finite mass distributed over its structure, typical in idealized fractal problems.

The final object is self-similar: it consists of eight smaller copies of itself, each scaled by a factor of $\text{s}=\frac{\displaystyle1}{\displaystyle3}$ (since each smaller square’s side is $\frac{\displaystyle1}{\displaystyle3}$of the previous). With a total mass $\text{m}$, each small copy has mass $\frac{\displaystyle\text{m}}{\displaystyle8}$. The moment of inertia $\text{I}$ about the central axis is the sum of contributions from these eight copies, each with its own moment of inertia about the central axis, accounting for their positions.

For a small copy:

• Mass: $\frac{\displaystyle\text{m}}{\displaystyle8}$;
• Scaling factor: $\text{s}=\frac{\displaystyle1}{\displaystyle3}$;
• Moment of inertia about its own center: If the entire object has moment of inertia $\text{I}$, a scaled copy’s moment of inertia about its own center scales with mass and distance squared. Since mass is $\frac{\displaystyle\text{m}}{\displaystyle8}$ and linear dimensions scale by $\frac{\displaystyle1}{\displaystyle3}$, the moment of inertia scales as:

$$\text{I}_\text{small}=\left(\frac{\frac{\text{m}}{8}}{\text{m}}\right)\left(\frac{1}{3}\right)^2\text{I}=\frac{\text{I}}{72}\tag1$$

However, we need the moment of inertia of each small copy about the central axis of the entire object, not its own center. Using the parallel axis theorem, $\text{I}=\text{I}_\text{cm}+\text{md}^2$, where $\text{d}$ is the distance from the small copy’s center to the central axis.

Place the original square from $(0,0)$ to $(\text{L},\text{L})$, with center at $\left(\frac{\displaystyle\text{L}}{\displaystyle2},\frac{\displaystyle\text{L}}{\displaystyle2}\right)$. After the first removal, the eight small squares (side $\frac{\displaystyle\text{L}}{\displaystyle3}$) are at:

• Corners: $(0,0)$ to $\left(\frac{\displaystyle\text{L}}{\displaystyle3},\frac{\displaystyle\text{L}}{\displaystyle3}\right)$, center $\left(\frac{\displaystyle\text{L}}{\displaystyle6},\frac{\displaystyle\text{L}}{\displaystyle6}\right)$; $\left(\frac{\displaystyle2\text{L}}{\displaystyle3},0\right)$ to $\left(\text{L},\frac{\displaystyle\text{L}}{\displaystyle3}\right)$, center $\left(\frac{\displaystyle5\text{L}}{\displaystyle6},\frac{\displaystyle\text{L}}{\displaystyle6}\right)$; etc.;
• Sides: $\left(\frac{\displaystyle\text{L}}{\displaystyle3},0\right)$ to $\left(\frac{\displaystyle2\text{L}}{\displaystyle3},\frac{\displaystyle\text{L}}{\displaystyle3}\right)$, center $\left(\frac{\displaystyle\text{L}}{\displaystyle2},\frac{\displaystyle\text{L}}{\displaystyle6}\right)$; etc.

Distances from $\left(\frac{\displaystyle\text{L}}{\displaystyle2},\frac{\displaystyle\text{L}}{\displaystyle2}\right)$:

• Corner squares (four such squares):

$$\text{d}=\sqrt{\left(\frac{\text{L}}{2}-\frac{\text{L}}{6}\right)^2+\left(\frac{\text{L}}{2}-\frac{\text{L}}{6}\right)^2}=\frac{\text{L}\sqrt{2}}{3}\tag2$$

• Side squares (four such squares):

$$\text{d}=\sqrt{\left(\frac{\text{L}}{2}-\frac{\text{L}}{2}\right)^2+\left(\frac{\text{L}}{2}-\frac{\text{L}}{6}\right)^2}=\frac{\text{L}}{3}\tag3$$

So, the total moment of inertia:

$$\text{I}=8\text{I}_\text{small}+4\left(\frac{\text{m}}{8}\left(\frac{\text{L}\sqrt{2}}{3}\right)^2\right)+4\left(\frac{\text{m}}{8}\left(\frac{\text{L}}{3}\right)^2\right)\tag4$$

Using $(1)$, we get:

$$\text{I}=8\cdot\frac{\text{I}}{72}+4\left(\frac{\text{m}}{8}\left(\frac{\text{L}\sqrt{2}}{3}\right)^2\right)+4\left(\frac{\text{m}}{8}\left(\frac{\text{L}}{3}\right)^2\right)\space\Longleftrightarrow\space\text{I}=\frac{5\text{mL}^2}{16}\tag5$$