Physics Friday 139: Moment of Inertia of a Fractal

What is the moment of inertia of a Sierpinski gasket of mass M and side length a about the axis through its center and perpendicular to its plane?

Dimensional analysis tells us that the moment of inertia, I, is proportional to Ma². Now, let us consider one-third of the gasket:

According to the parallel axis theorem, the moment of inertia of this part about our desired axis is
$i=i_{cm}+md^2$ ,
where i_cm is the moment of inertia of the piece about the axis parallel to our desired one through its center of mass, m is its mass, and d is the distance between the axes:

Geometry tells us that, since the distance from the corner to the center of an equilateral triangle of side length a is $\frac{a}{\sqrt3}$ , then the distance d between the axes is $d=\frac{a}{\sqrt3}-\frac{a}{2\sqrt3}=\frac{a}{2\sqrt3}$ . Since the piece we are considering is one of three congruent pieces that make up the overall figure, the mass is thus $m=\frac{M}3$ ,
so
$i=i_{cm}+\frac{M}{3}\left(\frac{a}{2\sqrt3}\right)^2=i_{cm}+\frac{Ma^2}{36}$ .
Since the gasket is made of three such sub-figures, and moment of inertia is additive, we then see
$I=3i=3i_{cm}+\frac{Ma^2}{12}$ .
But what is i_cm? Note that the sub-figure is just a smaller Sierpinski gasket, with mass $m=\frac{M}3$ and side length $\frac{a}2$ . Thus, by our dimensional analysis given earlier, this is proportional to I, simply reduced by a factor $\frac13\left(\frac12\right)^2=\frac1{12}$ , so
$i_{cm}=\frac{I}{12}$ ,
so we have
$I=\frac{I}{4}+\frac{Ma^2}{12}$ .
$\frac{3}{4}I=\frac{Ma^2}{12}$ .
$I=\frac{1}{9}Ma^2$ .

How about the moment of inertia about an axis in the plane, through the center and one vertex? Or about an axis in the plane, through the center and parallel to one of the sides?

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Tags: Fractal, Friday Physics, Moment of Inertia, Parallel Axis Theorem, physics, Sierpinski Gasket

This entry was posted on October 15, 2010 at 10:14 am and is filed under Math/Science. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

5 Responses to “Physics Friday 139: Moment of Inertia of a Fractal”

Dr.D Says:
October 15, 2010 at 10:55 am | Reply
Why does anybody care?
K. Says:
October 15, 2010 at 1:11 pm | Reply
I don’t think it’s a matter of caring, but the (quite nice) presentation of a physical and mathematical curiosity. If you don’t care, then you don’t have to read it, but some of us enjoy such questions (and solutions).
SH Says:
October 15, 2010 at 4:00 pm | Reply
Because it’s fun and cool! (stupid of me to rise to the troll bait…)
Also, it shows that even if fractals are involved the math can be quite simple which I didn’t expect. A good exercise.
Now going to try doing the two other inertias…
Physics Friday 140 « Twisted One 151's Weblog Says:
October 25, 2010 at 12:44 am | Reply
[...] answer the questions we left off with last week, we label last week’s solution as Iz, and denote with Iy the moment of inertia of the fractal [...]
Dr.D Says:
October 27, 2010 at 1:43 pm | Reply
It is a mathematical curiosity, but hardly a physical curiosity. Where might you expect to find such a thing is physical reality?

Twisted One 151's Weblog