## Physics Friday 140

To 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 about in-plane axis through one vertex, and Ix about an axis through the center of mass parallel to a side. Note that these three axes are all perpendicular.

Now, for Iy, we again divide into three smaller gaskets. The moment of inertia of each of these about an axis parallel to our axis through the piece’s center of mass is iy, which, as with our dimensional analysis in the previous part, is proportional to our overall Iy; .
Next, we again use the parallel axis theorem. Two of our sub-segments have a center of mass to axis distance of , while the third is already on our axis; so we have
.

For the x axis, we could perform this breakdown method again, though finding the parallel axis distances would be less simple in this case. However, instead we can recall that since our Sierpinski gasket is a plane figure, we can use the perpendicular axis theorem; here, we have
,
so then
.

Now, how about the moment of inertia about an axis perpendicular to the plane of a Sierpinski carpet of mass M and side length a? How about the moment of inertia about a face-centered axis of a Menger sponge of mass M and edge length a?