Physics Friday 140

To answer the questions we left off with last week, we label last week’s solution as I_z, and denote with I_y the moment of inertia of the fractal about in-plane axis through one vertex, and I_x about an axis through the center of mass parallel to a side. Note that these three axes are all perpendicular.

Now, for I_y, 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 i_y, which, as with our dimensional analysis in the previous part, is proportional to our overall I_y; $i_y=\frac{1}{12}I_y$ .
Next, we again use the parallel axis theorem. Two of our sub-segments have a center of mass to axis distance of $\frac{a}{4}$ , while the third is already on our axis; so we have
$\begin{eqnarray}I_y&=&2\left(\frac{1}{12}I_y+\frac{M}{3}\left(\frac{a}{4}\right)^2\right)+\frac{1}{12}I_y\\&=&2\left(\frac{1}{12}I_y+\frac{Ma^2}{48}\right)+\frac{1}{12}I_y\\I_y&=&\frac{1}{4}I_y+\frac{Ma^2}{24}\\\frac{3}{4}I_y&=&\frac{Ma^2}{24}\\I_y&=&\frac{1}{18}Ma^2\end{eqnarray}$ .

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
$I_z=I_x+I_y$ ,
so then
$I_x=I_z-I_y=\frac{1}{9}Ma^2-\frac{1}{18}Ma^2=\frac{1}{18}Ma^2$ .

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?

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

This entry was posted on October 22, 2010 at 12:10 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.

One Response to “Physics Friday 140”

Physics Friday 141 « Twisted One 151's Weblog Says:
October 29, 2010 at 1:44 am | Reply
[...] Friday 141 By twistedone151 To answer the questions left lingering last week, we begin with the Sierpinski carpet: We see that it is composed of eight smaller carpets, each [...]

Twisted One 151's Weblog