Archive for October, 2010

Physics Friday 141

October 29, 2010

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 with side length one-third that of the overall carpet; thus, if I is the moment of inertia of the total carpet of mass M and side length a, then each of the smaller carpets has moment about its center of
.
Now, we use the parallel axis theorem again. We see that four of the sub-carpets, the “edge” ones, have centers with distance from our axis of . The other four, the “corners”, have centers with distance .

Thus, we have
.

Now, for the Menger Sponge. We have 20 sub-units, each with edge-length one-third that of the overall sponge, so
.
Finding the distances, we see that there are 12 which lie along edges parallel to our axis (comparable to the “corners” in the carpet case), and thus have distance . The other 8 are analogous to the carpet “edge” units, with distance . Thus,
.

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Monday Math 140

October 25, 2010

What are the lengths of the diagonals of a regular n-sided polygon with sides of unit length?
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Physics Friday 140

October 22, 2010

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?

Monday Math 139

October 18, 2010

Prove that for positive integers k, m, and n, that
\sin\frac{k\pi}n\sin\frac{m\pi}n=\left(\sin\frac{\pi}n\right)\sum_{i=1}^{m}\sin\frac{(k-m+2i-1)\pi}n.
Solution:

Physics Friday 139: Moment of Inertia of a Fractal

October 15, 2010

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?

Solution:

Monday Math 138

October 11, 2010

Show that \sum\limits_{k=1}^{n}k(n+1-k)=\sum\limits_{k=1}^{n}\frac{k(k+1)}2.
Solutions:

Physics Friday 138

October 8, 2010

Here is a classic special relativity exercise: the “detonator paradox.” Suppose we have two objects, one shaped like a “T” and the other resembling a “U”

The T fits into the U, and comes just short of touching the “inside” of the U, where we have a pressure switch; this pressure switch is connected to the detonator for some explosives.

Now suppose we start with the T some distance away from the U, and send the T toward the U at a relativistic speed , so that γ=2. In the lab frame, the U frame, the U is at rest, and the T, moving at v, undergoes length contraction to half it’s rest length, leaving it clearly too short to reach the detonator.

However, in the T’s frame of reference, it is at rest and the U approaches at speed v. In this frame, the U is length contracted to half, and so the T can easily reach the switch, and trigger the explosion.

So, is there an explosion, or not?
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