# Math Problem - 9



## JBCM627 (Aug 14, 2008)

Hopefully this one is a bit harder, and hopefully Lucas hasn't seen it before. I looked around for a while for something like this  This is more of a physics problem, I guess.

Find the moment of inertia of an equilateral Sierpinski triangle of side length L and mass m about an axis through its circumcenter and perpendicular to the plane it is in. (Assume uniform distribution of mass to all points.)


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## badmephisto (Aug 14, 2008)

Spoiler



wow :s wait i thought the sirpinki triangle had 0 area... Doesnt that mean... its not 0 is it? haha
only way i could think of is to find ratio of moments of inertia of triangle in depth n and n+1.... 
I = integral(r^2dm)
dm=A*da (unifrom mass, A is constant)
=> I = integral(r^2Ada)
details are uglier but in any case looks like dependence on length is quadratic and on mass its linear, so from that I_big=I_small*4*3 because mass of smaller thingie is a third (since 1 out of 4 subtriangles is empty), and dimensions are halved, and area in triangle depends on dimension squared, so its (1/2)^2 = 1/4, so times 4 for that. So unless im mistaken the ratio is 12.

but now I'm thinking that this is invalid because each is calculated with its own axis around its own center... :s So the axis is not the same axis... I'm confused. Really great find JBCM :s /sarcasm. Maybe someone can find a symmetry argument to make it work?


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## Lucas Garron (Aug 14, 2008)

Spoiler



(13/72)*m*L^2.
Do you want an explanation? 

EDIT: Changed the L to an L^2. Thanks for pointing that out, badmephisto. By the way, though, there is no area here. But the determining radii still grow quadratically.
EDIT 2: Fine, you wanna give it arbitrary mass instead of 1? I'll throw in a factor of m.


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## badmephisto (Aug 14, 2008)

no Lucas we don't want an explanation. /sa... never mind



Spoiler



my gut feeling was that the answer would be quadratic in L since Area is the determining factor here, which depends on L quadratically :s


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## AvGalen (Aug 14, 2008)

Can I have an explanation of the question as well?


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## JBCM627 (Aug 14, 2008)

Explanation of the question? I tried to word it more concisely than what I initially found. I'll try and re-phrase a bit more simply: if you have a sierpinski triangle of mass m, all side lengths the same length L, and spun it around the center of the triangle, what would its moment of inertia be? So, if you threw a sierpinski triangle like a triangular frisbee, what would its moment of inertia be?

Area, while you can involve it in many cases, is actually not necessarily a factor. You can find the moment of inertia of an empty circle (which has 0 area), for example: Integral(r^2 dm). r is constant (radius of circle), and so the integral is really just r^2*integral(1dm), which is mr^2. Same with an orbiting point particle.

And yeah, units demand the end result will work out to be some variation of r^2. Actually, I should mention now that the entire structure should have mass "m", something I left out in the initial wording of the problem. I'll edit that now. The actual answer will also need to include "m".

And hrm I still have to work this out myself...


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## badmephisto (Aug 14, 2008)

Yeah physics problems are always nice because you always have the units sanity check at the end. it looks like Lucas will need another edit, because Inertia is in units of kg*m^2, not just m^2


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## Lucas Garron (Aug 14, 2008)

badmephisto said:


> Lucas will need another edit, because Inertia is in units of kg*m^2, not just m^2


Was editing as you wrote.

Anyhow, I'll just tell you what I did.


```
v1={0,Sqrt[3]/2};
v2=({1,0}-v1)/2;
v3=(-{1,0}-v1)/2;
lev=1; d={{0, 0}};
While[True, lev/=2;
d=Flatten[{#1+v1*lev, #1+v2*lev, #1+v3*lev}& /@ d,1];
Print[N[Total[(#1[[1]]^2 + #1[[2]]^2)& /@ d]/Length[d],20]];
];
```



Note:
PJK, can you make

```
s="ToExpression[s]"; ToExpression[s]
```
 display like 


Mathematica said:


> s="ToExpression"; ToExpression



, except with "Code: Mathematica"


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## AvGalen (Aug 14, 2008)

JBCM627 said:


> Explanation of the question? I tried to word it more concisely than what I initially found. I'll try and re-phrase a bit more simply: if you have a sierpinski triangle of mass m, all side lengths the same length L, and spun it around the center of the triangle, what would its moment of inertia be? So, if you threw a sierpinski triangle like a triangular frisbee, what would its moment of inertia be?
> 
> Area, while you can involve it in many cases, is actually not necessarily a factor. You can find the moment of inertia of an empty circle (which has 0 area), for example: Integral(r^2 dm). r is constant (radius of circle), and so the integral is really just r^2*integral(1dm), which is mr^2. Same with an orbiting point particle.
> 
> ...


Sorry, I should have added a smilie. This one :confused: to be exact.

I studied physics in university (Dutch version, basically the highest level of education) for 1.5 years. a lot of it was in English and a lot of it was math. But I had no idea what an "equilateral Sierpinski triangle" was and how "Assume uniform distribution of mass to all points" works.

I guess I should stick to "applied" physics/math and don't try to do these theoretical problems


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## Tim_Likes_Cubing (Aug 14, 2008)

I like maths (i'm doing a-levels 4 years early), but this is just hard. Glad to see some maths anyway, lol


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## Kenneth (Aug 14, 2008)

I'm afraid a Sierpinski can't have mass, it even has not got an surface 

Here is one that I made that has not got volume 

http://upload.wikimedia.org/wikipedia/commons/2/24/Sierpinski_tetrahedron.jpg

Pyraminx?


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## AvGalen (Aug 14, 2008)

No, megaultrasuperexaminx

(and I still wonder about this thing actually having mass myself)


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## JBCM627 (Aug 15, 2008)

Yeah, this object cannot actually exist, along with planes and other mathematical objects. (They are 2D... infinitely thin... not possible in real life. Well, maybe they could if brane theory were right...) And a Sierpinski triangle isn't even 2D... its ~1.58D, as explained in a remark on this solution. No physical object would ever be able to be as complex as this object, either. There are too many details at too small of a scale.

I didn't really make much progress on this myself. Apparently summations and integrals really don't come in handy here. One thing in this proof that would have really hindered people from solving this who haven't had a physics course would be the need to know something about the parallel axis theorem.

Original problem:
http://www.physics.harvard.edu/academics/undergrad/probweek/prob9.pdf
Solution:
http://www.physics.harvard.edu/academics/undergrad/probweek/sol9.pdf
Other problems from that site:
http://www.physics.harvard.edu/academics/undergrad/problems.html


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