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#1 2006-12-19 02:26:49

luca-deltodesco
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Registered: 2006-05-05
Posts: 1,470

double integrals

by wikipedia, the scalar moment of inertia for a solid body with continous mass-density function p, about a known axis can be calculated via:

where r maps x,y,z to the perpendicular distance from (x,y,z) to the axis of rotation and p maps x,y,z to the density of the solid at (x,y,z)

so in 2 dimensions, for a 2 dimensional solid, rather than a 3dimensional one you would have

now im trying to find this for a 2 dimensional rectangle.

if the rectangles centre is the axis of rotation, and the rectangle is centred on the origin with width = 2w, and height = 2h

then r would be:


if the mass-density is a constant m. then for all x,y

so i have:

evaluating the first integral


im pretty sure i must be getting something wrong along the way, because im fairly sure its not supposed to be 0 here.

Last edited by luca-deltodesco (2006-12-19 02:27:29)


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#2 2006-12-19 02:36:15

luca-deltodesco
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Registered: 2006-05-05
Posts: 1,470

Re: double integrals

oh wait ofcourse tongue i was adding it instead of taking it away


so i now have



anyone see any potential problems here?

Last edited by luca-deltodesco (2006-12-19 02:38:41)


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#3 2006-12-19 02:44:20

luca-deltodesco
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Registered: 2006-05-05
Posts: 1,470

Re: double integrals

according to a physics textbook, it should be more like:




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#4 2006-12-21 05:16:34

luca-deltodesco
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Re: double integrals

no-one?


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#5 2006-12-22 17:51:43

George,Y
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Registered: 2006-03-12
Posts: 1,379

Re: double integrals

Suppose you define W=2w, H=2h, and substitute w and h by W and H, the formula would be like the 3rd one.


X'(y-Xβ)=0

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#6 2006-12-22 20:40:21

luca-deltodesco
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Registered: 2006-05-05
Posts: 1,470

Re: double integrals

W = 2w
H = 2h



yes, i guess it does, apart from having an extra WH in it hmm


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