anonymous
  • anonymous
Need help with clarification on this. Using the moment of inertia equation I_x = ∫y^2dA I_y = ∫x^2dA to derive a a
Physics
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katieb
  • katieb
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anonymous
  • anonymous
using \[I_x = \int\limits_{a}^{?}y^2dA\] and \[I_y = \int\limits_{a}^{?}x^2dA\] derive the equations for a rectangle \[I_x = 1/12 * bh^3 , I_x = 1/3*bh^3\] and derive the equations for a right triangle \[I_x = 1/12 *bh^3, I_x = 1/36 * bh^3, I_y = 1/4 *b^3h\]
IrishBoy123
  • IrishBoy123
this is not hard; but approaching it a pre-defined equation makes me wonder. all we need to do is build a small element and integrate.....:p au naturale 🌿
anonymous
  • anonymous
there must be something I'm not seeing then. is it too obvious?

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mathmate
  • mathmate
|dw:1447123736118:dw| So for MI about centroidal axis \(I_x =\int\int y^2dA=\int_{-h/2}^{h/2}\int_{-b/2}^{b/2} y^2 ~dx~dy\) \(=b\int_{-h/2}^{h/2} y^2 ~dy=b[y^3/3]_{-h/2}^{h/2}=bh^3/12\) Similarly for Iy if you need it. I will leave it to you to derive the same about the base of the rectangle, which will give bh^3/3. [limts will be 0-b, and 0-h] For a right triangle, you need to find the equation of the hypotenuse, and be careful with the limits and order of integration.
IrishBoy123
  • IrishBoy123
hi @zuzu304 nah, not at all suggesting there is anything **obvious** about it. i was originally just thinking that, if you understand the idea behind the moment of inertia, then you don't need to remember equations, you just derive them as part of the process. however, looking again at mathmate's solution, i realise that maybe this is more of a mathematical exercise in double integrals. maybe the learning objective is getting comfortable with the double integration, in which case mathmate has provided the steer --see especially in the last para re limits of integration for the triangle. do come back if there is more you need :p
anonymous
  • anonymous
I read the book that the question was based on it and eventually got it. thanks though
IrishBoy123
  • IrishBoy123
\(\Huge{\Huge{\Huge 💥}}\)

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