How to determine the same moment of inertia in two different ways?

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Homework Help Overview

The discussion revolves around determining the moment of inertia of an area with respect to the y-axis using two different approaches involving rectangular elements. Participants are exploring the mathematical formulation and interpretation of the problem.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants are attempting to express the moment of inertia using integrals, with some suggesting specific forms involving area mass density. There is a question about the naming of a particular method used in the calculations.

Discussion Status

The discussion is ongoing, with participants sharing their attempts and questioning the appropriateness of certain equations. There is no explicit consensus yet, but various interpretations and approaches are being explored.

Contextual Notes

Participants note a potential misinterpretation of variables and are discussing the implications of using different thicknesses for the rectangular elements in their calculations. The original problem statement specifies the need to solve the moment of inertia using two distinct methods.

Tapias5000
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Homework Statement
determine the moment of inertia of the area with respect to the y-axis. with different rectangular elements, solve the problem in two ways: (a) with thickness dx, and (b) with thickness dy.
Relevant Equations
## I_x=\int _{ }^{ }y^2dA ##
Imagen2.png

My solution is

<math xmlns=http://www.w3.org/1998/Math/MathML display=block data-is-equatio=1 data-latex=\begin{array}{l}I_x=\int_{ }^{ }y^2dA\\
I_x=\int_0^4y^22xdy\ \left[y=4-4x^2,\ \textcolor{#E94D40}{\sqrt{\frac{4-y}{4}}=x}\right]\\
I_x=2\int_0^4y^2\sqrt{\frac{4-y}{4}}dy\ \left(u=\frac{4-y}{4},\ dy=-4du\right)\\
I_x=-8\int_0^4y^2\sqrt{u}du\ \left[u=\frac{4-y}{4},\ \textcolor{#E94D40}{4-4u=y}\right]\\
I_x=-8\int_0^4\left(4-4u\right)^2\sqrt{u}du\\
I_x=-8\int_0^4\left(4^2-2\cdot4\cdot4u+\left(4u\right)^2\right)\sqrt{u}du\\
I_x=-8\int_0^4\left(16-32u+16u^2\right)\sqrt{u}du\\
I_x=-8\int_0^4\left(16\sqrt{u}-32u\sqrt{u}+16u^2\sqrt{u}\right)du\\
I_x=-\frac{32\cdot8}{3}u^{\frac{3}{2}}+\frac{64\cdot8}{5}u^{\frac{5}{2}}-\frac{32\cdot8}{7}u^{\frac{7}{2}}\ \left[\textcolor{#E94D40}{u=\frac{4-y}{4}}\right]\\
I_x=-\frac{32\cdot8}{3}\left(\frac{4-y}{4}\right)^{\frac{3}{2}}+\frac{64\cdot8}{5}\left(\frac{4-y}{4}\right)^{\frac{5}{2}}-\frac{32\cdot8}{7}\left(\frac{4-y}{4}\right)^{\frac{7}{2}}\begin{bmatrix}4\\
0\end{bmatrix}\\
I_x=\cancel{\textcolor{#E94D40}{-\frac{32\cdot8}{3}\left(\frac{4-4}{4}\right)^{\frac{3}{2}}+\frac{64\cdot8}{5}\left(\frac{4-4}{4}\right)^{\frac{5}{2}}-\frac{32\cdot8}{7}\left(\frac{4-4}{4}\right)^{\frac{7}{2}}}}\\
-\left(-\frac{32\cdot8}{3}\left(\frac{4-0}{4}\right)^{\frac{3}{2}}+\frac{64\cdot8}{5}\left(\frac{4-0}{4}\right)^{\frac{5}{2}}-\frac{32\cdot8}{7}\left(\frac{4-0}{4}\right)^{\frac{7}{2}}\right)\\
I_x=19.5\mathit{\text{pulg}}^4\\
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5R0312grRjtCb0_DBMKABiZNFnXFZDNdnBYiWDqWz9Jf=s1600.png

now I am asked for the same result but in this form but I don't know where to start.
Lw0m9UUMPaizX9NMeeRmZaIPDWuUg1pKXf6t720gKKzE=s1600.png
 
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How about
2\sigma \int_0^1 x^2 y \ \ dx=32\sigma \int_0^{\pi/2} \sin^2\theta \cos^2\theta d\theta
where ##\sigma## is area mass density of board.
 
Last edited:
anuttarasammyak said:
How about
2\sigma \int_0^1 x^2 y \ \ dx=32\sigma \int_0^{\pi/2} \sin^2\theta \cos^2\theta d\theta
where ##\sigma## is area mass density of board.
waat, Can you tell me the name of this method?
 
I misinterpreted y so
2\sigma \int_0^1 x^2 y \ \ dx=2\int_0^1 x^2 (4-4x^2) \ \ dx
This a way (b).
Tapias5000 said:
Homework Statement:: determine the moment of inertia of the area with respect to the y-axis. with different rectangular elements, solve the problem in two ways: (a) with thickness dx, and (b) with thickness dy.
Relevant Equations:: ## I_x=\int _{ }^{ }y^2dA ##
This relevant equation seems inappropriate because x should be squared for y-axis rotation inertia.
For (a)

2\sigma\int_0^4 dy \int_0^{\frac{\sqrt{4-y}}{2}} x^2 dx
 
Last edited:

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