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

AI Thread Summary
The discussion focuses on determining the moment of inertia using two methods: one with thickness dx and the other with thickness dy. The user presents equations involving area mass density (σ) and integrals to calculate the moment of inertia regarding the y-axis. There is confusion about the appropriate use of variables and the correct form of the equations, particularly regarding the integration limits and the squared terms. The conversation highlights the importance of accurately interpreting variables in the context of the problem. Ultimately, the goal is to achieve consistent results through both methods of calculation.
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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