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

  • #1
Tapias5000
46
10
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\\
\ \end{array}><mtable columnalign=left columnspacing=1em rowspacing=4pt><mtr><mtd><msub><mi>I</mi><mi>x</mi></msub><mo>=</mo><msubsup><mo data-mjx-texclass=OP>∫</mo><mrow data-mjx-texclass=ORD/><mrow data-mjx-texclass=ORD/></msubsup><msup><mi>y</mi><mn>2</mn></msup><mi>d</mi><mi>A</mi></mtd></mtr><mtr><mtd><msub><mi>I</mi><mi>x</mi></msub><mo>=</mo><msubsup><mo data-mjx-texclass=OP>∫</mo><mn>0</mn><mn>4</mn></msubsup><msup><mi>y</mi><mn>2</mn></msup><mn>2</mn><mi>x</mi><mi>d</mi><mi>y</mi><mtext></mtext><mrow data-mjx-texclass=INNER><mo data-mjx-texclass=OPEN>[</mo><mi>y</mi><mo>=</mo><mn>4</mn><mo>−</mo><mn>4</mn><msup><mi>x</mi><mn>2</mn></msup><mo>,</mo><mtext></mtext><mstyle mathcolor=#E94D40><msqrt><mfrac><mrow><mn>4</mn><mo>−</mo><mi>y</mi></mrow><mn>4</mn></mfrac></msqrt><mo>=</mo><mi>x</mi></mstyle><mo data-mjx-texclass=CLOSE>]</mo></mrow></mtd></mtr><mtr><mtd><msub><mi>I</mi><mi>x</mi></msub><mo>=</mo><mn>2</mn><msubsup><mo data-mjx-texclass=OP>∫</mo><mn>0</mn><mn>4</mn></msubsup><msup><mi>y</mi><mn>2</mn></msup><msqrt><mfrac><mrow><mn>4</mn><mo>−</mo><mi>y</mi></mrow><mn>4</mn></mfrac></msqrt><mi>d</mi><mi>y</mi><mtext></mtext><mrow data-mjx-texclass=INNER><mo data-mjx-texclass=OPEN>(</mo><mi>u</mi><mo>=</mo><mfrac><mrow><mn>4</mn><mo>−</mo><mi>y</mi></mrow><mn>4</mn></mfrac><mo>,</mo><mtext></mtext><mi>d</mi><mi>y</mi><mo>=</mo><mo>−</mo><mn>4</mn><mi>d</mi><mi>u</mi><mo data-mjx-texclass=CLOSE>)</mo></mrow></mtd></mtr><mtr><mtd><msub><mi>I</mi><mi>x</mi></msub><mo>=</mo><mo>−</mo><mn>8</mn><msubsup><mo data-mjx-texclass=OP>∫</mo><mn>0</mn><mn>4</mn></msubsup><msup><mi>y</mi><mn>2</mn></msup><msqrt><mi>u</mi></msqrt><mi>d</mi><mi>u</mi><mtext></mtext><mrow data-mjx-texclass=INNER><mo data-mjx-texclass=OPEN>[</mo><mi>u</mi><mo>=</mo><mfrac><mrow><mn>4</mn><mo>−</mo><mi>y</mi></mrow><mn>4</mn></mfrac><mo>,</mo><mtext></mtext><mstyle mathcolor=#E94D40><mn>4</mn><mo>−</mo><mn>4</mn><mi>u</mi><mo>=</mo><mi>y</mi></mstyle><mo data-mjx-texclass=CLOSE>]</mo></mrow></mtd></mtr><mtr><mtd><msub><mi>I</mi><mi>x</mi></msub><mo>=</mo><mo>−</mo><mn>8</mn><msubsup><mo data-mjx-texclass=OP>∫</mo><mn>0</mn><mn>4</mn></msubsup><msup><mrow data-mjx-texclass=INNER><mo data-mjx-texclass=OPEN>(</mo><mn>4</mn><mo>−</mo><mn>4</mn><mi>u</mi><mo data-mjx-texclass=CLOSE>)</mo></mrow><mn>2</mn></msup><msqrt><mi>u</mi></msqrt><mi>d</mi><mi>u</mi></mtd></mtr><mtr><mtd><msub><mi>I</mi><mi>x</mi></msub><mo>=</mo><mo>−</mo><mn>8</mn><msubsup><mo data-mjx-texclass=OP>∫</mo><mn>0</mn><mn>4</mn></msubsup><mrow data-mjx-texclass=INNER><mo data-mjx-texclass=OPEN>(</mo><msup><mn>4</mn><mn>2</mn></msup><mo>−</mo><mn>2</mn><mo>⋅</mo><mn>4</mn><mo>⋅</mo><mn>4</mn><mi>u</mi><mo>+</mo><msup><mrow data-mjx-texclass=INNER><mo data-mjx-texclass=OPEN>(</mo><mn>4</mn><mi>u</mi><mo data-mjx-texclass=CLOSE>)</mo></mrow><mn>2</mn></msup><mo data-mjx-texclass=CLOSE>)</mo></mrow><msqrt><mi>u</mi></msqrt><mi>d</mi><mi>u</mi></mtd></mtr><mtr><mtd><msub><mi>I</mi><mi>x</mi></msub><mo>=</mo><mo>−</mo><mn>8</mn><msubsup><mo data-mjx-texclass=OP>∫</mo><mn>0</mn><mn>4</mn></msubsup><mrow data-mjx-texclass=INNER><mo data-mjx-texclass=OPEN>(</mo><mn>16</mn><mo>−</mo><mn>32</mn><mi>u</mi><mo>+</mo><mn>16</mn><msup><mi>u</mi><mn>2</mn></msup><mo data-mjx-texclass=CLOSE>)</mo></mrow><msqrt><mi>u</mi></msqrt><mi>d</mi><mi>u</mi></mtd></mtr
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
 

Answers and Replies

  • #2
anuttarasammyak
Gold Member
1,850
951
How about
[tex]2\sigma \int_0^1 x^2 y \ \ dx=32\sigma \int_0^{\pi/2} \sin^2\theta \cos^2\theta d\theta[/tex]
where ##\sigma## is area mass density of board.
 
Last edited:
  • #3
Tapias5000
46
10
How about
[tex]2\sigma \int_0^1 x^2 y \ \ dx=32\sigma \int_0^{\pi/2} \sin^2\theta \cos^2\theta d\theta[/tex]
where ##\sigma## is area mass density of board.
waat, Can you tell me the name of this method?
 
  • #4
anuttarasammyak
Gold Member
1,850
951
I misinterpreted y so
[tex]2\sigma \int_0^1 x^2 y \ \ dx=2\int_0^1 x^2 (4-4x^2) \ \ dx[/tex]
This a way (b).
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)

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

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