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Moment of inertia for a rhombus

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  • #26
OlderDan
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Hey I just did both integrals (mine and yours) and I get different answers. I have checked them and the same. Mine gives I = (8-3sqrt2)mc^2/(12sqrt2)
Yours is giving me a negative number which seems illogical. What is happening?
They are indeed the same.

Your way:

[tex] I = 2\sigma \int_0^{c/\sqrt 2 } {x^2 \int_x^{c\sqrt 2 - x} {dydx} } [/tex]

[tex] I = 2\sigma \int_0^{c/\sqrt 2 } {x^2 dx} \left( {c\sqrt 2 - 2x} \right) [/tex]

[tex] I = 2\sigma \int_0^{c/\sqrt 2 } {\left( {c\sqrt 2 x^2 - 2x^3 } \right)dx} [/tex]

[tex] I = 2\sigma \left( {c\sqrt 2 \frac{{\left( {c/\sqrt 2 } \right)^3 }}{3} - \frac{{\left( {c/\sqrt 2 } \right)^4 }}{2}} \right) = 2\sigma \left( {\frac{2}{3} - \frac{1}{2}} \right)\left( {c/\sqrt 2 } \right)^4 [/tex]

[tex] I = 2\sigma \left( { - \frac{{\left( {c/\sqrt 2 } \right)^4 }}{2} + c\sqrt 2 \frac{{\left( {c/\sqrt 2 } \right)^3 }}{3}} \right) = \sigma \left( {\frac{1}{{12}}} \right)c^4 = \frac{M}{{12}}c^2 [/tex]

My way:

[tex] I = 4\sigma \int_0^{c/\sqrt 2 } {x^2 \int_x^{c/\sqrt 2 } {dydx} } [/tex]

[tex] I = 4\sigma \int_0^{c/\sqrt 2 } {\left( {c/\sqrt 2 x^2 - x^3 } \right)dx} [/tex]

[tex] I = 4\sigma \left( {\frac{{\left( {c/\sqrt 2 } \right)^4 }}{3} - \frac{{\left( {c/\sqrt 2 } \right)^4 }}{4}} \right) = 4\sigma \left( {\frac{1}{3} - \frac{1}{4}} \right)\left( {c/\sqrt 2 } \right)^4 = \sigma \left( {\frac{1}{{12}}} \right)c^4 = \left( {\frac{M}{{12}}} \right)c^2 [/tex]
 
  • #27
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Oh it wasn't a problem with the limits; just the not being careful doing the integrals. Hey thanks a lot Dan. It's not the first time you help me. Appreciate it.
 
  • #28
OlderDan
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Oh it wasn't a problem with the limits; just the not being careful doing the integrals. Hey thanks a lot Dan. It's not the first time you help me. Appreciate it.
You're welcome. Anything about that result look familiar?
 
  • #29
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It is identical to the case of a rod with the axis of rotation through the center. Other rotational lines for the same square also give similar rotational inertias to other cases of axis for a rod. I can speculate why this happens but I'm not sure. Why is it that this happens?
 
  • #30
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sorry the comparison should have used at least a rectangle (more general), not a square. However, the results for the rectangle are identical to those of a rod.
 
  • #31
OlderDan
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It is identical to the case of a rod with the axis of rotation through the center. Other rotational lines for the same square also give similar rotational inertias to other cases of axis for a rod. I can speculate why this happens but I'm not sure. Why is it that this happens?
A moment of inertia about a symmetry axis is just a sum over all the dm mass elements times their distances from the axis squared. There are many ways to do such sums. Your division of the plate into two parts and my division of it into four parts are examples. A flat rectangular plate rotating about an in-plane axis that divides it into equal rectangles can be thought of as a series of thin rods stacked side by side. The total moment of inertia is just the sum of the moments of inertia of all the individual rods.

In your problem, you could have divided the plate into strips of height dy and treated each strip as a rod. The problem is more complicated than an axis that divides the plate into equal rectangles because the strips have variable length and mass. But you could have used dI = dmL²/12 as the moment of inertia of a strip of height dy, with dm and L being functions of y and dy and integrated over y to get the sum. In fact, if you take your integral and do the x integral first and the y integral second, that is effecively what you would be doing. You don't get I = M(2c²)/12, which would be the result for a rod whose length is the diagonal of the square because most of the "rods" you are adding are shorter than the diagonal and have proportionately less mass. But for an axis that divides the plate into equal rectangles, all the little rods are identical so the moment of inertia looks just like the moment of inertia of a rod.
 
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  • #32
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Hmm, I wonder why there is no explanation of this in my book. Thanks again.
 

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