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I have a trapezoidal and homogeneous lamina with mass m and ABCD vertices, on an xy plane. AB is the minor base, CD is the major base and AC is the height, with AB=AC=L and CD=2L. So basically I have a right-angled trapezoid, composed by a square of side L and a right isosceles triangle (measure of catheti is also L). I'm asked to find the moment of inertia with respect to an axis orthogonal to the lamina passing through the center of mass. My teacher suggests to find the moment of inertia with respect to a z-axis passing through D, and then to use the Huygens-steiner theorem using the distance between D and the COM. enter image description here

So I placed D in the origin and I calculated: \begin{equation} I_{zz}=\int_S{\sigma(x^2+y^2)dxdy} \end{equation} with $\sigma=2m/3L^2$ \begin{equation} \int_S{(x^2+y^2)dxdy}=\int_0^L{dx}\int_0^x{dy(x^2+y^2)}+\int_L^{2L}{dx}\int_0^L{dy(x^2+y^2)}=\frac{L^4}{3}+\frac{8L^4}{3}=3L^4 \end{equation} So \begin{equation} I_{zz}=2mL^2 \end{equation} But my teacher says that "the moment of inertia with respect to D is $J_D=37/18mL^2$". What am I doing wrong?

EDIT: Maybe I didn't explain myself correctly. I'm pretty sure my math is good, mine is not a math problem. But I'm not sure this is the correct way to proceed. My book always calculates the moment of inertia with respect to the center of mass and using the principal axes of inertia (I'm studying on Landau). So I'm not sure if I can calculate the moment of inertia with respect to a random point (and three random axes) and THEN use the parallel axes theorem to "shift" my moment of inertia to the center of mass. Am I proceeding right?

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  • $\begingroup$ I wonder if you would have better luck with this question on the math.stackexchange. "What is wrong with this" homework questions are considered off topic on this site. I have to admit I would solve the problem slightly differently - finding moment of inertia about X and Y separately, then using perpendicular axis theorem and finally parallel axis theorem. Turns the double integral into two single integrals. But your math doesn't look too wrong. $\endgroup$ – Floris May 8 '16 at 1:04
  • $\begingroup$ I edited my question, trying to explain my problem in a better way. You are suggesting to find the elements $I_{xx}$ and $I_{yy}$ of my inertia tensor and then apply $I_{zz}=I_{xx}+I_{yy}$? $\endgroup$ – Luthien May 8 '16 at 1:42
  • $\begingroup$ The edit seems like a different question that the original which points out that your teachers answer for I about D is different from yours by a tiny amount (1/18mL^2) $\endgroup$ – Floris May 8 '16 at 3:18
  • $\begingroup$ Yes, you're right, I didn't point out what was bothering me the most, not the integrals (they're really simple and I checked them multiple times), but the proceeding. Because of that tiny difference (1/18mL^2) I thought that I was doing something wrong and that maybe I couldn't calculate the moment of inertia using a point that was not the center of mass! $\endgroup$ – Luthien May 8 '16 at 11:26
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The parallel axis theorem should still be applicable - but note that it states that the moment of inertia about an arbitrary axis is always greater than the moment of inertia about the axis through the center of mass. So in your case after calculating the moment of inertia through D you have to subtract mr^2 as you move the axis to the center of mass.

Did I interpret your question correctly?

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  • $\begingroup$ That part is clear to me. I don't understand how to correctly calculate the moment through D. It seems to me that the orientation and the origin of x-y axes it's not important to calculate the moment with respect of an axis passing through D. So, i can choose whatever orientation and origin I want to calculate that moment of inertia? Am I doing right using that frame of reference? If I, for example, shift the origin of my reference frame my result will still be the same? (and I guess it should be, if I'm proceeding right) $\endgroup$ – Luthien May 8 '16 at 11:35
  • $\begingroup$ Thank for your effort and sorry if I'm not explaining myself very well but this is not my language and sometimes I find it hard to organize my thoughts in another language to explain myself in the best way! $\endgroup$ – Luthien May 8 '16 at 11:36
  • $\begingroup$ Yes there are multiple ways you can proceed. In principle you can pick any point you like (and point through which it is easy to calculate the moment of inertia $I_{easy}$), and then determine the distance $d$ from there to the COM. The answer is then $I_{easy}-md^2$ regardless of how you picked the initial axis. $\endgroup$ – Floris May 8 '16 at 11:59
  • $\begingroup$ Thank you, now it's clear to me! Plus you made me notice that I can check my result using $I_{zz}=I_{xx}+I_{yy}$, thanks! $\endgroup$ – Luthien May 9 '16 at 13:08

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