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My Mechanical textbook (Bedford & Fowler 4th Edition) has a worked out example for determining a couple using euler's equations. This is not a homework question (at least I don't think it is?), this is a request for some elaboration on a pre-worked example.

I am trying to work through the problem myself to come up with the same answer but I have inconsistencies with the values for the moment of inertia tensor. Here are the relevant pages of the book (apologies for the lack of quality on the pictures): enter image description here

enter image description here

As shown on page 2, the moment of inertia tensor is calculated as: $$ \begin{bmatrix} I_{xx} & -I_{xy} & -I_{xz} \\ -I_{yx} & I_{yy} & -I_{yz} \\ -Izx & -Izy & Izz \\ \end{bmatrix} = \begin{bmatrix} 0.48 & -0.18 & 0 \\ -0.18 & 0.12 & 0 \\ 0 & 0 & 0.6 \\ \end{bmatrix} $$

The values for the moments of inertia about the x, y and z axes for the plate seems to conflict with my understanding:

$$ I_{xy} = \int_{}^{}f(xy)\,dm = (0.150 * 0.300) * 4 kg = 0.18 kg.m^2 $$ $$ I_{yx} = \int_{}^{}f(yx)\,dm = (0.300 * 0.150) * 4 kg = 0.18 kg.m^2 $$ $$ I_{xz} = \int_{}^{}f(xz)\,dm = (0.150 * 0) * 4 kg = 0.00 kg.m^2 $$ $$ I_{zx} = \int_{}^{}f(zx)\,dm = (0 * 0.150) * 4 kg = 0.00 kg.m^2 $$ $$ I_{zy} = \int_{}^{}f(zy)\,dm = (0 * 0.300) * 4 kg = 0.00 kg.m^2 $$ $$ I_{yz} = \int_{}^{}f(yz)\,dm = (0.300 * 0) * 4 kg = 0.00 kg.m^2 $$ $$ I_{zz} = \int_{}^{}f(x^2 + y^2)\,dm = (0.150^2 + 0.300^2) * 4 kg = 0.45 kg.m^2 $$ $$ I_{xx} = \int_{}^{}f(y^2 + z^2)\,dm = (0.300^2 + 0^2) * 4 kg = 0.36 kg.m^2 $$ $$ I_{yy} = \int_{}^{}f(x^2 + z^2)\,dm = (0.150^2 + 0^2) * 4 kg = 0.09 kg.m^2 $$

Thus my inertia tensor is different from the book's: $$ \begin{bmatrix} 0.36 & -0.18 & 0 \\ -0.18 & 0.09 & 0 \\ 0 & 0 & 0.45 \\ \end{bmatrix} \ne \begin{bmatrix} 0.48 & -0.18 & 0 \\ -0.18 & 0.12 & 0 \\ 0 & 0 & 0.6 \\ \end{bmatrix} $$ Could someone explain to me what conceptual mistake I have made?

For a uniform plate with the reference rotation being at any corner, shouldn't the x,y and z distance (for the moment of inertia of the respective axis) be taken from the center of gravity of the plate to the rotation point?

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Be careful with your integration elements, as $dm=\rho dA$, where $\rho$ is the area density of the plate and $dA=dxdy$ is the area integration element. In addition you should integrate from the rotation point to the end of your plate. So for example;

$$\begin{equation}I_{xx}\\=\int_A(y^2+z^2)\rho dA\\=\rho\int_{x=0}^{0.3\text{ m}}dx\int_{y=0}^{0.6\text{ m}}y^2dy\\=\frac{4\text{ kg}}{0.3\text{ m}\cdot0.6\text{ m}}\cdot x|_0^{0.3\text{ m}}\cdot\tfrac{1}{3}y^3|_0^{0.6\text{ m}}=0.48 \text{ kg}\,\text{m}^2.\end{equation}$$

Note that maybe by accident the error you made in the $I_{xy}$ term cancels

$$I_{xy}=I_{yx}\\=\int_Axy\rho dA\\=\rho\int_xxdx\int_yydy\\=\rho\tfrac{1}{2}x^2|_0^{0.3}\tfrac{1}{2}y^2|_0^{0.6}=0.18\text{ kg}\,\text{m}^2.$$

Hope this helps.

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  • $\begingroup$ This helped immensely. I think I understand now as I was able to apply this understanding to a different problem involving a series of 3 connected "point beams" rotating about an axis where dm = ρ.dl with ρ representing the linear density of the beams (kg/m) and dl being the change in length which was equal to dx, dy or dz depending on the direction they were pointed in. $\endgroup$
    – user155876
    Nov 2 '15 at 17:19

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