[QUE/ME-14005] ME-PROBLEM

 Find the moment of inertia tensor of a uniform rectangular plate w.r.t. the centre \(O\) of the plate relative to the axes \(K\) shown in figure. Use parallel axes theorem to find moment of inertia tensor relative to a set of axes \(K_2\) with origin taken as the corner \(E\). The \(Z\)- axis for both systems is perpendicular to the plate and out of the plane of paper.
%FigBelow{10,-25}{40}{0} RectangularPlate

We shall do it by evaluating the double integral. Divide the plate by lines parallel to the coordinate axes as in %Figref MI-Plate
Consider one rectangular element of sides \((dx, dy)\) at position \((x,y)\). If the mass of the plate is \(M\), the density of the plate is \(\sigma\), then \(4ab\sigma =M\) and the contribution of the rectangular element to \(I_{xx}\) is \(\sigma (dx\,dy) y^2\). Summing over all elements means integrating over \(x, y\) over their respective ranges. Thus %FigBelow{10,-25}{40}{0} RectangularPlate \begin{eqnarray}\nonumber I_{xx} &=& \int_{-a}^a dx \int_{-b}^b dy \sigma y^2 =\int_{-a}^a dx \sigma \frac{2b^3}{3} = \sigma \frac{2b^3}{3} \int_{-a}^a dx\\ &=& \sigma \frac{2b^3}{3} (2a) = \frac{Mb^2}{3} \qquad \HighLight{\mbox{$\because 4\sigma ab=M$}} \end{eqnarray} % Similarly \begin{eqnarray}\nonumber I_{yy} &=& \int_{-a}^a dx \int_{-b}^b dy \sigma x^2 =\int_{-a}^a dx \sigma x^2 (2b) = \sigma \frac{2b^3}{3} \int_{-a}^a dx\qquad \HighLight{\mbox{$\because 4\sigma ab=M$}}\\ &=& \sigma \frac{2a^3}{3} (2b) = \frac{Ma^2}{3} \end{eqnarray} and appealing to the law of perpendicular axes we get \(I_{zz}= M\frac{a^2+b^2}{3}\) The off diagonal term \(I_{xy}\) is \begin{equation} I_{xy} = - \int_{-a}^a dx \int_{-b}^b dy \sigma (xy) \end{equation} which vanishes due to the symmetry of the problem. Also \begin{equation} I_{xz} = - \int_{-a}^a dx \int_{-b}^b dy \sigma (xz) = 0 \end{equation} because \(z=0\) for all rectangular elements of the plate. Thus the moment of inertia tensor w.r.t.the center of mass is given by \begin{equation} \underline{\Ibb} = \frac{M}{6} \begin{pmatrix} b^2 & 0 & 0\\ 0 & a^2 & 0\\ 0 & 0 & a^2+b^2 \end{pmatrix} \end{equation} \paragraph*{MI Tensor relative to a corner} The parallel axes for the inertia tensor states that \begin{equation} I_{jk} = I^\text{cm}_{jk} + \Delta_{jk} \end{equation} where \(\Delta_{jk} =|\vec{a}|^2 \delta_{jk} -a_ja_k\) and \(\vec{a}\) is the position vector of the corner relative to the origin \(O\). For the corner \(E\) we have \(\vec{a}=(-a,-b,0)\) and \(|\vec{a}|^2=(a^2+b^2)\). Therefore, \begin{eqnarray} &\Delta_{xx} = b^2,\qquad \qquad \qquad \Delta _{yy} = a^2,\quad \qquad \Delta_{zz} = (a^2+b^2)&\\ &\qquad\quad\Delta_{xy}=\Delta_{yx}=-ab\qquad \Delta_{yz}=\Delta_{zy}=0\qquad \Delta_{zx}=\Delta_{xz}=0.\qquad\qquad& \end{eqnarray} Thus the moment of inertia tensor w.r.t.the corner at \(E\) is given by \begin{equation} \underline{\Ibb} = \frac{M}{3} \begin{pmatrix} b^2 + 3a^2 & -3ab & 0\\ -3ab & a^2 + 3b^2 & 0\\ 0 & 0 & 4a^2+4b^2 \end{pmatrix}. \end{equation}