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 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}

Find rotation matrix for a rotation by an angle \(\alpha\) about the axis \(1,2,1\) where \(\cos\alpha=\frac{3}{5}, \sin\alpha =\frac{4}{5}\).

The unit vector along the direction \((1,2,1)\) is given by \(\hat{n}=\frac{1}{\sqrt{6}}(1,2,1)\).\\ Under a rotation by an angle \(\alpha\) about axis \(\hat{n}=(n_1,n_2,n_3)\), the new components \(\vec{X}\) are related to old components \(\vec{x}\) by equation \begin{equation} \vec{X} = \vec{x} -\sin \alpha (\hat{n}\times\vec{x}) + (1-\cos\alpha)\hat{n}\times(\hat{n}\times\vec{x})) \end{equation} We compute \[\hat{n}\times\vec{x}=\big({n_2} {x_3}-{n_3} {x_2},{n_3} {x_1}-{n_1} {x_3},{n_1} {x_2}-{n_2} {x_1}\big)\] \begin{eqnarray} (\hat{n}\times(\hat{n}\times\vec{x}))_1&=&{n_2} ({n_1} {x_2}-{n_2} {x_1})-{n_3} ({n_3} {x_1}-{n_1} {x_3})\\ (\hat{n}\times(\hat{n}\times\vec{x}))_2&=&{n_3} ({n_2} {x_3}-{n_3} {x_2})-{n_1} ({n_1} {x_2}-{n_2} {x_1})\\ (\hat{n}\times(\hat{n}\times\vec{x}))_3&=& {n_1} ({n_3} {x_1}-{n_1} {x_3})-{n_2} ({n_2} {x_3}-{n_3} {x_2}) \end{eqnarray} Therefore \begin{eqnarray}\nonumber X_1&=&(1-\cos \alpha ) ({n_2} ({n_1} {x_2}-{n_2} {x_1})-{n_3} ({n_3} {x_1}-{n_1} {x_3}))-\sin \alpha ({n_2} {x_3}-{n_3} {x_2})+{x_1}\\\nonumber X_2&=&(1-\cos \alpha ) ({n_3} ({n_2} {x_3}-{n_3} {x_2})-{n_1} ({n_1} {x_2}-{n_2} {x_1}))-\sin \alpha ({n_3} {x_1}-{n_1} {x_3})+{x_2}\\\nonumber X_3&=& (1-\cos \alpha ) ({n_1} ({n_3} {x_1}-{n_1} {x_3})-{n_2} ({n_2} {x_3}-{n_3} {x_2}))-\sin \alpha ({n_1} {x_2}-{n_2} {x_1})+{x_3} \end{eqnarray} Therefore \begin{equation} \begin{pmatrix} X_1\\X_2\\X_3 \end{pmatrix} = \underline{\Rbb} \begin{pmatrix} x_1\\x_2\\x_3 \end{pmatrix}. \end{equation} where the matrix \(\Rbb\) is given by \begin{eqnarray}\nonumber \begin{pmatrix} -(1-\cos\alpha)(n_2^2+n_3^2) + 1 & (1-\cos\alpha)n_1n_2 + \sin\alpha n_3 & (1-\cos\alpha)n_3n_1 - \sin\alpha n_2\\ (1-\cos\alpha)n_1 n_2 -n_3\sin\alpha & -(1-\cos\alpha)(n_3^2+n_1^2) +1 & (1-\cos\alpha)n_2n_3 + n_1\sin\alpha\\ (1-\cos \alpha)n_1n_3 + n_2 \sin\alpha & (1-\cos\alpha)n_2n_3 -n_1 \sin\alpha & -(1-\cos\alpha)(n_1^2+n_2^2) + 1 \end{pmatrix} \end{eqnarray} Substituting values \begin{equation*} \hat{n}=\frac{1}{\sqrt{6}}(1,2,1),\quad (1-\cos\alpha)=\frac{2}{5}, \quad\sin\alpha=\frac{4}{5} \end{equation*} and simplifying gives % \begin{equation} \left(\begin{array}{ccc} \frac{2}{3} & \frac{2}{15} \left(1+\sqrt{6}\right) & \frac{1}{15} \left(1-4 \sqrt{6}\right) \\[2mm] \frac{2}{15} \left(1-\sqrt{6}\right) & \frac{13}{15} & \frac{2}{15} \left(1+\sqrt{6}\right) \\[2mm] \frac{1}{15} \left(1+4 \sqrt{6}\right) & \frac{2}{15} \left(1-\sqrt{6}\right) & \frac{2}{3} \end{array} \right) \end{equation}