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Show that the moment of inertia tensor of a uniform triangular plate, see figure, with centre of mass chosen as origin is given by \begin{equation} I= \frac{mL^2}{18} \begin{pmatrix} 1 & -\frac{1}{2} & 0\\ -\frac{1}{2} & 1 & 0\\ 0 & 0 & 2 \end{pmatrix}\end{equation} 

In % Figref{me-fig-14012} OAB is the right angle triangle lamina with two sides of equal length \(L\). We choose the coordinate axes as shown in the figure. The corners of the plate are \((0,0,0),(L,0,0),(l,l,0)\) and the centre of mass has coordinates \((\frac{2L}{3},\frac{L}{3}, 0)\). It is easy to calculate the moment of inertia tensor w.r.t. this set of axes. The inertia tensor relative to the centre of mass will then be found by making use of parallel axes theorem.
% FigBelow{10,-14}{40}{00}{me-fig-14012} Choice of axes
Simplest way to solve the problem is to use double integral. Divide the plate into small rectangular elements as shown, obtain contribution of one such element and perform integrations over a range of \(x\) and \(y\) so as to get sum over all the rectangular elements of the plate. Let the mass per unit area of the plate be \(\sigma\), then \(M=\frac{1}{2}\sigma L^2.\) We first record the expression for moment of inertia tensor for a rigid body consisting of masses \(m_\alpha\) at position \(\vec{r}_\alpha\). \begin{eqnarray} I_{xx} &=& \sum_\alpha m_\alpha \big( \vec{r}\,^2- x_\alpha^2\big) =\sum_\alpha m_\alpha \big( y_\alpha^2+ z_\alpha^2\big)\\ I_{yy} &=& \sum_\alpha m_\alpha \big( \vec{r}\,^2_\alpha- y_\alpha^2\big) =\sum_\alpha m_\alpha \big(x_\alpha^2+ z_\alpha^2\big)\\ I_{zz} &=& \sum_\alpha m_\alpha \big( \vec{r}\,^2_\alpha- z_\alpha^2\big) =\sum_\alpha m_\alpha \big(y_\alpha^2+ z_\alpha^2\big) \end{eqnarray} The contribution of the element \(dx\, dy\) at position \(x,y\) to \(I_{xx}\) is easily written down as, remembering that for points on the plate \(z=0\), we get \begin{eqnarray} dI_{xx}= (\sigma dx\,dy)(\vec{r}^2-x^2) = (\sigma dx\,dy) y^2. \end{eqnarray} For a fixed \(x\), the variable \(y\) takes values from 0 to \(x\) and for the plate \(x\) ranges from 0 to \(L\). Therefore \(I_{xx}\) is given by \begin{eqnarray} I_{xx} &=& \int_0^L dx \int_0^x dy \sigma y^2 = \sigma\int_0^L dx \frac{x^3}{3} = \frac{\sigma L^4}{12} \\ &=& \frac{1}{6}M L^2, \qquad \qquad \because \sigma L^2 =2M. \end{eqnarray} The expression for moment of inertia \(I_{yy}\) \begin{equation} I_{yy} = \sum_\alpha m_\alpha (\vec{r}_\alpha^2 -y_\alpha^2) \end{equation} for the planer lamina problem takes the form \begin{eqnarray} I_{yy} &=& \int_0^L dx \int_0^x dy (\sigma x^2) = \int_0^L dx\, (\sigma {x^2})\times y\Big|_{y=0}^{y=x} = \frac{\sigma L^4}{4}\\ &=& \frac{ML^2}{2}. \end{eqnarray} and \(I_{zz}\) is given by \begin{eqnarray} I_{zz} &=& \int_0^L dx \int_0^x dy \sigma (x^2 + y^2) = \int_0^L \sigma (x^2 y + \frac{y^3}{3})\Big|_{y=0}^{y=x}\\ &=& \int_0^L dx \sigma \frac{4x^3}{3}= \sigma \frac{L^4}{3}\\ &=& \frac{2ML^2}{3}. \end{eqnarray} The off diagonal term \(I_{xy}\) is nonzero and is given by \begin{eqnarray} I_{xy} &=& -\int_0^L dx \int_0^L dy \sigma (\sigma x)\,y = -\int_0^L dx (\sigma x) \times \frac{y^2}{2}\Big|_{y=0}^{y=x}\\ & =& -\int_0^L dx \sigma \frac{x^3}{8} = - \sigma \frac{L^2}{8}\\ &=& - \frac{M L^2}{4}. \end{eqnarray} The parallel axis theorem gives \begin{eqnarray} I^\text{cm}_{jk}= I_{jk} - M (|\vec{a}|^2 \delta_{jk}-a_j a_k) \end{eqnarray} where \(\vec{a}\) is the position vector of the origin \(O\) relative to the centre of mass. In the present case the centre of mass has coordinates \((2L/3,L/3,0)\). Therefore \(\vec{a} = (-2L/3, -L/3,0)\). Thus we get \begin{eqnarray} I^\text{cm}_{xx} &=& I_{xx} - M (L^2/9) = (\frac{1}{6}-\frac{1}{9})ML^2 =\frac{ML^2}{18}\\ I^\text{cm}_{yy} &=& I_{yy} - M (4 L^2/9) = (\frac{1}{2}-\frac{4}{9})ML^2 =\frac{ML^2}{18}\\ I^\text{cm}_{zz}&=& I_{zz} - ML^2{(1+4)L^3}{9} = (\frac{2}{3}-\frac{5}{9})ML^2 = \frac{ML^2}{9}\\ I^\text{cm}_xy &=& I_{xy} -a_xa_y = - \frac{ML^2}{4} -\frac{2ML^2}{9} =(\frac{1}{4}-\frac{2}{9}) ML^2\\ &=& -\frac{ML^2}{36}. \end{eqnarray} Thus the moment of inertia tensor relative to the centre of mass is given by \begin{equation*} \underline{\sf I}=\frac{ML^2}{18} \begin{pmatrix} 1 & -\frac{1}{2} & 0 \\ -\frac{1}{2} & 1 & 0 \\ 0 & 0 & 2 \end{pmatrix} \end{equation*}

 Using the definition of Levi-Civita and Kronecker delta symbols to prove the following identity \begin{eqnarray}\label{EQ01} \epsilon_{i\,j\,k}\,\epsilon_{l\,m\,n}\ &=&\ \begin{vmatrix} \delta_{il}&\delta_{im}&\delta_{in}\\ \delta_{jl}&\delta_{jm}&\delta_{jn}\\ \delta_{kl}&\delta_{km}&\delta_{kn} \end{vmatrix} \end{eqnarray} 

Interchanging \(i\leftrightarrow j\) in the right hand side amounts to interchange of the first and second rows and therefore the right hand side changes sign. Similar statement holds for exchange of \(i\leftrightarrow k\) and \(i\leftrightarrow k\). Thus the hand side is completely antisymmetric in indices \(i,j,k\). Repeating the above argument we see that the right hand side is totally antisymmetrc in indices \(\ell, m, n \) also. Thus the right hand side must be proportional to \(\epsilon_{i\,j\,k}\,\epsilon_{l\,m\,n}\). Thus \begin{equation} \text{R.H.S. of \eqRef{EQ01}} = K \epsilon_{i\,j\,k}\,\epsilon_{l\,m\,n}\label{EQ02} \end{equation} The constant of proportionality \(K\) is fixed by substituting \(i=1,j=2,k=3, \ell=1, m=2, n=3\) in \eqRef{EQ02}. The right hand side of \eqRef{EQ01} becomes determinant of unit matix and hence equal to 1. The right hand side of \eqRef{EQ02} becomes \(K \epsilon_{123}\epsilon_{123}=K\). This gives \(K=1\) and the proof of \eqRef{EQ01} is complete.