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Algebra Difficulty 6.4 National olympiad Prove it

[ Moment of Inertia ]

a) Prove that the moment of inertia relative to the center of mass of a system of points with unit masses is equal to 1n∑i<j\frac{1}{n} \sum_{i<j} aij2a_{\mathrm{ij}}^{2}, where n−n- is the number of points, aij−a_{\mathrm{ij}}- is the distance between points with numbers ii and jj.

b) Prove that the moment of inertia relative to the center of mass of a system of points with masses m1,…,mnm_{1}, \ldots, m_{\mathrm{n}}, is equal to 1m∑i<j\frac{1}{m} \sum_{i<j} mimjaij2m_{\mathrm{i}} m_{\mathrm{j}} a_{\mathrm{ij}}^{2}, where m=m1+…+mn,aij−m=m_{1}+\ldots+m_{\mathrm{n}}, a_{\mathrm{ij}}- is the distance between points with numbers ii and jj.

Solution

a) Let xi\mathbf{x}_{i} be a vector with its origin at the center of mass OO and its end at the point with number ii. Then ∑i,j(xi−xj)2=∑i,j(xi2+xj2)−\sum_{i, j}\left(\mathbf{x}_{\mathbf{i}}-\mathbf{x}_{\mathrm{j}}\right)^{2}=\sum_{i, j}\left(\mathbf{x}_{\mathrm{i}}{ }^{2}+\mathbf{x}_{\mathrm{j}}{ }^{2}\right)- 2∑i,j(xi,xj)2 \sum_{i, j}\left(\mathbf{x}_{i}, \mathbf{x}_{\mathrm{j}}\right), where the summation is over all possible pairs of point numbers. Clearly, ∑i,j(xi2+xj2)=2n∑i\sum_{i, j}\left(\mathbf{x}_{\mathrm{i}}{ }^{2}+\mathbf{x}_{\mathrm{j}}{ }^{2}\right)=2 n \sum_{\boldsymbol{i}} xi2=2nIO\mathbf{x}_{\mathrm{i}}^{2}=2 n I_{\mathrm{O}} and ∑i,j(xi,xj)=∑i(xi,∑jxj)=0\sum_{i, j}\left(\mathbf{x}_{\mathrm{i}}, \mathbf{x}_{\mathrm{j}}\right)=\sum_{i}\left(\mathbf{x}_{\mathrm{i}}, \sum_{j} \mathbf{x}_{\mathrm{j}}\right)=0. Therefore, 2nIO=∑i,j(xi−xj)2=2∑i<jaij22 n I_{\mathrm{O}}=\sum_{i, j}\left(\mathbf{x}_{\mathrm{i}}-\mathbf{x}_{\mathrm{j}}\right)^{2}=2 \sum_{i<j} a_{\mathrm{ij}}{ }^{2}.

b) Let xi\mathbf{x}_{\mathrm{i}} be a vector with its origin at the center of mass OO and its end at the point with number ii. Then ∑i,jmimj(xi−xj)2=∑i,j\sum_{\mathbf{i}, j} m_{\mathrm{i}} m_{\mathrm{j}}\left(\mathbf{x}_{\mathrm{i}}-\mathbf{x}_{\mathrm{j}}\right)^{2}=\sum_{i, j} mimj(xi2+xj2)−2∑i,jmimj(xi,xj)m_{\mathrm{i}} m_{\mathrm{j}}\left(\mathbf{x}_{\mathrm{i}}^{2}+\mathbf{x}_{\mathrm{j}}{ }^{2}\right)-2 \sum_{i, j} m_{\mathrm{i}} m_{\mathrm{j}}\left(\mathbf{x}_{\mathrm{i}}, \mathbf{x}_{\mathrm{j}}\right). Clearly, ∑i,jmimj(xi2+xj2)=∑imi∑j(mjxi2+mjxj2)=∑imi(mxi2+IO)=2mIO\sum_{i, j} m_{\mathrm{i}} m_{\mathrm{j}}\left(\mathbf{x}_{\mathrm{i}}^{2}+\mathbf{x}_{\mathrm{j}}{ }^{2}\right)=\sum_{\boldsymbol{i}} m_{\mathrm{i}} \sum_{\boldsymbol{j}}\left(m_{\mathrm{j}} \mathbf{x}_{\mathrm{i}}^{2}+m_{\mathrm{j}} \mathbf{x}_{\mathrm{j}}{ }^{2}\right)=\sum_{\boldsymbol{i}} m_{\mathrm{i}}\left(m \mathbf{x}_{\mathrm{i}}{ }^{2}+I_{\mathrm{O}}\right)=2 m I_{\mathrm{O}} and ∑i,jmimj(xi,xj)=∑imi(xi,∑jmjxj)=0\sum_{i, j} m_{\mathrm{i}} m_{\mathrm{j}}\left(\mathbf{x}_{\mathrm{i}}, \mathbf{x}_{\mathrm{j}}\right)=\sum_{i} m_{\mathrm{i}}\left(\mathbf{x}_{\mathrm{i}}, \sum_{j} m_{\mathrm{j}} \mathbf{x}_{\mathrm{j}}\right)=0. Therefore, 2mIO=∑i,jmimj(xi−xj)2=2∑i<jmimjai22 m I_{\mathrm{O}}=\sum_{i, j} m_{\mathrm{i}} m_{\mathrm{j}}\left(\mathbf{x}_{\mathrm{i}}-\mathbf{x}_{\mathrm{j}}\right)^{2}=2 \sum_{i<j} m_{\mathrm{i}} m_{\mathrm{j}}{a_{\mathrm{i}}}^{2}.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.