6 cosmetic tidying edited Sep 2 '18 at 14:51 user197851 This is actually a nice example of tensors and minimization using Lagrange multipliers. TheFor rotation about the COM, the inertia tensor $$\mathbf{I}$$ is defined as a symmetric $$3\times3$$ matrix with elements such as $$I_{xx} = \sum_k m_k (y_k^2+z_k^2), \quad I_{xy} = I_{yx} = -\sum_k m_k x_k y_k, \quad \ldots$$$$I_{xx} = \sum_k m_k (y_k^2+z_k^2), \quad I_{xy} = I_{yx} = -\sum_k m_k x_k y_k, \quad I_{xz} = I_{zx} = -\sum_k m_k x_k z_k, \quad \ldots$$ where the position vectors $$(x_k,y_k,z_k)$$ are relative to the COM. Even a 2D arrangement of particles will, in general, have a $$3\times3$$ inertia tensor: you can rotate them about any axis in 3D space. Because it is a tensor, the moment of inertia associated with rotation about any axis through the COM, represented by a unit vector $$\mathbf{n}$$, will have a value $$\mathbf{n}\cdot\mathbf{I}\cdot\mathbf{n}$$ So we can seek the valuevector $$\mathbf{n}$$ that minimizes this quadratic form. However, we must remember the constraint that $$\mathbf{n}$$ is a unit vector, i.e. satisfies $$\mathbf{n}\cdot\mathbf{n}=1$$. So we can apply the method of Lagrange undetermined multipliers, and minimize without constraints the function $$\Phi(\mathbf{n}) = \mathbf{n}\cdot\mathbf{I}\cdot\mathbf{n} - \lambda \mathbf{n}\cdot\mathbf{n}$$ This minimum (or maximum) occurs when the gradient of the function with respect to $$\mathbf{n}$$ vanishes, and this will happen when $$\mathbf{I}\cdot\mathbf{n} = \lambda \mathbf{n}$$ This is an eigenvalue problem. So the answer to your question is Diagonalize the inertia tensor, to give its three principal eigenvalues $$I_1$$, $$I_2$$, $$I_3$$. Pick the smallest of these. The corresponding eigenvector is the axis you want. As mentioned above, provided you calculate the inertia tensor as a $$3\times3$$ matrix, it makes no difference whether the arrangement of masses is in 2D or 3D. If the particles are all in the $$xy$$ plane, though, it is easy to show that the $$z$$ axis is an eigenvector of the inertia tensor, and also (because of the perpendicular axis theorem) that the moment of inertia about the $$z$$ axis is larger than about any of the axes that lie in the $$xy$$ plane. Essentially, the problem becomes a $$2\times2$$ matrix eigenvalue problem. This is actually a nice example of tensors and minimization using Lagrange multipliers. The inertia tensor $$\mathbf{I}$$ is defined as a symmetric $$3\times3$$ matrix with elements such as $$I_{xx} = \sum_k m_k (y_k^2+z_k^2), \quad I_{xy} = I_{yx} = -\sum_k m_k x_k y_k, \quad \ldots$$ Even a 2D arrangement of particles will, in general, have a $$3\times3$$ inertia tensor: you can rotate them about any axis in 3D space. Because it is a tensor, the moment of inertia associated with rotation about any axis, represented by a unit vector $$\mathbf{n}$$, will have a value $$\mathbf{n}\cdot\mathbf{I}\cdot\mathbf{n}$$ So we can seek the value that minimizes this quadratic form. However, we must remember the constraint that $$\mathbf{n}$$ is a unit vector, i.e. satisfies $$\mathbf{n}\cdot\mathbf{n}=1$$. So we can apply the method of Lagrange undetermined multipliers, and minimize without constraints the function $$\Phi(\mathbf{n}) = \mathbf{n}\cdot\mathbf{I}\cdot\mathbf{n} - \lambda \mathbf{n}\cdot\mathbf{n}$$ This minimum (or maximum) occurs when the gradient of the function with respect to $$\mathbf{n}$$ vanishes, and this will happen when $$\mathbf{I}\cdot\mathbf{n} = \lambda \mathbf{n}$$ This is an eigenvalue problem. So the answer to your question is Diagonalize the inertia tensor, to give its three principal eigenvalues $$I_1$$, $$I_2$$, $$I_3$$. Pick the smallest of these. The corresponding eigenvector is the axis you want. As mentioned above, provided you calculate the inertia tensor as a $$3\times3$$ matrix, it makes no difference whether the arrangement of masses is in 2D or 3D. If the particles are all in the $$xy$$ plane, though, it is easy to show that the $$z$$ axis is an eigenvector of the inertia tensor, and also (because of the perpendicular axis theorem) that the moment of inertia about the $$z$$ axis is larger than about any of the axes that lie in the $$xy$$ plane. Essentially, the problem becomes a $$2\times2$$ matrix eigenvalue problem. This is actually a nice example of tensors and minimization using Lagrange multipliers. For rotation about the COM, the inertia tensor $$\mathbf{I}$$ is defined as a symmetric $$3\times3$$ matrix with elements such as $$I_{xx} = \sum_k m_k (y_k^2+z_k^2), \quad I_{xy} = I_{yx} = -\sum_k m_k x_k y_k, \quad I_{xz} = I_{zx} = -\sum_k m_k x_k z_k, \quad \ldots$$ where the position vectors $$(x_k,y_k,z_k)$$ are relative to the COM. Even a 2D arrangement of particles will, in general, have a $$3\times3$$ inertia tensor: you can rotate them about any axis in 3D space. Because it is a tensor, the moment of inertia associated with rotation about any axis through the COM, represented by a unit vector $$\mathbf{n}$$, will have a value $$\mathbf{n}\cdot\mathbf{I}\cdot\mathbf{n}$$ So we can seek the vector $$\mathbf{n}$$ that minimizes this quadratic form. However, we must remember the constraint that $$\mathbf{n}$$ is a unit vector, i.e. satisfies $$\mathbf{n}\cdot\mathbf{n}=1$$. So we can apply the method of Lagrange undetermined multipliers, and minimize without constraints the function $$\Phi(\mathbf{n}) = \mathbf{n}\cdot\mathbf{I}\cdot\mathbf{n} - \lambda \mathbf{n}\cdot\mathbf{n}$$ This minimum (or maximum) occurs when the gradient of the function with respect to $$\mathbf{n}$$ vanishes, and this will happen when $$\mathbf{I}\cdot\mathbf{n} = \lambda \mathbf{n}$$ This is an eigenvalue problem. So the answer to your question is Diagonalize the inertia tensor, to give its three principal eigenvalues $$I_1$$, $$I_2$$, $$I_3$$. Pick the smallest of these. The corresponding eigenvector is the axis you want. As mentioned above, provided you calculate the inertia tensor as a $$3\times3$$ matrix, it makes no difference whether the arrangement of masses is in 2D or 3D. If the particles are all in the $$xy$$ plane, though, it is easy to show that the $$z$$ axis is an eigenvector of the inertia tensor, and also (because of the perpendicular axis theorem) that the moment of inertia about the $$z$$ axis is larger than about any of the axes that lie in the $$xy$$ plane. Essentially, the problem becomes a $$2\times2$$ matrix eigenvalue problem. 5 cosmetic tidying edited Sep 2 '18 at 14:46 user197851 This is actually a nice example of tensors and minimization using Lagrange multipliers. The inertia tensor $$\mathbf{I}$$ is defined as a symmetric $$3\times3$$ matrix with elements such as $$I_{xx} = \sum_k m_k (y_k^2+z_k^2), \quad I_{xy} = I_{yx} = -\sum_k m_k x_k y_k, \quad \ldots$$ Even a 2D arrangement of particles will, in general, have a $$3\times3$$ inertia tensor: you can rotate them about any axis in 3D space. Because it is a tensor, the moment of inertia associated with rotation about any axis, represented by a unit vector $$\mathbf{n}$$, will have a value $$\mathbf{n}\cdot\mathbf{I}\cdot\mathbf{n}$$ So we can seek the value that minimizes this quadratic form. However, we must remember the constraint that $$\mathbf{n}$$ is a unit vector, i.e. satisfies $$\mathbf{n}\cdot\mathbf{n}=1$$. So we can apply the method of Lagrange undetermined multipliers, and minimize without constraints the function $$\Phi(\mathbf{n}) = \mathbf{n}\cdot\mathbf{I}\cdot\mathbf{n} - \lambda \mathbf{n}\cdot\mathbf{n}$$ This minimum (or maximum) occurs when the gradient of the function with respect to $$\mathbf{n}$$ vanishes, and this will happen when $$\mathbf{I}\cdot\mathbf{n} = \lambda \mathbf{n}$$ This is an eigenvalue problem. So the answer to your question is Diagonalize the inertia tensor, to give its three principal eigenvalues $$I_1$$, $$I_2$$, $$I_3$$. Pick the smallest of these. The corresponding eigenvector is the axis you want. As mentioned above, provided you calculate the inertia tensor as a $$3\times3$$ matrix, it makes no difference whether the arrangement of masses is in 2D or 3D. If the particles are all in the $$xy$$ plane, though, it is easy to show that the $$z$$ axis is an eigenvector of the inertia tensor, and also (because of the perpendicular axis theorem) that the moment of inertia about the $$z$$ axis is larger than about any of the axes that lie in the $$xy$$ plane. Essentially, the problem becomes a $$2\times2$$ matrix eigenvalue problem. This is actually a nice example of tensors and minimization using Lagrange multipliers. The inertia tensor $$\mathbf{I}$$ is defined as a symmetric $$3\times3$$ matrix with elements such as $$I_{xx} = \sum_k m_k (y_k^2+z_k^2), \quad I_{xy} = I_{yx} = -\sum_k m_k x_k y_k, \quad \ldots$$ Even a 2D arrangement of particles will, in general, have a $$3\times3$$ inertia tensor: you can rotate them about any axis in 3D space. Because it is a tensor, the moment of inertia associated with rotation about any axis, represented by a unit vector $$\mathbf{n}$$, will have a value $$\mathbf{n}\cdot\mathbf{I}\cdot\mathbf{n}$$ So we can seek the value that minimizes this quadratic form. However, we must remember the constraint that $$\mathbf{n}$$ is a unit vector, i.e. satisfies $$\mathbf{n}\cdot\mathbf{n}=1$$. So we can apply the method of Lagrange undetermined multipliers, and minimize without constraints the function $$\Phi(\mathbf{n}) = \mathbf{n}\cdot\mathbf{I}\cdot\mathbf{n} - \lambda \mathbf{n}\cdot\mathbf{n}$$ This minimum (or maximum) occurs when the gradient of the function with respect to $$\mathbf{n}$$ vanishes, and this will happen when $$\mathbf{I}\cdot\mathbf{n} = \lambda \mathbf{n}$$ This is an eigenvalue problem. So the answer to your question is Diagonalize the inertia tensor, to give its three principal eigenvalues $$I_1$$, $$I_2$$, $$I_3$$. Pick the smallest of these. The corresponding eigenvector is the axis you want. As mentioned above, provided you calculate the inertia tensor as $$3\times3$$ matrix, it makes no difference whether the arrangement of masses is in 2D or 3D. If the particles are all in the $$xy$$ plane, though, it is easy to show that the $$z$$ axis is an eigenvector of the inertia tensor, and also (because of the perpendicular axis theorem) that the moment of inertia about the $$z$$ axis is larger than about any of the axes that lie in the $$xy$$ plane. Essentially, the problem becomes a $$2\times2$$ matrix eigenvalue problem. This is actually a nice example of tensors and minimization using Lagrange multipliers. The inertia tensor $$\mathbf{I}$$ is defined as a symmetric $$3\times3$$ matrix with elements such as $$I_{xx} = \sum_k m_k (y_k^2+z_k^2), \quad I_{xy} = I_{yx} = -\sum_k m_k x_k y_k, \quad \ldots$$ Even a 2D arrangement of particles will, in general, have a $$3\times3$$ inertia tensor: you can rotate them about any axis in 3D space. Because it is a tensor, the moment of inertia associated with rotation about any axis, represented by a unit vector $$\mathbf{n}$$, will have a value $$\mathbf{n}\cdot\mathbf{I}\cdot\mathbf{n}$$ So we can seek the value that minimizes this quadratic form. However, we must remember the constraint that $$\mathbf{n}$$ is a unit vector, i.e. satisfies $$\mathbf{n}\cdot\mathbf{n}=1$$. So we can apply the method of Lagrange undetermined multipliers, and minimize without constraints the function $$\Phi(\mathbf{n}) = \mathbf{n}\cdot\mathbf{I}\cdot\mathbf{n} - \lambda \mathbf{n}\cdot\mathbf{n}$$ This minimum (or maximum) occurs when the gradient of the function with respect to $$\mathbf{n}$$ vanishes, and this will happen when $$\mathbf{I}\cdot\mathbf{n} = \lambda \mathbf{n}$$ This is an eigenvalue problem. So the answer to your question is Diagonalize the inertia tensor, to give its three principal eigenvalues $$I_1$$, $$I_2$$, $$I_3$$. Pick the smallest of these. The corresponding eigenvector is the axis you want. As mentioned above, provided you calculate the inertia tensor as a $$3\times3$$ matrix, it makes no difference whether the arrangement of masses is in 2D or 3D. If the particles are all in the $$xy$$ plane, though, it is easy to show that the $$z$$ axis is an eigenvector of the inertia tensor, and also (because of the perpendicular axis theorem) that the moment of inertia about the $$z$$ axis is larger than about any of the axes that lie in the $$xy$$ plane. Essentially, the problem becomes a $$2\times2$$ matrix eigenvalue problem. 4 minor clarification edited Sep 2 '18 at 14:37 user197851 This is actually a nice example of tensors and minimization using Lagrange multipliers. The inertia tensor $$\mathbf{I}$$ is defined as a symmetric $$3\times3$$ matrix with elements such as $$I_{xx} = \sum_k m_k (y_k^2+z_k^2), \quad I_{xy} = I_{yx} = -\sum_k m_k x_k y_k, \quad \ldots$$ Even a 2D arrangement of particles will, in general, have a $$3\times3$$ inertia tensor: you can rotate them about any axis in 3D space. Because it is a tensor, the moment of inertia associated with rotation about any axis, represented by a unit vector $$\mathbf{n}$$, will have a value $$\mathbf{n}\cdot\mathbf{I}\cdot\mathbf{n}$$ So we can seek the value that minimizes this quadratic form. However, we must remember the constraint that $$\mathbf{n}$$ is a unit vector, i.e. satisfies $$\mathbf{n}\cdot\mathbf{n}=1$$. So we can apply the method of Lagrange undetermined multipliers, and minimize without constraints the function $$\Phi(\mathbf{n}) = \mathbf{n}\cdot\mathbf{I}\cdot\mathbf{n} - \lambda \mathbf{n}\cdot\mathbf{n}$$ This minimum (or maximum) occurs when the gradient of the function with respect to $$\mathbf{n}$$ vanishes, and this will happen when $$\mathbf{I}\cdot\mathbf{n} = \lambda \mathbf{n}$$ This is an eigenvalue problem. So the answer to your question is Diagonalize the inertia tensor, to give its three principal eigenvalues $$I_1$$, $$I_2$$, $$I_3$$. Pick the smallest of these. The corresponding eigenvector is the axis you want. As mentioned above, provided you calculate the inertia tensor as $$3\times3$$ matrix, it makes no difference whether the arrangement of masses is in 2D or 3D. If the particles are all in the $$xy$$ plane, though, it is easy to show that the $$z$$ axis is an eigenvector of the inertia tensor, and also (because of the perpendicular axis theorem) that the moment of inertia about the $$z$$ axis is larger than about any of the axes that lie in the $$xy$$ plane. Essentially, the problem becomes a $$2\times2$$ matrix eigenvalue problem. This is actually a nice example of tensors and minimization using Lagrange multipliers. The inertia tensor $$\mathbf{I}$$ is defined as a symmetric $$3\times3$$ matrix with elements such as $$I_{xx} = \sum_k m_k (y_k^2+z_k^2), \quad I_{xy} = I_{yx} = -\sum_k m_k x_k y_k, \quad \ldots$$ Even a 2D arrangement of particles will, in general, have a $$3\times3$$ inertia tensor: you can rotate them about any axis in 3D space. Because it is a tensor, the moment of inertia associated with rotation about any axis, represented by a unit vector $$\mathbf{n}$$, will have a value $$\mathbf{n}\cdot\mathbf{I}\cdot\mathbf{n}$$ So we can seek the value that minimizes this quadratic form. However, we must remember the constraint that $$\mathbf{n}$$ is a unit vector, i.e. satisfies $$\mathbf{n}\cdot\mathbf{n}=1$$. So we can apply the method of Lagrange undetermined multipliers, and minimize without constraints the function $$\Phi(\mathbf{n}) = \mathbf{n}\cdot\mathbf{I}\cdot\mathbf{n} - \lambda \mathbf{n}\cdot\mathbf{n}$$ This minimum (or maximum) occurs when the gradient of the function with respect to $$\mathbf{n}$$ vanishes, and this will happen when $$\mathbf{I}\cdot\mathbf{n} = \lambda \mathbf{n}$$ This is an eigenvalue problem. So the answer to your question is Diagonalize the inertia tensor, to give its three principal eigenvalues $$I_1$$, $$I_2$$, $$I_3$$. Pick the smallest of these. The corresponding eigenvector is the axis you want. As mentioned above, provided you calculate the inertia tensor as $$3\times3$$ matrix, it makes no difference whether the arrangement of masses is in 2D or 3D. This is actually a nice example of tensors and minimization using Lagrange multipliers. The inertia tensor $$\mathbf{I}$$ is defined as a symmetric $$3\times3$$ matrix with elements such as $$I_{xx} = \sum_k m_k (y_k^2+z_k^2), \quad I_{xy} = I_{yx} = -\sum_k m_k x_k y_k, \quad \ldots$$ Even a 2D arrangement of particles will, in general, have a $$3\times3$$ inertia tensor: you can rotate them about any axis in 3D space. Because it is a tensor, the moment of inertia associated with rotation about any axis, represented by a unit vector $$\mathbf{n}$$, will have a value $$\mathbf{n}\cdot\mathbf{I}\cdot\mathbf{n}$$ So we can seek the value that minimizes this quadratic form. However, we must remember the constraint that $$\mathbf{n}$$ is a unit vector, i.e. satisfies $$\mathbf{n}\cdot\mathbf{n}=1$$. So we can apply the method of Lagrange undetermined multipliers, and minimize without constraints the function $$\Phi(\mathbf{n}) = \mathbf{n}\cdot\mathbf{I}\cdot\mathbf{n} - \lambda \mathbf{n}\cdot\mathbf{n}$$ This minimum (or maximum) occurs when the gradient of the function with respect to $$\mathbf{n}$$ vanishes, and this will happen when $$\mathbf{I}\cdot\mathbf{n} = \lambda \mathbf{n}$$ This is an eigenvalue problem. So the answer to your question is Diagonalize the inertia tensor, to give its three principal eigenvalues $$I_1$$, $$I_2$$, $$I_3$$. Pick the smallest of these. The corresponding eigenvector is the axis you want. As mentioned above, provided you calculate the inertia tensor as $$3\times3$$ matrix, it makes no difference whether the arrangement of masses is in 2D or 3D. If the particles are all in the $$xy$$ plane, though, it is easy to show that the $$z$$ axis is an eigenvector of the inertia tensor, and also (because of the perpendicular axis theorem) that the moment of inertia about the $$z$$ axis is larger than about any of the axes that lie in the $$xy$$ plane. Essentially, the problem becomes a $$2\times2$$ matrix eigenvalue problem. 3 removed mention of linear special case, as irrelevant to the question edited Sep 2 '18 at 14:18 user197851 2 cosmetic tidying edited Sep 2 '18 at 14:11 user197851 1 answered Sep 2 '18 at 14:06 user197851