6 cosmetic tidying
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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

  1. Diagonalize the inertia tensor, to give its three principal eigenvalues $I_1$, $I_2$, $I_3$.
  2. Pick the smallest of these.
  3. 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

  1. Diagonalize the inertia tensor, to give its three principal eigenvalues $I_1$, $I_2$, $I_3$.
  2. Pick the smallest of these.
  3. 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

  1. Diagonalize the inertia tensor, to give its three principal eigenvalues $I_1$, $I_2$, $I_3$.
  2. Pick the smallest of these.
  3. 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
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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

  1. Diagonalize the inertia tensor, to give its three principal eigenvalues $I_1$, $I_2$, $I_3$.
  2. Pick the smallest of these.
  3. 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

  1. Diagonalize the inertia tensor, to give its three principal eigenvalues $I_1$, $I_2$, $I_3$.
  2. Pick the smallest of these.
  3. 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

  1. Diagonalize the inertia tensor, to give its three principal eigenvalues $I_1$, $I_2$, $I_3$.
  2. Pick the smallest of these.
  3. 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
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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

  1. Diagonalize the inertia tensor, to give its three principal eigenvalues $I_1$, $I_2$, $I_3$.
  2. Pick the smallest of these.
  3. 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

  1. Diagonalize the inertia tensor, to give its three principal eigenvalues $I_1$, $I_2$, $I_3$.
  2. Pick the smallest of these.
  3. 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

  1. Diagonalize the inertia tensor, to give its three principal eigenvalues $I_1$, $I_2$, $I_3$.
  2. Pick the smallest of these.
  3. 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
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2 cosmetic tidying
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1
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