If a system of particles is given, in a 2D plane, with particles having masses $M_1$, $M_2$, $M_3, \ldots M_n$ and coordinates $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3), \ldots (x_n y_n)$,
then how can one find the axis about which the system has the minimum moment of inertia ?
I know that among parallel axes, the one that passes through the center of mass has minimum inertia, but which axis among those that pass through COM has least inertia about it ?
Bonus: if possible please explain how one can find this axis for a system of particles in 3D space.
Best Answer
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
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.