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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.

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    $\begingroup$ You can find a general equation for the moment of inertia based on the angles the axis of rotation makes with the z-axis and the x-y plane. Then it becomes an optimization problem. Of course this is easier said than done. This would work in both 2D and 3D. Also, I think in 2D the axis needs to lie in the plane of the particles, so you would only have to optimize with respect to one angle rather than two. $\endgroup$ – Aaron Stevens Sep 2 '18 at 13:16
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    $\begingroup$ Related meta post: physics.meta.stackexchange.com/q/10793/2451 $\endgroup$ – Qmechanic Sep 2 '18 at 17:02
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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.

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    $\begingroup$ because the symmetry of the inertia tensor all eigenvalues are real! $\endgroup$ – Eli Sep 2 '18 at 21:19
  • $\begingroup$ I am still in school and haven't studied many of those topics yet (eigenvalue, tensors etc) so I don't fully understand your solution. Maybe I will understand more once I study those topics.Anyway thanks for your answer. $\endgroup$ – Nilaksh Singh Sep 4 '18 at 15:14
  • $\begingroup$ I understand; sorry about that. In the UK, these would typically be covered in the second year of a university physics course. The $\mathbf{n}\cdot\mathbf{I}\cdot\mathbf{n}$ expression is calculating the moment of inertia along any desired axis; then we want to minimize this quantity by finding where its derivative (with respect to all possible rotations of $\mathbf{n}$) is zero. The Lagrange multipliers help us do that. Finding the eigenvalues/eigenvectors is equivalent to finding the "special" orientations of the axes that make the inertia tensor diagonal. One of these is the one you seek. $\endgroup$ – LonelyProf Sep 4 '18 at 15:28
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    $\begingroup$ Anyway, each of these topics (inertia tensor, Lagrange multipliers, eigenvalues) can be studied separately, and hopefully the links given in the answer will be suitable jumping-off points. Good luck! $\endgroup$ – LonelyProf Sep 4 '18 at 15:30
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I am solving this for particles distributed in a 2D space. First we know that the moment of inertia of a particle about an axis is given by

$$I=Mr^2$$

And we know that the axes passing through COM have the minimum moment of inertia. Our job is to find the slope of the axes with min Inertia, then we can use the slope point form to find the equation of the required axis :

$$y - y_0 = m ( x - x_0 )$$

(Where m is the slope of the line and $(x_0, y_0)$ is the point through which it passes)

Now let us consider the distribution of masses, the coordinate of COM is given by: $$ x_{cm} = \frac{1}{M} \sum_i M_i x_i $$ $$ y_{cm} = \frac{1}{M} \sum_i M_i y_i $$

And now let us shift the origin to COM to make our calculations easy. If the original coordinate of mass $M_i$ was $(x_i, y_i)$ then the new shifted coordinates are: \begin{align} x_i' & = x_i - x_{cm} \\ y_i' & = y_i - y_{cm} \end{align} Hence the axis that we seek now passes to the new shifted origin $(x_0, y_0)$.

Now let the equation of the line be

$$y = mx$$ (as this line passes through origin, $c = 0$).

Then the distance $r_i$ of the $i$th particle from this line, is given by : \begin{align} r_i = \frac{|m x_i' -y_i'|}{\sqrt{1+m^2}} \end{align}

Hence the total moment of Inertia of the system is :

\begin{align} I = \sum M_i r_i^2 \end{align}

Now we can differentiate it with respect to $m$ (slope) and equate it to zero to find the minima: $$ \frac{dI}{dm} = \sum \frac{2M_i(mx_i'-y_i)(x_i'+my_i)}{(1+m^2)^2} $$

Equating it to zero gives :

$$ \sum M_i(mx_i'-y_i)(x_i'+my_i)=0 $$

Solving for m gives the following quadratic equation :

$$ \left(\sum M_i x_i'y_i'\right)m^2 + \left(\sum M_i (x_i'^2-y_i'^2)\right)m - \left(\sum M_i x_i'y_i'\right) = 0 $$

Note that the product of roots (that is slopes) is -1 i.e. this gives two two axes perpendicular to each other and one among them has minimum momentum and other has max about it (they can be found out by differentiating the above quadratic equation again)

Thus the the axis we seek, in the original coordinate system maybe be written as: $$ y-y_{cm} = m(x-x_{cm}). $$ where $m$ is the slope which corresponds to minimum inertia.

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