Uniqueness of Mass Moment of Inertia tensor

Question

I was curious to know if there can exist two different objects (shape and/or mass distribution) that can have the same inertia tensor.

Answer 1

Yes, inertia tensor depends on the distribution of mass, and there are an infinite number of different distributions of mass (objects) which have the same inertia tensor. The reason is that inertia tensor (like mass, volume, momentum, kinetic energy, etc) is an extensive property of an object : it is the sum of the same property for individual parts of the object (point masses), and there are many different ways of generating the same sum.

For example :

object A consists of 2 particles with masses 9, 9 at positions x=-1, +1 along the x axis. Its mass is 18 units and its principal moments of inertia are $I_x=0, I_y=I_z=9*(-1)^2+9*(+1)^2=18$ units.

object B consists of 4 particles with masses 2, 1, 1, 2 at positions x=-2, -1, +1, +2 along the x axis. Its mass is 6 units and its principal moments of inertia are $I_x=0, I_y=I_z=2*(-2)^2+1*(-1)^2+1*(+1)^2+2*(+2)^2=18$ units.

object C consists of 3 particles with masses 1, 1, 1 at positions -3, 0, +3 along the x axis. Its mass is 3 units and its principal moments of inertia are $I_x=0, I_y=I_z=1*(-3)^2+1*(0)^2+1*(+3)^2=18$ units.

object D consists of 3 particles with masses 1, 1, 0.5 at positions x=-3, +1, +2 along the x axis. Its mass is 2.5 units and its principal moments of inertia are $I_x=0, I_y=I_z=1*(-4)^2+1*(+1)^2+0.5*(+2)^2=18$ units.

Note: it is not necessary for objects with the same inertia tensor to have the same mass, nor is it necessary for them to be symmetrical.

Answer 2

Yes indeed. By definition, the MMOI tensor reduces the infinite variations of shape into a single equivalent one.

In the simplest example a solid disk of radius $10\, \rm mm$ and mass $1\,\rm kg$ as the same mass moment of inertia as a thin ring of the same mass and radius of $5 \sqrt{2} \,\rm mm$.

• Solid Disk of radius $R=10\,\text{mm}$ and mass $m=1\,\text{kg}$ $$ I = \frac{m}{2} R^2 = \frac{1}{2} 10^2 = 50\,\text{kg mm}^2$$

• Thin Ring of radius $R=5 \sqrt{2}\,\text{mm}$ and mass $m=1\,\text{kg}$ $$ I = m R^2 = (5 \sqrt{2})^2 = 50\,\text{kg mm}^2$$