What is the moment of inertia tensor? How is it derived?

Let’s talk about where the moment of inertia tensor came from. One of the confusing aspects, I think, is that it seems like a completely separate quantity. What really happens is that when you compute the kinetic energy of a rigid body rotating through space, you’ll find that a specific choice of coordinates allows you to split up the total energy into two components:translationaland rotational. Note, I’ve borrowed a lot of the mathematical rigor from Hand & Finch.

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Figure 1: A diagram of a rigid body. By definition, a rigid body is such that (r⃗ i−r⃗ j)2= constant  (the distance between any two points is a constant).

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The velocity of the ith  point in the rigid body as expressed in the space system is 
            v⃗ i|space=R⃗ ˙+ω⃗ ×(r⃗ i|body) 
Remember, the vector r⃗ i=0 when the point is located at the origin of the body system K. I'm going to drop the body subscript since I'll only be using the r⃗ i from the body frame K.

Total Kinetic Energy

To an outside observer (in the space frame K′), the total kinetic energy, by definition

Remember, the total mass M is equal to the sum of the point masses of the rigid body, so

                     M=∑imi              ∑imir⃗ i=Mr⃗ cm
The last expression should be incredibly familiar. It follows straight from how we compute the location of the center of mass (weighted average).

The interesting thing about our rigid body is that we’ve generalized it completely. We can make some simplifying assumptions about r⃗ cm. In particular, we can choose a symmetry such that the center of mass of the rigid body lies in the origin of K space, such that r⃗ cm=0. Then, the above expression for kinetic energy has two pieces.
            T=12MR⃗ ˙2translational+12∑imi(ω⃗ ×r⃗ i)2rotational
The translational kinetic energy is the same as if all the mass were located at the center of mass of the rigid body. Thus from now on, we assume the origin of K is located at the center of mass.

The Moment of Inertia Tensor

Tensors are just a fancy way of really saying "matrices and vectors" (at least for the level of this answer). When we talk about a tensor, they can be m by ndimensional. They contain properties and help us talk about maps between spaces. For now, when we talk about a moment of inertia tensor, just think about a matrix.

Imagine a pen thrown up in the air. Any of its points will follow a rather complicated motion, but if we talk about the motion in terms of the center of mass of the pen... we see that the center of mass follows a parabolic trajectory. It’s just like the motion of a point particle, this is boring. So let’s focus on the rotational part.
            (ω⃗ ×r⃗ i)⋅(ω⃗ ×r⃗ i)=ω⃗ ⋅(r⃗ i×(ω⃗ ×r⃗ i))
from the scalar triple product (it is invariant under a circular shift of its operands). Then
            ω⃗ ⋅(r⃗ i×(ω⃗ ×r⃗ i))=ω⃗ ⋅(r2iω⃗ −(ω⃗ ⋅r⃗ i)r⃗ i)
from the vector triple product. This is known as the triple product expansion. Now, let's expand out this product in terms of the components. Since we'll be talking about 3D space, it'll be useful to just write with indices α,β=1,2,3for x,y,z components respectively. That is
            x⃗ =(x1,x2,x3)=∑3α=1xαe⃗ α=xαe⃗  α
And likewise, the dot product between two vectors
            a⃗ ⋅b⃗ =∑αaαbα=aαbα
Simplify that expression we obtained earlier by rewriting it in a component form:
            ω⃗ ⋅(⋯)=∑αr2i,α∑βω2β−∑αri,αωα∑βri,βωβ
The first term is from the dot product between ω⃗ ⋅ω⃗  and the second term is from the dot product between ω⃗ ⋅r⃗ i. Use the Feynman method, think really hard and write down the answer:
    Trotation=12∑imi(ω⃗ ×r⃗ i)2
                = 12∑imi(∑αr2i,α∑βω2β−∑αri,αωα∑βri,βωβ)
                   =12∑α∑β∑imi(r2i,αω2β−ri,αri,βωαωβ)
                  =12∑α∑βωαωβIαβ
where we defined the moment of inertia tensor I with components Iαβ given by
             Iαβ=∑imi(r2iδαβ−ri,αri,β)
and
            δαβ={10if α=βotherwise

In matrix form

Notice some interesting features: Iαβ=Iβα which is a symmetric tensor (we often state this by saying I[αβ]=0. Because of this, we can write the total kinetic energy in a more compact form!

            T=12MR⃗ ˙2+12ωTIω