Derive formula for mass moment of inertia

Mass moment of inertia is derived from the angular momentum of a system of particles that are stuck together rotating. Each particle contributes a small part of angular momentum, and when summed up the rotational motion can be factored out of the expression leaving the mass moment of inertia in between.

Consider a planar case with particle $m_i$ rotating about the center of mass, and hence having speed $v_i = r_i omega$ where $r_i = sqrt{x_i^2+y_i^2}$ is the radial distance to the center.

The total angular momentum of the system of particles is derived from the momentum $p_i = m_i v_i$ and the moment arm $r_i$:

$$ L = sum_i r_i (m_i v_i) = underbrace{ sum_i m_i r_i^2 }_{rm mmoi}; omega = I ,omega$$

Full development in 3D of this idea is given in this answer. Also read this similar answer here.

The proof can found in any elementary-level textbook, here I'm giving a quick idea. Consider the following fig

Consider a body rotating around the z-axis so that $v_j=dot{r}_j=omega rho_j.$ The angular momentum of the jth particle, $mathbf{L}_j$, is
$$mathbf{L}_j=mathbf{r}_jtimes m_j mathbf{v}_j$$ Here we are concerned only with $L_z$, the component of angular momentum along the axis of rotation. Since $mathbf{v}_j$ lies in the xy-plane, $$L_{j,z}=m_jv_jrho_j$$ $$L_{j,z}=m_jrho_j^2omega$$ The z component of the total angular momentum of the body $L_z$ is the sum of the individual $z$ components $$L_z=sum_j L_{j,z}=sum_jm_jrho_j^2omega$$ which can be written as $$L_z=Iomega$$ where $$I equiv sum_jm_jrho_j^2omega$$

Note that here we are talking about angular momentum about axis , in general when we take it about a point, momentum inertia turn out to be tensor of rank 3. More on this here.