Derive formula for mass moment of inertia

Question

I always wonder how the formula for moment of inertia is actually derived. Some say that moment of inertia is simply equals to $MR^2$ but some derive it saying moment of inertia is directly proportional to mass and distance squared, but I want a killer explanation why is it proportional to distance squared from the axis of rotation?

Answer 1

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.

Answer 2

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
$$\mathbf{L}_j=\mathbf{r}_j\times m_j \mathbf{v}_j$$ 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, $$L_{j,z}=m_jv_j\rho_j$$ $$L_{j,z}=m_j\rho_j^2\omega$$ The z component of the total angular momentum of the body $L_z$ is the sum of the individual $z$ components $$L_z=\sum_j L_{j,z}=\sum_jm_j\rho_j^2\omega$$ which can be written as $$L_z=I\omega$$ where $$I \equiv \sum_jm_j\rho_j^2\omega$$

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.

Answer 3

Consider a small mass, m, attached to the end of a thin mass-less rod of length, r. The other end of the rod is attached to a friction-less perpendicular axle. Apply a force, F, perpendicular to the rod and axle, at a distance, R, from the axle. In a short time, the force causes a rotation and does work: W = F(Rθ). This work is transmitted by the rod to the mass, doing work W = (ma)(rθ). Equating and dividing by θ yields: FR = (mr)(rα) = (m$r^2$)α. In other words: τ = Iα where I = m$r^2$. This analysis can be extended to any distribution of masses or forces in a plane (or any other plane) which is perpendicular to an axle.