# Derive formula for mass moment of inertia

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?

• Have you considered looking up a textbook where rotational dynamics is discussed? Oct 27, 2020 at 12:19
• See bullet #5 of this answer Oct 27, 2020 at 12:19
• What do you mean by a killer explanation? that sounds quite terrifying honestly Oct 27, 2020 at 12:42

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.

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.

• Wow thanks, this actually made good sense in least physical terms than other complex answers . Great 👌 proof. Oct 27, 2020 at 14:27

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.

• Did you refer kleppner and kolenkow for writing this? :-) Oct 27, 2020 at 12:43
• that's my course standard text book. so yes Oct 27, 2020 at 12:44
• I see but the tensor concepts came out a bit abruptly, is that discussed in the book? From a quick glance, I thought it was the book due to the notation Oct 27, 2020 at 12:45
• No tensor thing I added for my self as it's necessary for OP to know. Oct 27, 2020 at 13:00