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?
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1$\begingroup$ Have you considered looking up a textbook where rotational dynamics is discussed? $\endgroup$– ZeroTheHeroOct 27, 2020 at 12:19
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$\begingroup$ See bullet #5 of this answer $\endgroup$– John AlexiouOct 27, 2020 at 12:19
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1$\begingroup$ What do you mean by a killer explanation? that sounds quite terrifying honestly $\endgroup$– tryst with freedomOct 27, 2020 at 12:42
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3 Answers
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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.
answered Oct 27, 2020 at 12:27
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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.
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$\begingroup$ Wow thanks, this actually made good sense in least physical terms than other complex answers . Great 👌 proof. $\endgroup$– AdityaOct 27, 2020 at 14:27
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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.
answered Oct 27, 2020 at 12:36
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$\begingroup$ Did you refer kleppner and kolenkow for writing this? :-) $\endgroup$ Oct 27, 2020 at 12:43
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$\begingroup$ that's my course standard text book. so yes $\endgroup$ Oct 27, 2020 at 12:44
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$\begingroup$ 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 $\endgroup$ Oct 27, 2020 at 12:45
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$\begingroup$ No tensor thing I added for my self as it's necessary for OP to know. $\endgroup$ Oct 27, 2020 at 13:00