Moment of inertia doubt While calculating moment of inertia of a disc, we divide the disc into many rings of infinitesimal thickness and then calculate using integration. But in case of a thin rod, we approximate each point by a segment $dx$ and then calculate because moment of inertia is defined for point mass. But in case of a disc, we are taking rings. What difference does it make if we take big rings or small rings? Why does infinitesimal thickness need to be taken? I didn't understand this since moment of inertia is defined for point mass and here it is a ring.
 A: A ring also consists of many point masses. The fact is the distance from the axis to the each point mass is the same. Let's say that point masses are $m_1, m_2, m_3,......$ and the distance from the axis to each is $r$, then find the moment of inertia of each particle about the axis, $m_1r^2, m_2r^2, m_3r^2,......$.Sum of all of these is the moment of inertia of the ring. $$m_1r^2+m_2r^2+m_3r^2+......+m_nr^2$$ $$\Rightarrow (m_1+m_2+m_3+......+m_n)r^2$$ $$\Rightarrow Mr^2$$ Total mass of the ring $M$ is the sum of masses of all particles. Eventually, you can see moment of inertia is still defined for a point mass.
 You are probably being misled because $Mr^2$ gives the moi of a ring which is similar to the moi of a point mass. But I think now it is obvious that it is derived from the definition of moi, which you use directly in the calculations.
A: A ring represents a collection of point masses that share the same MMOI. The detailed procedure to calculate MMOI is a triple integral over a domain ${\rm d}V$ and it makes sense to take advantage of geometric symmetries to simplify the integral down to a single one.
Consider each particle in a domain ${\rm d}V$ with position $\vec{\rm pos} = \pmatrix{x \\ y \\ z}$. The process to derive the center of mass and mass moment of inertia is as follows

*

*Mass $$m = \int \rho\, {\rm d}V$$

*Center of mass $$\vec{\rm cm} = \frac{1}{m} \int  \pmatrix{x\\y\\z} \,\rho\, {\rm d}V$$

*Mass moment of inertia tensor $$\mathbf{I} = \int \begin{bmatrix}y^2+z^2 & -x y & -x z \\ -x y & x^2+z^2 & -y z \\ -x z & -y z & x^2+y^2\end{bmatrix}\,\rho\,{\rm d}V $$
Now in the case for a cylinder (thick disk), the position of each particle is parametrized with three variables, the radius from the center $r$, the angle $\varphi$ about the axis and the distance $z$ along the axis.
The volume element in cylindrical coordinates is $${\rm d}V = r\, {\rm d}r\, {\rm d}\varphi\, {\rm d}z$$
But the integral about the axis ${\rm d}\varphi$ can be pre-calculated to make
$${\rm d}V = (2 \pi r)\, {\rm d}r\, {\rm d}z$$
This means the MMOI integral is simplified to
$$\mathbf{I} = \int_{-h/2}^{h/2} \int_0^R 2\pi \begin{bmatrix}
 r^2/2 + z^2 & & \\ & r^2/2+z^2 & \\ & & r^2
\end{bmatrix}\,\rho\,{\rm d}r\, {\rm d}z = \begin{bmatrix}
 m/12 ( 2 R^2+h^2) & & \\ & m/12 ( 2 R^2+h^2) & \\ & & m R^2/3
\end{bmatrix} $$
The case of the thin disk arrives when the thickness is zero $h=0$, and the case for the long slender rod when the radius is zero $R=0$
$$\mathbf{I}_{\rm disk} = \begin{bmatrix}
 m R^2/6 & & \\ & m R^2/6 & \\ & & m R^2/3
\end{bmatrix}$$
$$\mathbf{I}_{\rm rod} = \begin{bmatrix}
 m h^2/12 & & \\ & m h^2/12 & \\ & & 0
\end{bmatrix}$$
A: The problem with taking a big ring ​(i.e. an annulus) is that the distance from a point on the big ring to the axis of rotation is not constant. Namely, if we take an annulus with inner radius $r_1$ and outer radius $r_2$, then the distances of points on the annulus to the axis range through $[r_1,r_2]$.
However, if we take an annulus of infinitesimal thickness with $r_1=r$ and $r_2 = r+dr$ then the distances of points on the annulus to the axis range through $[r,r+dr]$ which can be approximated with $r$. Which means that the annulus indeed becomes a ring of radius $r$.
You can arrive at the same conclusion by explicit manipulation of differentials. For the moment of inertia of a disk $R$ of uniform area density $\sigma$, we have
\begin{align}
\int_{\text{disk}} r^2\,dm &= \int_{r=0}^R \left(\int_{\text{annulus from $r$ to $r+dr$}} s^2\,dm\right) = \int_{r=0}^R \left(\int_{s=r}^{r+dr} s^2\sigma\cdot2\pi s\,ds\right)\\
&= 2\pi \sigma \int_{r=0}^R \left(\int_{s=r}^{r+dr} s^3\,ds\right) = 2\pi \sigma \int_{r=0}^R \frac14\left((r+dr)^4-r^4\right)\\
&= 2\pi \sigma \int_{r=0}^R \frac14\left(4r^3\,dr + 6r^2(dr)^2 + 4r\,(dr)^3+(dr)^4\right)\\
&= 2\pi \sigma \int_{r=0}^R r^3\,dr = \frac12R^4\pi\sigma = \frac12(R^2\pi\sigma)R^2\\
\end{align}
since higher powers of $dr$ are equal to $0$.
