Question

A ring has a greater moment of inertia than a circular disc of the same mass and radius, about an axis passing through its center of mass perpendicular to its plane, why?

Answer 1

Step 1: Moment of inertia

1. It is the measure of mass distribution around the axis of rotation.
2. the moment of inertia of a point mass about an axis in space is defined as the product of its mass m and the square of its perpendicular distance x from the axis, i.e, Moment of inertia, $I=m{x}^2$.

Step 2: Moment of inertia of a circular disc about an axis passing through its center of mass perpendicular to its plane.

Let the disc be divided into concentric circular rings (fig). We choose the ring between $r$ and $r+dr$ as the mass element $dm$.

So,

$$\begin{gathered} dm=areaofthering\times massperunitarea \\ ordm=\left(2\mathrm{\pi rdr}\right)\times \frac{m}{{\mathrm{\pi R}}^2} \end{gathered}$$

The moment of inertia of the ring under the reference about the axis is $dI=r^{2}dm$. So the moment of inertia of the disc is

$$\begin{gathered} I_d=\int dI=\int r^{2}dm=\frac{2m}{R^2}{\int }_0^R{r}^{3}dr=\frac{2m}{R^2}.\frac{R^4}{4}=\frac{1}{2}m{R}^2 \\ Or{I}_d=\frac{1}{2}m{R}^2.................\left(1\right) \end{gathered}$$

Step 3: Diagram

Step 4: The moment of inertia of a circular ring about an axis passing through its center of mass perpendicular to its plane is

We consider a thin uniform ring of radius a. Let us consider an element of the length $dx$ of the ring and let $\rho$ be the mass per unit length of the ring. The moment of inertia of the element $dx$ about the given axis is $dI=\rho dx{R}^2$.

The moment of inertia of the ring about an axis under consideration is given by

$$\begin{gathered} I_r=\underset{}{\overset{}{\sum \rho dx{R}^2}}=\rho R^2\times 2\mathrm{\pi a}=M{R}^2 \\ \mathrm{or}{\mathrm{I}}_{\mathrm{r}}=M{R}^2 \end{gathered}$$

Step 5: Diagram

Hence,

1. From the above two results, we can say that the moment of inertia of the circular ring is more than the circular disc of the same mass and radius.
2. This is because the mass of the circular ring is distributed at the maximum distance on the rim of the ring but in the case of a circular disc, mass is distributed all along the surface from $\mathrm{r}=0$ to $\mathrm{r}=\mathrm{R}$.