7.3 Circular ring

2.1 Pure translational motion
2.2 Pure rotational motion
2.3 Plane motion

3.1 Angular velocity $\omega $
3.2 Angular acceleration $\left( \alpha \right)$
3.3 Kinematics equation for rotational motion
3.4 Analogy between translational motion & rotational motion

6.1 Parallel-axis theorem
6.2 Perpendicular axis theorem

7.1 Thin rod
7.2 Rectangular lamina
7.3 Circular ring
7.4 Circular disc
7.5 Solid cylinder
7.6 Cylindrical shell
7.7 Solid sphere
7.8 Hollow sphere
7.9 Spherical shell
7.10 Solid cone
7.11 Hollow cone
7.12 Hollow hemisphere
7.13 Parallelopiped
7.14 List of moment of inertia $(I)$ and radius of gyration $(K)$ of different bodies

Let us consider a circular ring of mass $M$ and radius $R$. Assume an axis passing through its centre of mass $(COM)$ as shown in the figure.

Mass per unit length $\left( {{\lambda _L}} \right) = \frac{M}{{2\pi R}}$

Consider an infinitesimally small section $Rd\theta $ of mass $dm$ at an angle $\theta $ with the $x$-axis as shown in the figure.

$$\begin{equation} \begin{aligned} dm = {\lambda _L}(Rd\theta ) = \frac{M}{{2\pi R}}Rd\theta \\ dm = \frac{M}{{2\pi }}d\theta \quad ...(i) \\\end{aligned} \end{equation} $$

As we know that the moment of inertia $(I)$ of a continious rigid body is given by, $$I = \int_m {{R^2}dm\quad ...(ii)} $$

where,

$\int_m : $ Integrating over the entire mass

$R$: Perpendicular distance from an axis of rotation (Radius of the ring)

$dm$: mass of infinitesimally small section

From equation $(i)$ & $(ii)$, we get, $$I = \int_m {{R^2}\left( {\frac{M}{{2\pi }}d\theta } \right)} $$

Integrating the above equation for the whole ring is given by, $$\begin{equation} \begin{aligned} I = \frac{{M{R^2}}}{{2\pi }}\int\limits_0^{2\pi } {d\theta } \\ I = \frac{{M{R^2}}}{{2\pi }}\left[ \theta \right]_0^{2\pi } \\ I = \frac{{M{R^2}}}{{2\pi }}(2\pi - 0) \\ I = M{R^2}\quad ...(iii) \\\end{aligned} \end{equation} $$

Radius of gyration $(K)$ is given by, $$\begin{equation} \begin{aligned} I = M{K^2} \\ M{R^2} = M{K^2} \\ K = R \\\end{aligned} \end{equation} $$

Moment of inertia about an axis along the diameter of the ring.

From the perpendicular axis theorem, $${I_z} = {I_x} + {I_y}\quad ...(iv)$$

As circular ring is symmetric in all directions in $x-y$ plane.

So, $${I_x} = {I_y}\quad ...(v)$$

From equation $(iii),(iv)$ & $(v)$ we get, $$\begin{equation} \begin{aligned} M{R^2} = 2{I_x} \\ {I_x} = \frac{{M{R^2}}}{2} \\\end{aligned} \end{equation} $$ Similarly, $${I_y} = \frac{{M{R^2}}}{2}$$

Radius of gyration $(K)$ is given as, $$\begin{equation} \begin{aligned} I = M{K^2} \\ \frac{{M{R^2}}}{2} = M{K^2} \\ K = \frac{R}{{\sqrt 2 }} \\\end{aligned} \end{equation} $$

From parallel axis theorem we get, $$\begin{equation} \begin{aligned} {I_1} = {I_z} + M{R^2} \\ {I_1} = M{R^2} + M{R^2}\quad (As,\;{I_z} = M{R^2}) \\ {I_1} = M{R^2} \\\end{aligned} \end{equation} $$

Radius of gyration $(K)$ is given as, $$\begin{equation} \begin{aligned} I = M{K^2} \\ 2M{R^2} = M{K^2} \\ K = \sqrt 2 R \\\end{aligned} \end{equation} $$

From parallel axis theorem we get,

$$\begin{equation} \begin{aligned} {I_2} = {I_x} + M{R^2} \\ {I_2} = \frac{{M{R^2}}}{2} + M{R^2} \\ {I_2} = \frac{{3M{R^2}}}{2} \\\end{aligned} \end{equation} $$

Radius of gyration $(K)$ is given as, $$\begin{equation} \begin{aligned} I = M{K^2} \\ \frac{{3M{R^2}}}{2} = M{K^2} \\ K = \sqrt {\frac{3}{2}} R \\\end{aligned} \end{equation} $$