7.1 Thin rod

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 rod of length $L$ and mass $M$. Assume an axis passing though its centre of mass $(COM)$ as shown in the figure.

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

Consider an infinitesimally small section $dx$ of mass $dm$ at a distance $x$ from an axis of rotation (AOR) as shown in the figure. As we know that the moment of inertia $(I)$ of a continious rigid body is given by, $$I = \int_m {{x^2}dm\quad ...(ii)} $$

where,

$\int_m : $ Integrating over the entire mass

$x$: perpendicular distance from an axis of rotation

$dm$: mass of infinitesimally small section

From equation $(i)$ & $(ii)$, we get, $$I = \int_m {{x^2}\left( {\frac{M}{L}dx} \right)} $$

Integrating the above equation for the whole rod is given by, $$\begin{equation} \begin{aligned} I = \frac{M}{L}\int\limits_{ - \frac{L}{2}}^{\frac{L}{2}} {{x^2}dx} \\ I = \frac{M}{L}\left[ {\frac{{{x^3}}}{3}} \right]_{ - \frac{L}{2}}^{\frac{L}{2}} \\ I = \frac{M}{L}\left\{ {\frac{{{L^3}}}{{24}} - \left( { - \frac{{{L^3}}}{{24}}} \right)} \right\} \\ I = \frac{{M{L^2}}}{{12}} \\\end{aligned} \end{equation} $$

So, the moment of inertia about an axis passing through centre of mass is, ${I_{COM}} = \frac{{M{L^2}}}{{12}}$

Radius of gyration $(K)$ is given as, $$\begin{equation} \begin{aligned} I = M{K^2} \\ \frac{{M{L^2}}}{{12}} = M{K^2} \\ K = \frac{L}{{\sqrt {12} }} \\\end{aligned} \end{equation} $$

So, the radius of gyration about an axis passing through centre of mass is, ${K_{COM}} = \frac{L}{{\sqrt {12} }}$

From the parallel axis theorem, $$I = {I_{COM}} + M{d^2}$$

where,

$I$: Moment of inertia about any given axis

$d = \frac{L}{2}$: Perpendicular distance between the two parallel axes $$\begin{equation} \begin{aligned} I = \frac{{M{L^2}}}{{12}} + M{\left( {\frac{L}{2}} \right)^2} \\ I = \frac{{M{L^2}}}{{12}} + \frac{{M{L^2}}}{4} \\ I = \frac{{M{L^2}}}{3} \\\end{aligned} \end{equation} $$

So, moment of inertia about an axis passing though an end of the rod is, $I = \frac{{M{L^2}}}{3}$

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

So, radius of gyration about an axis passing through end of the rod is, $K = \frac{L}{{\sqrt 3 }}$