7.2 Rectangular lamina

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

Consider a rectangular lamina of length $a$, breadth $b$ and mass $m$.

Mss per unit area $$\left( \lambda \right) = \frac{M}{{ab}}$$

$I_z$ is the moment of inertia about an axis passing through its centre and perpendicular to its plane.

So, from perpendicular axis theorem, $${I_z} = {I_x} + {I_y}\quad ...(i)$$

So, for finding $I_z$, we have to find $I_x$ & $I_y$

Consider an infinitesimal small section $dy$ and small area $dA$ of mass $dm$ at a distance $y$ from an axis of rotation. $$dA = ady$$

$$\begin{equation} \begin{aligned} dm = {\lambda _A}dA \\ dm = \left( {\frac{M}{{ab}}} \right)(ady) \\ dm = \frac{{Mdy}}{b}\quad ...(ii) \\\end{aligned} \end{equation} $$

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

where,

$\int_m : $ Integrating over the entire mass

$y$: perpendicular distance from an axis of rotation

$dm$: mass of infinitesimally small section

From equation $(ii)$ & $(iii)$ we get, $$\begin{equation} \begin{aligned} {I_x} = \int_m {{y^2}\left( {\frac{{Mdy}}{b}} \right)} \\ {I_x} = \frac{M}{b}\int_m {{y^2}dy} \\\end{aligned} \end{equation} $$

Integrating the above equation for the whole rectangular lamina is given by, $$\begin{equation} \begin{aligned} {I_x} = \frac{M}{b}\int\limits_{ - \frac{b}{2}}^{\frac{b}{2}} {{y^2}dy} \\ {I_x} = \frac{M}{b}\left[ {\frac{{{y^3}}}{3}} \right]_{ - \frac{b}{2}}^{\frac{b}{2}} \\ {I_x} = \frac{M}{b}\left\{ {\frac{{{b^3}}}{{24}} - \left( { - \frac{{{b^3}}}{{24}}} \right)} \right\} \\ {I_x} = \frac{{M{b^2}}}{{12}}\quad ...(iv) \\\end{aligned} \end{equation} $$

Consider an infinitesimal small section $dx$ and small area $dA$ of mass $dm$ at a distance $x$ from an axis of rotation. $$dA = bdx$$

$$\begin{equation} \begin{aligned} dm = {\lambda _A}dA \\ dm = \left( {\frac{M}{{ab}}} \right)(bdx) \\ dm = \frac{{Mdx}}{a}\quad ...(v) \\\end{aligned} \end{equation} $$

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

where,

$\int_m : $ Integrating over the entire mass

$x$: perpendicular distance from an axis of rotation

$dm$: mass of infinitesimally small section

From equation $(v)$ & $(vi)$ we get, $$\begin{equation} \begin{aligned} {I_y} = \int_m {{x^2}\left( {\frac{{Mdx}}{a}} \right)} \\ {I_y} = \frac{M}{a}\int_m {{x^2}dx} \\\end{aligned} \end{equation} $$

Integrating the above equation for the whole rectangular lamina is given by, $$\begin{equation} \begin{aligned} {I_y} = \frac{M}{a}\int\limits_{ - \frac{a}{2}}^{\frac{a}{2}} {{x^2}dy} \\ {I_y} = \frac{M}{a}\left[ {\frac{{{x^3}}}{3}} \right]_{ - \frac{a}{2}}^{\frac{a}{2}} \\ {I_y} = \frac{M}{a}\left\{ {\frac{{{a^3}}}{{24}} - \left( { - \frac{{{a^3}}}{{24}}} \right)} \right\} \\ {I_y} = \frac{{M{a^2}}}{{12}}\quad ...(vii) \\\end{aligned} \end{equation} $$

From equation $(i),(iv)$ & $(vii)$ we get, $${I_z} = \frac{{M{b^2}}}{{12}} + \frac{{M{a^2}}}{{12}}$$ or $${I_z} = \frac{{M({a^2} + {b^2})}}{{12}}$$

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