Position of the axis of rotation : (a) About an axis passing through centre of mass and perpendicular to its length (b)


Moment of Inertia (I) = $\frac{{\mathrm{ML}}^2}{12}$

Radius of gyration (K) = $\frac{\mathrm{L}}{\sqrt{12}}$

Position of the axis of rotation : About an axis passing through one end and perpendicular to length of the rod


Moment of Inertia (I) = $\frac{{\mathrm{ML}}^2}{3}$

Radius of gyration (K) = $\frac{\mathrm{L}}{\sqrt{3}}$

The angular momentum $\vec{L}$ of a system remains constant if the net external torque acting on the system is zero: $ \vec{L} =\text { a constant } \text { (isolated system) } $

$ \text { or } \vec{L}_i =\vec{L}_f \quad \text { (isolated system). } $

This is the law of conservation of angular momentum.

For a rigid body rotating about a fixed axis, the component of its angular momentum parallel to the rotation axis is

$L=I\omega \text{ (rigid body, fixed axis). }$

The angular momentum $\vec{L}$ of a system of particles is the vector sum of the angular momenta of the individual particles:

$$ \vec{L}=\vec{\ell}_1+\vec{\ell}_2+\cdots+\vec{\ell}_n=\sum\limits_{i=1}^n \vec{\ell}_i . $$

The time rate of change of this angular momentum is equal to the net external torque on the system (the vector sum of the torques due to interactions with particles external to the system):

$$ \vec{\tau}_{\text {net }}=\frac{d \vec{L}}{d t} \quad \text { (system of particles). } $$

Newton's second law for a particle can be written in angular form as

$$ \vec{\tau}_{\text {net }}=\frac{d \vec{\ell}}{d t}, $$

where $\vec{\tau}_{\text {net }}$ is the net torque acting on the particle and $\vec{\ell}$ is the angular momentum of the particle.

The angular momentum $\vec{\ell}$ of a particle with linear momentum $\vec{p}$, mass $m$, and linear velocity $\vec{v}$ is a vector quantity defined relative to a fixed point (usually an origin) as

$$ \vec{\ell}=\vec{r} \times \vec{p}=m(\vec{r} \times \vec{v}) $$

The magnitude of $\vec{\ell}$ is given by

$$ \begin{aligned} \ell &=r m v \sin \phi \\\\ &=r p_{\perp}=r m v_{\perp} \\\\ &=r_{\perp} p=r_{\perp} m v, \end{aligned} $$

where $\phi$ is the angle between $\vec{r}$ and $\vec{p}, p_{\perp}$ and $v_{\perp}$ are the components of $\vec{p}$ and $\vec{v}$ perpendicular to $\vec{r}$, and $r_{\perp}$ is the perpendicular distance between the fixed point and the extension of $\vec{p}$. The direction of $\ell$ is given by the right-hand rule for cross products.

Consider a system of point masses rotating with angular velocity $\omega$ about a fixed axis through $\mathrm{O}$. Kinetic energy of a point mass $m_1$ located at $P$ distant $r_1$ from the centre is given by:

$\frac{1}{2}{m}_1{v}_1^2=\frac{1}{2}{m}_1{r}_1^2{\omega }^2$

Kinetic energy of the system

$\begin{array}{rl}& =\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2+\dots \dots \\ & \\ & =\frac{1}{2}{m}_1{r}_1^2{\omega }^2+\frac{1}{2}{m}_2{r}_2^2{\omega }^2+\dots \dots ..\\ & \\ & =\frac{1}{2}\left(m_1{r}_2^2+m{r}_2^2+m_3{r}_3^2+\dots \dots \right){\omega }^2\end{array}$

The expression $\left(m_1{r}_1^2+m_2{r}_2^2+\dots ..\right)$ is known as moment of inertia of the system about the corresponding axis. It is denoted by $\mathrm{I}$.

$\begin{array}{rl}& \mathrm{I}=m_1{r}_1^2+m_2{r}_2^2+m_3{r}_3^2+\dots \dots \dots \dots \\ & \\ & \mathrm{I}=\sum m{r}^2\end{array}$

Hence kinetic energy of the system = $\frac{1}{2}I{\omega }^2$

The rotational analog of Newton's second law is

$$ \tau_{\text {net }}=I \alpha, $$

where $\tau_{\text {net }}$ is the net torque acting on a particle or rigid body, $I$ is the rotational inertia of the particle or body about the rotation axis, and $\alpha$ is the resulting angular acceleration about that axis.

If a rigid body is in rotational equilibrium under the action of several coplanar forces, the resultant torque of all the forces about any axis perpendicular to the plane containing the forces must be zero. In the figure a body is shown under the action of several external coplanar forces $F_1, F_2, \ldots . . F_i$ and $F_n$.

$$\sum {\overrightarrow \tau } $$ = 0

Torque about point : $\vec{\tau }=\overrightarrow{\mathrm{r}}\times \overrightarrow{\mathrm{F}}$



Magnitude of torque $=$ Force $\times$ perpendicular distance of line of action of force from the axis of rotation.

$\tau =\mathrm{rF}\sin \theta$

Direction of torque can be determined by using right hand thumb rule.

Position of the axis of rotation : About its geometrical axis, which is along its length


Moment of Inertia (I) = $\frac{{\mathrm{Ma}}^2}{6}$

Radius of gyration (K) = $\frac{\mathrm{a}}{\sqrt{6}}$

Position of the axis of rotation : About an axis passing through centre of mass and perpendicular to side $\mathrm{b}$ in its plane


Moment of Inertia (I) = $\frac{{\mathrm{Mb}}^2}{12}$

Radius of gyration (K) = $\frac{\mathrm{b}}{2\sqrt{3}}$

Position of the axis of rotation : About an axis passing through centre of mass and perpendicular to side a in its plane.


Moment of Inertia (I) = $\frac{{\mathrm{Ma}}^2}{12}$

Radius of gyration (K) = $\frac{\mathrm{a}}{2\sqrt{3}}$

Position of the axis of rotation : About an axis passing throught centre of mass and perpendicular to plane


Moment of Inertia (I) = $\frac{\mathrm{M}\left({\mathrm{a}}^2+{\mathrm{b}}^2\right)}{12}$

Radius of gyration (K) = $\sqrt{\frac{{\mathrm{a}}^2+{\mathrm{b}}^2}{12}}$

The moment of inertia of a particles with respect to an axis of rotation is equal to the product of mass of the particle and square of distance from rotational axis.

$\mathrm{I}={\mathrm{mr}}^2$

$\mathrm{r}=$ perpendicular distance from axis of rotation

Position of the axis of rotation : About its geometrical axis, which is along its length


Moment of Inertia (I) = $\frac{{\mathrm{MR}}^2}{2}$

Radius of gyration (K) = $\frac{\mathrm{R}}{\sqrt{2}}$

Position of the axis of rotation : About an axis tangential to the cylinderical surface and parallel to its geometrical axis is perpendicular to its length and passes through its centre of mass


Moment of Inertia (I) = $\frac{3}{2}{\mathrm{MR}}^2$

Radius of gyration (K) = $\sqrt{\frac{3}{2}}\mathrm{R}$

Position of the axis of rotation : About an axis passing through the centre of mass and perpendicular to its length


Moment of Inertia (I) = $\frac{{\mathrm{ML}}^2}{12}+\frac{{\mathrm{MR}}^2}{4}$

Radius of gyration (K) = $\sqrt{\frac{{\mathrm{L}}^2}{12}+\frac{{\mathrm{R}}^2}{4}}$

Position of the axis of rotation : About its geometrical axis which is parallel to its length


Moment of Inertia (I) = ${\mathrm{MR}}^2$

Radius of gyration (K) = $R$

Position of the axis of rotation : About an axis which is perpendicular to its length and passes through its centre of mass

Moment of Inertia (I) = $\frac{{\mathrm{MR}}^2}{2}+\frac{{\mathrm{ML}}^2}{12}$

Radius of gyration (K) = $\sqrt{\frac{{\mathrm{R}}^2}{2}+\frac{{\mathrm{L}}^2}{12}}$

Position of the axis of rotation : About diametric axis passing through centre of mass


Moment of Inertia (I) = $\frac{2}{3}{\mathrm{MR}}^2$

Radius of gyration (K) = $\sqrt{\frac{2}{3}}\mathrm{R}$

Position of the axis of rotation : About a tangent to the surface


Moment of Inertia (I) = $\frac{5}{3}{\mathrm{MR}}^2$

Radius of gyration (K) = $\sqrt{\frac{5}{3}}\mathrm{R}$

Position of the axis of rotation : About its diametric axis which passes through its centre of mass


Moment of Inertia (I) = $${2 \over 5}M{R^2}$$

Radius of gyration (K) = $$\sqrt {{2 \over 5}} R$$

Position of the axis of rotation : About a tangent to the sphere


Moment of Inertia (I) = $${7 \over 5}M{R^2}$$

Radius of gyration (K) = $$\sqrt {{7 \over 5}} R$$

Position of the axis of rotation : About an axis perpendicular to the plane and passes through the centre


Moment of Inertia (I) = $\frac{\mathrm{M}}{2}\left[{\mathrm{R}}_1^2+{\mathrm{R}}_2^2\right]$

Radius of gyration (K) = $\sqrt{\frac{{\mathrm{R}}_1^2+{\mathrm{R}}_2^2}{2}}$

Position of the axis of rotation : About the diametric axis

Moment of Inertia (I) = $\frac{\mathrm{M}}{4}\left[{\mathrm{R}}_1^2+{\mathrm{R}}_2^2\right]$

Radius of gyration (K) = $\frac{\sqrt{R_1^2+R_2^2}}{2}$

Position of the axis of rotation : About an axis perpendicular to the plane and passes through the centre


Moment of Inertia (I) = $\frac{1}{2}M{R}^2$

Radius of gyration (K) = $\frac{R}{\sqrt{2}}$

Position of the axis of rotation : About the diametric axis


Moment of Inertia (I) = $\frac{1}{4}M{R}^2$

Radius of gyration (K) = $\frac{R}{2}$

Position of the axis of rotation : About an axis perpendicular to the plane and passes through the centre


Moment of Inertia (I) = MR²

Radius of gyration (K) = R

Position of the axis of rotation : About the diametric axis


Moment of Inertia (I) = $\frac{1}{2}M{R}^2$

Radius of gyration (K) = $\frac{R}{\sqrt{2}}$

The parallel-axis theorem relates the rotational or moments of inertia $I$ of a body about any axis to that of the same body about a parallel axis through the center of mass :

$I=I_{\text{com }}+M{h}^2.$

Here $h$ is the perpendicular distance between the two axes.

$I_z=I_x+I_y$

(body lies on the $x-y$ plane)

Note : (Valid only for 2-dimensional body)

$\mathrm{K}$ has no meaning without axis of rotation.

$\mathrm{I}={\mathrm{MK}}^2\mathrm{K}$ is a scalar quantity

Radius of gyaration $:K=\sqrt{\frac{I}{M}}$

Moment of inertia of a rigid body about any axis of rotation.

$\mathrm{I}=\int \mathrm{dm}{\mathrm{r}}^2$

$\mathrm{I}={\mathrm{m}}_1{\mathrm{r}}_1^2+{\mathrm{m}}_2{\mathrm{r}}_2^2+{\mathrm{m}}_3{\mathrm{r}}_3^2+\dots \dots$