3.3 Kinematics equation for rotational motion

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

Kinematics equation for uniformly accelerated rotational motion is as follows,

where,

$\overrightarrow \omega $: angular velocity at any time $t$

${\overrightarrow \omega _0}$: initial angular velocity

$\overrightarrow \alpha $: angular acceleration

As we know, $$\overrightarrow \alpha = \frac{{d\overrightarrow \omega }}{{dt}}$$ or $$\overrightarrow \alpha dt = d\overrightarrow \omega $$

Integrating the above limits with proper limits we get, $$\begin{equation} \begin{aligned} \int\limits_{{\omega _0}}^\omega {d\overrightarrow \omega } = \overrightarrow \alpha \left[ t \right]_0^t \\ \left[ {\overrightarrow \omega } \right]_{{\omega _0}}^\omega = \overrightarrow \alpha \left[ t \right]_0^t \\ \left( {\overrightarrow \omega - {{\overrightarrow \omega }_0}} \right) = \overrightarrow \alpha \left( {t - 0} \right) \\ \overrightarrow \omega - {\overrightarrow \omega _0} = \overrightarrow \alpha t \\\end{aligned} \end{equation} $$ or $$\overrightarrow \omega = {\overrightarrow \omega _0} + \overrightarrow \alpha t$$

As we know, $$\begin{equation} \begin{aligned} \overrightarrow \omega = {\overrightarrow \omega _0} + \overrightarrow \alpha t\quad ...(i) \\ \overrightarrow \omega = \frac{{d\overrightarrow \theta }}{{dt}}\quad ...(ii) \\\end{aligned} \end{equation} $$ From equation $(i)$ & $(ii)$ we get, $$\frac{{d\overrightarrow \theta }}{{dt}} = {\overrightarrow \omega _0} + \overrightarrow \alpha t$$ or $$d\overrightarrow \theta = {\overrightarrow \omega _0}dt + \overrightarrow \alpha dt$$ Integrating the above equation with proper limits we get, $$\begin{equation} \begin{aligned} \int\limits_0^\theta {d\overrightarrow \theta = {{\overrightarrow \omega }_0}\int\limits_0^t {dt} + } \overrightarrow \alpha \int\limits_0^t {tdt} \\ \left[ {\overrightarrow \theta } \right]_0^\theta = {\overrightarrow \omega _0}\left[ t \right]_0^t + \overrightarrow \alpha \left[ {\frac{{{t^2}}}{2}} \right]_0^t \\ \left( {\overrightarrow \theta - 0} \right) = {\overrightarrow \omega _0}(t - 0) + \frac{1}{2}\overrightarrow \alpha \left( {{t^2} - 0} \right) \\ \overrightarrow \theta = {\overrightarrow \omega _0}t + \frac{1}{2}\overrightarrow \alpha {t^2} \\\end{aligned} \end{equation} $$

As we know, $$\overrightarrow \alpha = \frac{{d\overrightarrow \omega }}{{dt}}$$ or $$\begin{equation} \begin{aligned} \overrightarrow \alpha = \frac{{d\overrightarrow \omega }}{{d\overrightarrow \theta }}\frac{{d\overrightarrow \theta }}{{dt}}\quad ...(i) \\ \overrightarrow \omega = \frac{{d\overrightarrow \theta }}{{dt}}\quad ...(ii) \\\end{aligned} \end{equation} $$ From equation $(i)$ & $(ii)$ we get, $$\overrightarrow \alpha = \overrightarrow \omega \frac{{d\overrightarrow \omega }}{{d\overrightarrow \theta }}$$ or $$\overrightarrow \omega d\overrightarrow \omega = \overrightarrow \alpha d\overrightarrow \theta $$

Integrating the above equation with proper limits we get, $$\begin{equation} \begin{aligned} \int\limits_{{\omega _0}}^\omega {\overrightarrow \omega d\overrightarrow \omega = \overrightarrow \alpha \int\limits_0^\theta {d\overrightarrow \theta } } \\ \left[ {\frac{{{\omega ^2}}}{2}} \right]_{{\omega _0}}^\omega = \overrightarrow \alpha \left[ {\overrightarrow \theta } \right]_0^\theta \\ \frac{1}{2}\left( {{\omega ^2} - \omega _0^2} \right) = \overrightarrow \alpha \left( {\overrightarrow \theta - 0} \right) \\ {\omega ^2} - \omega _0^2 = 2\overrightarrow {\alpha \overrightarrow \theta } \\\end{aligned} \end{equation} $$

Linear velocity $\left( {\overrightarrow v } \right)$ is the vector cross-product of the angular velocity $\left( {\overrightarrow \omega } \right)$ and the radius vector $\left( {\overrightarrow r } \right)$ from point $O$.

Mathematically, $$\overrightarrow v = \overrightarrow \omega \times \overrightarrow r \quad ...(i)$$

where,

$\overrightarrow v $: linear velocity

$\overrightarrow \omega $: angular velocity

$\overrightarrow r $: perpendicular distance from the axis of rotation

From the property of the vector cross product we know, $$\overrightarrow v \bot \overrightarrow \omega \quad \& \quad \overrightarrow v \bot \overrightarrow r $$

Consider a rigid body rotating with angular velocity $\left( {\overrightarrow \omega } \right)$ about an axis of rotation (AOR).

Assume a particle on the rigid body which moves in a circle of radius $r$ as shown in the figure.

Let, $$\begin{equation} \begin{aligned} \overrightarrow \omega = \omega \widehat k\quad ...(ii) \\ \overrightarrow r = r\widehat i\quad ...(iii) \\\end{aligned} \end{equation} $$

From equation $(i),(ii)$ & $(iii)$ we get, $$\overrightarrow v = \omega \widehat k \times r\widehat i$$ or \[\overrightarrow v = \left| {\begin{array}{c} {\widehat i}&{\widehat j}&{\widehat k} \\ 0&0&\omega \\ r&0&0 \end{array}} \right|\]

$$\begin{equation} \begin{aligned} \overrightarrow v = \widehat i(0 - 0) - \widehat j(0 - r\omega ) + \widehat k(0 - 0) \\ \overrightarrow v = r\omega \widehat j \\\end{aligned} \end{equation} $$

Linear acceleration $\left( {\overrightarrow a } \right)$ is the vector cross product of the angular acceleration $\left( {\overrightarrow \alpha } \right)$ and the radius vector $\left( {\overrightarrow r } \right)$ from point $O$.

Mathematically, $$\overrightarrow a = \overrightarrow \alpha \times \overrightarrow r \quad ...(i)$$

where,

$\overrightarrow a $: linear acceleration

$\overrightarrow \alpha $: angular acceleration

$\overrightarrow r $: perpendicular distance from the axis of rotation

From the property of the vector cross product, we know, $$\overrightarrow a \bot \overrightarrow \alpha \quad \& \quad \overrightarrow a \bot \overrightarrow r $$

Consider a rigid body rotating with angular acceleration $\left( {\overrightarrow \alpha } \right)$ about an axis of rotation (AOR).

Assume a particle on the rigid bod y which moves in a circle of radius $r$ as shown in the figure.

Let, $$\begin{equation} \begin{aligned} \overrightarrow a = \alpha \widehat k\quad ...(ii) \\ \overrightarrow r = r\widehat i\quad ...\left( {iii} \right) \\\end{aligned} \end{equation} $$

From equation $(i),(ii)$ & $(iii)$ we get, $$\overrightarrow a = \alpha \widehat k \times r\widehat i$$ or \[\overrightarrow a = \left| {\begin{array}{c} {\widehat i}&{\widehat j}&{\widehat k} \\ 0&0&\alpha \\ r&0&0 \end{array}} \right|\]

$$\begin{equation} \begin{aligned} \vec a = \hat i(0 - 0) - \hat j(0 - r\alpha ) + \hat k(0 - 0) \\ \vec a = r\alpha \hat j \\\end{aligned} \end{equation} $$