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**Hint:**The terms angular velocity and angular acceleration are related to the circular motion. So, to define or understand these terms we must study the concept of circular motion of any object.

**Complete step by step answer:**The circular motion is a motion of a particle along the circumference of a circle.

Angular velocity: The angular velocity is the rate at which displacement is changing with respect to the time in circular motion.

The instantaneous angular velocity is given by,

$\begin{array}{c}

\overrightarrow \omega = \mathop {\lim }\limits_{\delta t \to 0} \dfrac{{\delta \theta }}{{\delta t}}\\

= \dfrac{{d\overrightarrow \theta }}{{dt}}

\end{array}$

The finite angular velocity is give by,

$\omega = \dfrac{\theta }{t}$

Here, $\omega $ is angular velocity, $\theta $ is angular displacement, and $t$ is time.

The unit of angular velocity is ${\rm{radian/s}}$. The dimension of angular acceleration is $\left[ {{{\rm{M}}^{\rm{0}}}{{\rm{L}}^{\rm{0}}}{{\rm{T}}^{ - 1}}} \right]$.

Angular acceleration: The angular acceleration is rate at which the angular velocity is changing with respect to the time. It is denoted by the symbol $\overrightarrow \alpha $.

If ${\overrightarrow \omega _0}$ and $\overrightarrow \omega $ are the angular velocities of an particle performing circular motion at time ${t_0}$ and $t$ then angular acceleration is given by,

$\begin{array}{c}

\overrightarrow \alpha = \dfrac{{\delta \overrightarrow \omega }}{{\delta t}}\\

= \dfrac{{\overrightarrow \omega - {{\overrightarrow \omega }_0}}}{{t - {t_0}}}

\end{array}$

The unit of angular acceleration is ${\rm{radian/}}{{\rm{s}}^2}$. The dimension of angular acceleration is $\left[ {{{\rm{M}}^{\rm{0}}}{{\rm{L}}^{\rm{0}}}{{\rm{T}}^{ - 2}}} \right]$.

The relationship between linear velocity and angular velocity can be obtained as follows.

Consider a particle is travelling along a circular path of radius $r$.

Let $\overrightarrow \omega $ be angular velocity, and $\overrightarrow v $ is linear velocity.

The liner displacement in vector form is given by,

$\overrightarrow {\delta s} = \overrightarrow {\delta \theta } \times \overrightarrow r $

Dividing both sides of the above equation by $\delta t$

$\dfrac{{\overrightarrow {\delta s} }}{{\delta t}} = \dfrac{{\overrightarrow {\delta \theta } }}{{\delta t}} \times \overrightarrow r $

Taking limiting value of the equation $\dfrac{{\overrightarrow {\delta s} }}{{\delta t}} = \dfrac{{\overrightarrow {\delta \theta } }}{{\delta t}} \times \overrightarrow r $

\[\begin{array}{c}

\mathop {\lim }\limits_{\delta t \to 0} \dfrac{{\overrightarrow {\delta s} }}{{\delta t}} = \mathop {\lim }\limits_{\delta t \to 0} \dfrac{{\overrightarrow {\delta \theta } }}{{\delta t}} \times \overrightarrow r \\

\dfrac{{\overrightarrow {ds} }}{{dt}} = \dfrac{{\overrightarrow {d\theta } }}{{dt}} \times \overrightarrow r

\end{array}\]

But it is known that \[\dfrac{{\overrightarrow {ds} }}{{dt}} = \overrightarrow v \] and \[\dfrac{{\overrightarrow {d\theta } }}{{dt}} = \overrightarrow \omega \]. So,

$\overrightarrow v = \overrightarrow \omega \times \overrightarrow r $

**Note:**e know that the angular velocity is also a ratio of displacement and time, and the angular acceleration is a ratio of angular velocity and time. The relationship between angular velocity and linear velocity can be obtained by two methods; first is the Analytical method and the other one is the Calculus method.

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