Angular displacement is the angle made by radius vector at the centre within fixed interval of time.

Suppose a particle is moving in a circular path of radius $r$ and centred at $O$. Let $A$ be its position at any instant $t$ and suppose it reaches to a point $B$ at $t+∆t$.

If $θ$ is the angle made by $OB$ with $OA$, then $θ$ is the angular displacement of the particle in time $∆t$. Then, angular displacement $θ$ is, \[θ=\frac{\text{Arc}}{\text{Radius}}\] \[θ=\frac{s}{r}\] where, arc $OA = s$ and radius $OA = r$.

The unit of angular displacement is radian. It is a dimensionless quantity [Dimensional Analysis].

Here, $π \text{ radian} = 180°$.

The rate of change of angular displacement with respect to time is known as angular velocity. It is denoted by $ω$.

Suppose the particle moving in a circular path is at position $B$ in time $t_1$ making angular displacement $theta_1$ and at position $C$ in time $t_2$ making angular displacement $θ_2$.

Then, angular velocity is given by, \[ω_{av}=\frac{\text{Change in angular displacement}}{\text{Time taken}}\] \[ω_{av}=\frac{θ_2-θ_1}{t_2-t_1}\] \[ω_{av}=\frac{Δθ}{Δt}\] The time rate of angular displacement when time interval approaches zero is known as instantaneous angular displacement. Instantaneous angular velocity is given by, \[ω=\lim_{Δt \to 0}\frac{Δθ}{Δt}\] \[ω=\frac{dθ}{dt}\] Its unit is $\text{rad s}^{-1}$.

Since its dimensional formula is $[M^0L^0T^{-1}]$, it is also known as angular frequency.

Time period is the time taken by the particle to complete one revolution. It is denoted by $T$.

In time $T$, the particle completes one revolution that is of angular displacement $2π$ $\text{rad}$. Thus, angular velocity $ω$ is, \[ω=\frac{2π}{T}\]

Frequency is the number of revolutions completed by the particle in one second. \[\text{Since, } f=\frac{1}{T}\] \[∴ω=2πf\]

The rate of change of angular velocity with respect to time is called angular acceleration. It is denoted by $α$.

If the particle has angular velocity $ω_1$ at time $t_1$ and angular velocity $ω_2$ at time $t_2$, then, Average angular acceleration is given by, \[α_{av}=\frac{\text{Change in Angular Velocity}}{\text{Time taken}}\] \[α_{av}=\frac{ω_2-ω_1}{t_2-t_1}\] \[α_{av}=\frac{Δω}{Δt}\] The limiting value of average angular acceleration when time interval approaches zero is called instantaneous angular acceleration. Instantaneous angular acceleration is given by, \[α=\lim_{Δt\to 0}\frac{Δω}{Δt}\] \[α=\frac{dω}{dt}\] \[α=\frac{d}{dt}\left(\frac{dθ}{dt}\right)\] \[α=\frac{d²θ}{dt²}\] The unit of angular acceleration is $\text{rad s}^{-2}$.

Here, Angular displacement $θ$ is, \[θ=\frac{Arc}{Radius}\] \[θ=\frac{s}{r}\] \[s=rθ\] Differentiating with respect to time $t$, \[\frac{ds}{dt}=r\frac{dθ}{dt}\text{ __(1)}\] Here, \[\frac{ds}{dt}=v\] $v$ is instantaneous linear velocity of the particle. At any point of the circular path, it acts along the tangent to the path of that point. and, \[\frac{dθ}{dt}=ω\] $ω$ is instantaneous angular velocity of the particle. Thus, from $\text{(1)}$, \[v=rω\]

we have, \[v=rω\] Differentiating with respect to time $t$, \[\frac{dv}{dt}=r\frac{dω}{dt}\] \[a=rα\] where, $a$ is the linear acceleration, $r$ is the radius and $α$ is the angular acceleration of the particle.

**More on Circular Motion**

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