`text((a))` `text(Angular Position :)`

The angle made by the position vector of a particle undergoing circular motion with a given line (reference line) is called angular position.

Suppose a particle `P` is moving in a circle of radius `r` and centre `O`.

The position of the particle Pat a given instant may be described by the angle `theta` between `OP` and `OX`. This angle `theta` is called the angular position of the particle.

`text((b))` `text(Angular Displacement :)`

Angle rotated by a position vector of the moving particle with some reference line is called angular displacement.

(i) Infinitesimal angular displacement is a vector quantity but finite angular displacement is not.
(ii) Direction of infinitesimal angular displacement is decided by right hand thumb rule.

`text((c))` `text(Angular Velocity)` (`omega`) :

(i) Average Angular Velocity :

If `theta_1` and `theta_2` are angular position of the particle at time `t_1` and `t_2`

`omega_(avg)=text(Total Angle of Rotation)/text(Total time taken) `

`omega_(avg)=(theta_2-theta_1)/(t_2-t_1)=(Delta theta)/(Delta t)`

(ii) Instantaneous Angular Velocity :

The rate at which the position vector of a particle w.r.t. the centre rotates at a given instant, is called its instantaneous angular velocity.

`:.` instantaneous angular velocity ` omega=lim_(Delta t->0) (Delta theta)/(Delta t)`

`:. omega= (d theta)/(dt)`

`text((d))` `text(Angular Acceleration)` (`alpha`)

(i) Average Angular Acceleration :

Let `omega_1` and `omega_2` be the instantaneous angular speeds at times `t_1` and `t_2` respectively, then the average angular acceleration (`alpha`) is defined as

`alpha_(avg)=(omega_2-omega_1)/(t_2-t_1)=(Delta omega)/(Delta t)`

(ii) Instantaneous Angular Acceleration :

The rate of change of angular velocity at a given instant

`alpha=lim_(Delta t-> 0) (Delta omega)/(Delta t)=(domega)/(dt)=(d^2theta)/(dt^2)=omega(domega)/(d theta)`

`text(Important points :)`

(i) If `alpha=0` circular motion is said to be uniform.
(ii) If the angular acceleration `alpha` is constant, we have

`theta=omega_0t+1/2alphat^2`..............(1)

`omega_t=omega_0+alphat`.................(2)

`omega_t^2=omega_0^2+2alphatheta`...........(3)

where `omega_0` and `omega_t` are the angular velocities at `t = 0` and at time `t` and is the angular displacement in time interval `t`.

`text((e))` `text(Relation Between Linear Speed and Angular Speed)`

The linear distance travelled by the particle in time `Deltat` is

`Deltas=rDeltatheta`

`lim_(Deltat->0)(Deltas)/(Deltat)=lim_(Deltat->0)r(Deltatheta)/(Deltat)`

`v=romega`........(4)

If the particle moves in the circle with a uniform speed, we call it a uniform circular motion. In this case , `(dv)/(dt)=0` and

`alpha=-omega^2 r hat e_r`

Thus, the acceleration of the particle is in the direction of `-hat e_r`, that is, towards the centre. The magnitude of the acceleration is

`alpha=omega^2r=v^2/r`..............(5)

Thus, if a particle moves in a circle of radius `r` with a constant speed `v`, its acceleration is `v^2//r` directed towards the centre. This acceleration is called centripetal acceleration.