Some important terms in Circular Motion

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Circle is a polygon having an infinite number of sides. So, when a particle moves in a circular path, it changes its direction at each point ie. an infinite number of times.This note provides us an information about some important terms in circular motion
Some important terms in Circular Motion

Circle is a polygon having an infinite number of sides. So, when a particle moves in a circular path, it changes its direction at each point ie. an infinite number of times.

Motion of a point particle along a circle.
The motion of a point particle along a circle.

Angular displacement: It is the angle described by the radius when a particle moves from one point to another in a circular path.

Here, θ is the angular displacement.

The relation between angular displacement and linear displacement

θ =, where, s is the arc length.

\(\theta = \frac Sr\) where s is the arc length.

Angular velocity in a circular motion.
Angular velocity in a circular motion.

Angular velocity: The rate of change of angular displacement is known as angular velocity.

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$$ \omega = \frac {\Delta \theta}{\Delta t} $$

Unit of angular velocity is radian/sec.

Angular velocity in a circular motion.
The relation between linear velocity and angular velocity in the circular motion.

Relation between angular velocity and linear velocity

$$ \omega = \frac {\Delta \theta}{\Delta t} $$

$$\text {As} \: \theta =\frac Sr \Delta \theta = \frac {\Delta S}{\Delta r} $$

We know,

$$ \omega = \frac {\Delta \theta}{\Delta t} $$

$$ \text {or,}\:\omega = \frac {\Delta \theta}{\Delta t} \times \frac 1r $$

$$ \text {or,}\:\omega = V \times \frac 1r $$

$$ \boxed {\therefore\: V = \omega r} $$

Instantaneous angular velocity: It is the velocity at a particular instant of time.

$$\omega = \Delta t \rightarrow 0 = \frac {d \theta}{dt}$$

Angular Acceleration: The rate of change of angular velocity with respect to time.

$$\alpha = \frac {\Delta \omega}{\Delta t}$$

The unit of angular acceleration is rad/sec2.

Relation between angular acceleration and linear acceleration

We know,

$$V = \omega \times r$$

$$ \text {or,} \Delta V = \Delta \omega \times r $$

$$ \text {or,} \frac {\Delta V} {\Delta t} = \frac {\Delta \omega}{\Delta t} \times r $$

$$ \boxed {\therefore a = \infty \times r}$$

Frequency: Number of rotation completed in one second is known as frequency. Its notation is 'f' and the unit is 'Hertz'.

$$ f =\frac {\text {cycle}}{\text {frequency}} $$

Time period: Time taken to complete one rotation is known as time period. It is denoted by 'T' and its unit is second.

Relation between f and T

f rotation in 1 second

1 rotation in 1/f second

$$\therefore T = \frac 1F$$

Expression for Centripetal Acceleration

Consider a particle moving with constant speed v in a circular path of radius r with centre at O. At a point A on its path, its velocity \(\overrightarrow A\) which is acted along tangent AC. After short interval of time \(\Delta t\) body is at B where velocity of body is \(\overrightarrow B\) acted along tangent BD since the direction of velocity is different from A to B. Hence there is change in velocity from A to B which has to find change in velocity from A to B.

(a) Acceleration in a circle (b) Change in velocity
(a) Acceleration in a circle (b) Change in velocity

$$\Delta V = \overrightarrow {V_B} - \overrightarrow {V_A} = \overrightarrow {V_B} - (-\overrightarrow {V_A})$$

Now the magnitude and direction of change in velocity is calculated by using triangle law of vector addition such the magnitude and direction on \(\vec V_B\) is represented by PQ, -\(\vec V_A\) is represented by QR side when magnitude and direction of resultant are represented by PR side of PQR as shown in figure b.

$$\text {i.e} \overrightarrow {PR} = \overrightarrow {PQ} + \overrightarrow {QR}$$

$$\Delta V = \overrightarrow {V_B} - \overrightarrow {V_A}$$

Since PR vector is acted towards the centre of circle hence change in velocity as well as acceleration acted towards the centre.

Using triangle law of vector addition,

$$R = \sqrt {P^2 +Q^2 + 2PQ\cos\theta }$$

$$\Delta V = \sqrt {V_B^2 + V_A^2 + 2V_AV_B \cos (180 - \Delta \theta)}$$

$$ = \sqrt {V_B^2 + V_A^2 + 2V^2\cos \Delta \theta } $$

$$= \sqrt {2V^2 (1 - \cos\Delta\theta)}$$

$$=\sqrt {2V^2 \times 2\sin^2 \frac {\Delta \theta}{2}}$$

$$\Delta V = 2V\sin^2 \frac {\Delta \theta}{2}\dots i$$

Now centripetal acceleration

$$\frac {\Delta V}{\Delta t} = \frac {2V\sin \frac {\Delta \theta}{2}}{\Delta t} $$

$$= 2.V\frac { \frac {\sin \frac {\Delta \theta}{2}}{\frac {\Delta \theta}{2}}}{\Delta t} . \frac {\Delta \theta}{2}$$

A and B are neighbourhood points

$$\Delta t \rightarrow 0, \Delta \theta \rightarrow 0, \frac {\Delta \theta}{2} \rightarrow 0$$

$$\therefore a = \lim \limits_{\Delta t \to 0} \frac {2 V \Delta \theta}{2 \times \Delta t }$$

$$a = V\omega $$

$$= V \times\frac {V}{r}$$

$$= \frac {V^2}{r} \dots (ii)$$

forΔt→0 B is verry close to A andΔθ becomes verry small .

since v=rω,the centripetal acceleration can be written as

$$\boxed {\therefore a = \omega ^2 r}$$

Things to remember
  • It is the angle described by the radius when a particle moves from one point to another in a circular path.
  • The rate of change of angular displacement is known as angular velocity.

    $$ \omega = \frac {\Delta \theta}{\Delta t} $$

  •  centripetal acceleration can be written as

    $$\boxed {\therefore a = \omega ^2 r}$$

  • Time taken to complete one rotation is known as the time period. It is denoted by 'T' and its unit is second.
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  • There can be more than one community in a society. Community smaller than society.
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