Subject: Physics
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.
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: The rate of change of angular displacement is known as angular velocity.
$$ \omega = \frac {\Delta \theta}{\Delta t} $$
Unit of angular velocity is radian/sec.
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/sec^{2}.
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
$$\therefore T = \frac 1F$$
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.
$$\Delta V = \overrightarrow {V_B} - \overrightarrow {V_A} = \overrightarrow {V_B} - (-\overrightarrow {V_A})$$
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
$$\boxed {\therefore a = \omega ^2 r}$$
The rate of change of angular displacement is known as angular velocity.
$$ \omega = \frac {\Delta \theta}{\Delta t} $$
centripetal acceleration can be written as
$$\boxed {\therefore a = \omega ^2 r}$$
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