Let us fix the co-ordinate system at the point O as shown in...

Let us fix the co-ordinate system at the point $O$ as shown in the figure, such that the radius vector $\vec{r}$ of point $A$ makes an angle $\theta$ with $x$ axis at the moment shown. Note that the radius vector of the particle $A$ rotates clockwise and we here take line $o x$ as reference line, so in this case obviously the angular velocity $\omega=\left(-\frac{d \theta}{d t}\right)$ taking anticlockwise sense of angular displacement as positive. Also from the geometry of the triangle $O A C$ $\frac{R}{\sin \theta}=\frac{r}{\sin (\pi-2 \theta)}$ or, $r=2 R \cos \theta$ Let us write, $\vec{r}=r \cos \theta \vec{i}+r \sin \theta \vec{j}=2 R \cos ^{2} \theta \vec{i}+R \sin 2 \theta \vec{j}$ Differentiating with respect to time. $\frac{\overrightarrow{d r}}{d t}$ or $\vec{v}=2 R 2 \cos \theta(-\sin \theta) \frac{d \theta}{d t} \vec{i}+2 R \cos 2 \theta \frac{d \theta}{d t} \vec{j}$ or, $\vec{v}=2 R\left(\frac{-d \theta}{d t}\right)[\sin 2 \theta \vec{i}-\cos 2 \theta \vec{j}]$ or, $\vec{v}=2 R \omega\left(\sin 2 \theta \vec{i}-\cos ^{2} \theta \vec{j}\right)$ So, $|\vec{v}|$ or $v=2 \omega R=0 \cdot 4 \mathrm{~m} / \mathrm{s}$ As $\omega$ is constant, $v$ is also constant and $w_{t}=\frac{d v}{d t}=0$, So, $w=w_{n}=\frac{v^{2}}{R}=\frac{(2 \omega R)^{2}}{R}=4 \omega^{2} R=0.32 \mathrm{~m} / \mathrm{s}^{2}$ Alternate : From the Fig. the angular velocity of the point $A$, with respect to centre of the circle $C$ becomesThus we have the problem of finding the velocity and acceleration of a particle moving along a circle of radius $R$ with constant angular velocity $2 \omega$. Hence $\quad v=2 \omega R$ and $$ w=w_{n}=\frac{v^{2}}{R}=\frac{(2 \omega R)^{2}}{R}=4 \omega^{2} R $$

Let us fix the co-ordinate system at the point $O$ as shown in the figure, such that the radius vector $\vec{r}$ of point $A$ makes an angle $\theta$ with $x$ axis at the moment shown. Note that the radius vector of the particle $A$ rotates clockwise and we here take line $o x$ as reference line, so in this case obviously the angular velocity $\omega=\left(-\frac{d \theta}{d t}\right)$ taking anticlockwise sense of angular displacement as positive. Also from the geometry of the triangle $O A C$ $\frac{R}{\sin \theta}=\frac{r}{\sin (\pi-2 \theta)}$ or, $r=2 R \cos \theta$
Let us write, $\vec{r}=r \cos \theta \vec{i}+r \sin \theta \vec{j}=2 R \cos ^{2} \theta \vec{i}+R \sin 2 \theta \vec{j}$
Differentiating with respect to time. $\frac{\overrightarrow{d r}}{d t}$ or $\vec{v}=2 R 2 \cos \theta(-\sin \theta) \frac{d \theta}{d t} \vec{i}+2 R \cos 2 \theta \frac{d \theta}{d t} \vec{j}$
or, $\vec{v}=2 R\left(\frac{-d \theta}{d t}\right)[\sin 2 \theta \vec{i}-\cos 2 \theta \vec{j}]$
or, $\vec{v}=2 R \omega\left(\sin 2 \theta \vec{i}-\cos ^{2} \theta \vec{j}\right)$
So, $|\vec{v}|$ or $v=2 \omega R=0 \cdot 4 \mathrm{~m} / \mathrm{s}$
As $\omega$ is constant, $v$ is also constant and $w_{t}=\frac{d v}{d t}=0$,
So, $w=w_{n}=\frac{v^{2}}{R}=\frac{(2 \omega R)^{2}}{R}=4 \omega^{2} R=0.32 \mathrm{~m} / \mathrm{s}^{2}$
Alternate : From the Fig. the angular velocity of the point $A$, with respect to centre of the circle $C$ becomesThus we have the problem of finding the velocity and acceleration of a particle moving along a circle of radius $R$ with constant angular velocity $2 \omega$.
Hence $\quad v=2 \omega R$ and
$$
w=w_{n}=\frac{v^{2}}{R}=\frac{(2 \omega R)^{2}}{R}=4 \omega^{2} R
$$

Key Concepts

Uniform Circular Motion

This concept describes the motion of a particle travelling along a circular path with a constant rate of rotation. In uniform circular motion, the speed's magnitude remains constant while the direction continuously changes, which necessitates a centripetal (or normal) acceleration directed toward the center of the circle.

Angular Velocity

Angular velocity is a measure of how fast the angular position of a particle changes with time. It indicates the rate at which a particle sweeps out an angle along its circular path. The sign of angular velocity is determined by the chosen orientation (for example, anticlockwise may be taken as positive), and it plays a crucial role in relating linear speed to the radius of the circle.

Vector Differentiation in Kinematics

In kinematics, the velocity and acceleration vectors of a moving particle are obtained by differentiating the position vector with respect to time. This involves applying the chain rule when the position components are expressed in terms of a time-dependent angle, allowing one to capture both the magnitude and direction changes in the motion.

Centripetal (Normal) Acceleration

Centripetal acceleration is the acceleration experienced by a particle moving in a circular path that keeps it moving along that path. It is given by the formula v²/R, where v is the tangential speed and R is the radius of the circle. This acceleration is always directed towards the center of the circle and is essential for maintaining circular motion.

Geometric Relationships in Circular Motion

Geometric relationships are used to relate the position and displacement of a particle in circular motion. For instance, by using properties of triangles and trigonometric relations, one can connect the radius vector, the distance along the circle, and the angle of rotation. These relationships are key to expressing the particle's coordinates and subsequently differentiating them to obtain velocity and acceleration.

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