Angular Variables and Basic of Uniform Circular Motion Questions in English

(B) Given the position vector: $\vec{r} = (a \cos \omega t)\hat{i} + (a \sin \omega t)\hat{j}$.
To find the velocity vector $\vec{v}$,we differentiate $\vec{r}$ with respect to time $t$:
$\vec{v} = \frac{d\vec{r}}{dt} = \frac{d}{dt}[(a \cos \omega t)\hat{i} + (a \sin \omega t)\hat{j}] = -a \omega \sin \omega t \hat{i} + a \omega \cos \omega t \hat{j}$.
Now,we calculate the dot product of $\vec{r}$ and $\vec{v}$ to check their orientation:
$\vec{r} \cdot \vec{v} = [(a \cos \omega t)\hat{i} + (a \sin \omega t)\hat{j}] \cdot [(-a \omega \sin \omega t)\hat{i} + (a \omega \cos \omega t)\hat{j}]$
$\vec{r} \cdot \vec{v} = (a \cos \omega t)(-a \omega \sin \omega t) + (a \sin \omega t)(a \omega \cos \omega t)$
$\vec{r} \cdot \vec{v} = -a^2 \omega \sin \omega t \cos \omega t + a^2 \omega \sin \omega t \cos \omega t = 0$.
Since the dot product is $0$,the velocity vector is perpendicular to the position vector.

(C) The angular speed $\omega$ of an object moving in a circular path is defined as the rate of change of angular displacement,given by the formula $\omega = \frac{2\pi}{T}$,where $T$ is the time period of one complete revolution.
Since both cars complete their respective circles in the same duration of time $T$,their time periods are equal.
Therefore,the ratio of their angular speeds is $\frac{\omega_1}{\omega_2} = \frac{2\pi / T}{2\pi / T} = 1:1$.
Thus,the correct option is $C$.

(C) When a particle moves in a circle describing equal angles in equal times,it is performing $Uniform$ $Circular$ $Motion$ $(UCM)$.
In $UCM$,the speed of the particle remains constant,but the direction of the velocity vector changes continuously.
The velocity vector is always directed along the tangent to the circular path at any given point.
Since the direction of the tangent changes as the particle moves along the circle,the velocity vector changes in direction even though its magnitude remains constant.
Therefore,the correct option is $C$.

(C) When a body moves in a circular path with a constant speed $v$,it undergoes uniform circular motion.
In uniform circular motion,the velocity vector changes continuously because its direction changes at every point,even though its magnitude remains constant.
Since acceleration is the rate of change of velocity,the body experiences a centripetal acceleration directed towards the center of the circle.
The magnitude of this centripetal acceleration is given by $a_c = \frac{v^2}{r}$,where $v$ is the constant speed and $r$ is the radius of the circular path.
Since both $v$ and $r$ are constant,the magnitude of the acceleration is constant,but its direction changes continuously as the body moves along the path.
Therefore,the body has an acceleration of constant magnitude.

(C) In uniform circular motion,the speed of the object remains constant,but the direction of motion changes at every point along the path.
Since linear velocity is a vector quantity (having both magnitude and direction),it changes continuously due to the change in direction.
Acceleration (centripetal acceleration) is directed towards the center of the circle,and its direction changes as the object moves.
Force (centripetal force) is also directed towards the center and changes direction continuously.
However,the angular velocity $\omega = \frac{v}{r}$ remains constant because both speed $v$ and radius $r$ are constant.
Therefore,the correct option is $(c)$.

(B) The angular velocity $\omega$ of a particle moving with velocity $v$ at a distance $r_{\perp}$ from a point is given by $\omega = \frac{v_{\perp}}{r_{\perp}}$,where $v_{\perp}$ is the component of velocity perpendicular to the position vector.
At point $B$,the velocity $v$ is perpendicular to the diameter $AB$.
For point $A$,the distance is $AB = 2a$. Thus,the angular velocity about $A$ is $\omega_A = \frac{v}{2a}$.
For point $C$,the distance is $CB = a$. Thus,the angular velocity about $C$ is $\omega_C = \frac{v}{a}$.
The ratio of angular velocities is $\frac{\omega_A}{\omega_C} = \frac{v/2a}{v/a} = \frac{1}{2}$.

(D) In uniform circular motion,the speed of the body is constant because the magnitude of the velocity vector $\vec{v}$ is given by $|\vec{v}| = |\vec{\omega}| r$,where $|\vec{\omega}|$ is the constant angular velocity and $r$ is the radius of the circular path.
Since the speed $|\vec{v}|$ is constant,the kinetic energy $K = \frac{1}{2} m |\vec{v}|^2$ remains constant.
However,the velocity vector $\vec{v}$ changes direction continuously,so velocity is not constant.
The momentum $\vec{p} = m\vec{v}$ also changes direction continuously,so momentum is not constant.
The centripetal acceleration $\vec{a}_c = \vec{\omega} \times \vec{v}$ changes direction as the body moves,so acceleration is not constant.
Therefore,only the kinetic energy remains constant throughout the motion.

(C) In uniform circular motion,the speed $v$ of the particle remains constant.
Kinetic energy is given by the formula $K.E. = \frac{1}{2}mv^2$.
Since mass $m$ and speed $v$ are constant,the kinetic energy remains constant.
Velocity,acceleration,and displacement are vector quantities that change direction continuously as the particle moves along the circular path.

(D) The angular speed $\omega$ is given by the formula $\omega = 2\pi n$,where $n$ is the frequency in revolutions per second.
Given,$n = 120 \, \text{rev/min} = \frac{120}{60} \, \text{rev/s} = 2 \, \text{rev/s}$.
Substituting the value of $n$ into the formula:
$\omega = 2\pi \times 2 = 4\pi \, \text{rad/s}$.
Therefore,the correct option is $D$.

(C) In uniform circular motion,the speed of the particle remains constant,but the direction of velocity changes continuously. This change in velocity gives rise to centripetal acceleration. This acceleration is always directed towards the centre of the circular path,which is along the radius.

(D) The length of the second's hand is $r = 1 \, cm$. The time period of the second's hand is $T = 60 \, s$.
The angular velocity is $\omega = \frac{2\pi}{T} = \frac{2\pi}{60} = \frac{\pi}{30} \, rad/s$.
The linear velocity of the tip is $v = r\omega = 1 \times \frac{\pi}{30} = \frac{\pi}{30} \, cm/s$.
In $15 \, seconds$, the second's hand rotates through an angle $\theta = 90^\circ$ (since $60 \, s$ corresponds to $360^\circ$).
The change in velocity is given by $\Delta v = |\vec{v_2} - \vec{v_1}| = 2v \sin(\theta/2)$.
Substituting the values: $\Delta v = 2 \times \left(\frac{\pi}{30}\right) \times \sin(90^\circ/2) = 2 \times \frac{\pi}{30} \times \sin(45^\circ) = 2 \times \frac{\pi}{30} \times \frac{1}{\sqrt{2}} = \frac{\pi\sqrt{2}}{30} \, cm/s$.

(B) In uniform circular motion,the speed of the particle remains constant,but the direction of velocity changes continuously.
Since the velocity vector is always tangent to the circular path,the particle possesses tangential velocity.
Because the direction of the velocity vector changes,there must be an acceleration.
In uniform circular motion,this acceleration is directed towards the center of the circle,which is known as radial acceleration (or centripetal acceleration).
Since the speed is constant,there is no tangential acceleration.
Therefore,the particle has tangential velocity and radial acceleration.

(B) The relationship between linear velocity $(v)$,radius $(r)$,and angular velocity $(\omega)$ is given by the formula: $v = r \times \omega$.
Given:
Radius $r = 20 \,cm = 0.2 \,m$.
Angular velocity $\omega = 10 \,rad/s$.
Substituting the values into the formula:
$v = 0.2 \,m \times 10 \,rad/s = 2 \,m/s$.
Therefore,the linear velocity is $2 \,m/s$.

(A) The angular speed $\omega$ is given by the formula $\omega = \frac{2\pi}{T}$.
For the seconds needle of a watch,the time period $T$ required to complete one full rotation is $60 \, s$.
Substituting the value of $T$ into the formula:
$\omega = \frac{2\pi}{60} \, rad/s = \frac{\pi}{30} \, rad/s$.
Therefore,the correct option is $A$.

(B) The angular velocity $\omega$ is given by the formula $\omega = 2\pi n$,where $n$ is the frequency of rotation in revolutions per second.
Given,the particle rotates $100$ times per minute,so the frequency $n = \frac{100}{60} \, rev/s$.
Substituting the values into the formula:
$\omega = 2 \times \pi \times \frac{100}{60} \, rad/s$
$\omega = \frac{200\pi}{60} \, rad/s = \frac{10\pi}{3} \, rad/s$
$\omega \approx \frac{10 \times 3.14159}{3} \, rad/s \approx 10.47 \, rad/s$.
Thus,the correct option is $B$.

(C) The radius of the wheel is $r = 0.4\ m$.
The time period of one revolution is $T = 1\ s$.
The frequency of revolution is $n = 1/T = 1\ Hz$.
The angular velocity is $\omega = 2\pi n = 2\pi(1) = 2\pi\ rad/s$.
The centripetal acceleration $a_c$ of a point on the rim is given by $a_c = \omega^2 r$.
Substituting the values,we get $a_c = (2\pi)^2 \times 0.4 = 4\pi^2 \times 0.4 = 1.6\pi^2\ m/s^2$.

(D) The kinetic energy $(K.E.)$ of a particle is given by $K.E. = 1/2 mv^2$.
Since the particle moves with a constant speed $v$,the magnitude of its velocity remains unchanged.
Therefore,the kinetic energy at the initial point and the final point is the same.
Change in kinetic energy $\Delta K.E. = K.E._{final} - K.E._{initial} = 1/2 mv^2 - 1/2 mv^2 = 0$.
Thus,the change in kinetic energy is zero.

(A) The relationship between linear speed $v$,angular speed $\omega$,and radius $r$ for an object in circular motion is given by the formula: $v = r\omega$.
Rearranging for angular speed: $\omega = \frac{v}{r}$.
Given values are $v = 100\, m/s$ and $r = 100\, m$.
Substituting these values: $\omega = \frac{100\, m/s}{100\, m} = 1\, rad/s$.
Therefore,the correct option is $A$.

(C) Given: Radius $r = 4 \, m$,Time period $T = 2 \, s$.
Frequency $n = \frac{1}{T} = \frac{1}{2} \, Hz$.
Since the wheel is rotating,the acceleration of a point on the rim is the centripetal acceleration $a_c$.
The formula for centripetal acceleration is $a_c = \omega^2 r$,where $\omega = 2\pi n$.
Substituting the values: $\omega = 2\pi \times \frac{1}{2} = \pi \, rad/s$.
Therefore,$a_c = (\pi)^2 \times 4 = 4\pi^2 \, m/s^2$.

(A) In uniform circular motion,the speed of the particle remains constant,but the direction of velocity changes continuously.
The velocity vector is always directed along the tangent to the circular path at any point.
The acceleration in uniform circular motion is centripetal acceleration,which is always directed towards the center of the circle along the radius.
Since the tangent to a circle is always perpendicular to the radius at the point of contact,the velocity vector and the centripetal acceleration vector are always perpendicular to each other.

(D) Since the car covers equal angles in equal intervals of time,its angular velocity $\omega$ is constant.
Given the relation between linear velocity $v$ and angular velocity $\omega$ is $v = r\omega$,where $r$ is the radius of the circular path.
Since both $r$ and $\omega$ are constant,the magnitude of the linear velocity $v$ remains constant.
However,in circular motion,the direction of the velocity vector is always tangent to the path at any point,which changes continuously.
Therefore,the magnitude of the velocity is constant,but its direction changes.

(B) The relationship between linear speed $v$,angular speed $\omega$,and radius $r$ is given by the formula $v = r\omega$.
Rearranging for angular speed,we get $\omega = \frac{v}{r}$.
Given values are $v = 10 \,m/s$ and $r = 100 \,m$.
Substituting these values into the formula: $\omega = \frac{10}{100} = 0.1 \,rad/s$.
Thus,the angular speed of the scooter is $0.1 \,rad/s$.

(A) For a particle moving in a circular path with constant speed $v$,the acceleration is the centripetal acceleration,given by $a = v^2/r$.
Since the particle completes one circle of circumference $2\pi r$ in time $T$,the speed is $v = (2\pi r)/T$.
From this,we can write $r = (vT)/(2\pi)$.
Substituting the value of $r$ into the acceleration formula: $a = v^2 / ((vT)/(2\pi)) = v^2 \cdot (2\pi) / (vT) = (2\pi v)/T$.
Therefore,the acceleration of the particle is $(2\pi v)/T$.

(C) In uniform circular motion $(U.C.M.)$,the speed of the particle is constant,and the radius of the path is constant.
Since the angular velocity $\omega = v/r$,it remains constant in both magnitude and direction (perpendicular to the plane of motion).
The angular momentum is given by $L = I\omega$,where $I$ is the moment of inertia and $\omega$ is the angular velocity.
For a particle in $U.C.M.$ about the center of the circle,$I = mr^2$ is constant and $\omega$ is constant.
Therefore,both the angular velocity and the angular momentum remain constant.

(B) For a particle moving in a circle with constant angular speed (Uniform Circular Motion),the velocity vector is always tangent to the circular path at any given point.
Since the motion is uniform,the tangential acceleration is zero,and the net acceleration is purely centripetal.
The centripetal acceleration vector always points towards the centre of the circle.
Because the velocity vector is tangent (along the circumference) and the acceleration vector is radial (towards the centre),they are always perpendicular to each other.
Therefore,the statement that the acceleration vector is tangent to the circle is false.

(C) In uniform circular motion,the speed of the particle remains constant.
Tangential acceleration ${a_t}$ is responsible for the change in the magnitude of velocity (speed). Since the speed is constant in uniform circular motion,${a_t} = 0$.
Radial (or centripetal) acceleration ${a_r}$ is responsible for the change in the direction of velocity. Since the particle is moving in a circle,its direction changes continuously,so ${a_r} \neq 0$.
Therefore,the condition for uniform circular motion is ${a_r} \neq 0$ and ${a_t} = 0$.

(B) The formula for centripetal force $F$ is given by $F = m\omega^2 R$,where $m$ is the mass,$\omega$ is the angular velocity,and $R$ is the radius of the orbit.
Since the angular velocity $\omega = \frac{2\pi}{T}$,where $T$ is the time period,we can write the force as $F = m(\frac{2\pi}{T})^2 R = m \frac{4\pi^2}{T^2} R$.
Given that the masses $m$ and the time periods $T$ are the same for both bodies,the expression $\frac{m 4\pi^2}{T^2}$ is constant.
Therefore,the centripetal force is directly proportional to the radius,$F \propto R$.
Thus,the ratio of the centripetal forces is $\frac{F_1}{F_2} = \frac{R_1}{R_2}$.

(B) In uniform circular motion,the magnitude of velocity (speed) remains constant,but the direction of velocity changes continuously at every point along the circular path.
Since momentum is defined as $\vec{p} = m\vec{v}$,and the direction of the velocity vector $\vec{v}$ changes,the momentum vector $\vec{p}$ also changes continuously.
Therefore,momentum is not a constant quantity.
Speed,kinetic energy $(K = \frac{1}{2}mv^2)$,and mass are scalar quantities that remain constant in uniform circular motion.

(C) The length of the string is the radius of the circular path,$r = 1\,m$.
The frequency of revolution $n$ is given by $n = \frac{\text{number of revolutions}}{\text{time}} = \frac{22}{44} = 0.5\,Hz$.
The angular velocity $\omega$ is given by $\omega = 2\pi n = 2\pi(0.5) = \pi\,rad/s$.
The centripetal acceleration $a$ is given by $a = \omega^2 r$.
Substituting the values,$a = (\pi)^2 \times 1 = \pi^2\,m/s^2$.
In uniform circular motion,the acceleration is centripetal,meaning its direction is always along the radius and towards the centre.

(A) The angular velocity $\omega$ is given by the formula $\omega = \frac{2\pi}{T}$,where $T$ is the time period of rotation of the Earth.
For the Earth,the time period of one complete rotation is $24 \text{ hours}$.
Converting this time into seconds: $T = 24 \times 60 \times 60 \text{ s} = 86400 \text{ s}$.
Therefore,the angular velocity is $\omega = \frac{2\pi}{86400} \text{ rad/s}$.
Thus,the correct option is $A$.

(B) The angular velocity $\omega$ is given by $\omega = 2\pi n$,where $n$ is the frequency in revolutions per second.
Initial frequency $n_1 = 600 \text{ rpm} = \frac{600}{60} \text{ rev/s} = 10 \text{ rev/s}$.
Final frequency $n_2 = 1200 \text{ rpm} = \frac{1200}{60} \text{ rev/s} = 20 \text{ rev/s}$.
Initial angular velocity $\omega_1 = 2\pi n_1 = 2\pi(10) = 20\pi \text{ rad/s}$.
Final angular velocity $\omega_2 = 2\pi n_2 = 2\pi(20) = 40\pi \text{ rad/s}$.
The increase in angular velocity is $\Delta\omega = \omega_2 - \omega_1 = 40\pi - 20\pi = 20\pi \text{ rad/s}$.

(A) The velocity at point $A$ is directed along the tangent,which is $\vec{v}_A = v\hat{j}$.
The velocity at point $B$ is directed along the tangent,which is $\vec{v}_B = -v\hat{i}$.
The change in velocity is $\Delta \vec{v} = \vec{v}_B - \vec{v}_A = -v\hat{i} - v\hat{j}$.
The magnitude of the change in velocity is $|\Delta \vec{v}| = \sqrt{(-v)^2 + (-v)^2} = \sqrt{2v^2} = v\sqrt{2}$.
Alternatively,using the formula for change in velocity for an angle $\theta$ between two velocity vectors: $|\Delta \vec{v}| = 2v \sin(\theta/2)$.
Here,the angle $\theta$ between the radii to $A$ and $B$ is $90^\circ$.
$|\Delta \vec{v}| = 2v \sin(90^\circ/2) = 2v \sin(45^\circ) = 2v \times (1/\sqrt{2}) = v\sqrt{2}$.

(B) When a particle moves in a circle with uniform speed,it covers equal distances in equal intervals of time along the circumference.
Since it returns to the same position after a fixed time interval (the time period),the motion is periodic.
However,simple harmonic motion $(SHM)$ requires a restoring force proportional to the displacement from the mean position $(F = -kx)$,which is not the case for uniform circular motion.
Therefore,the motion is periodic but not simple harmonic.

(C) The angular speed $\omega$ is given by the formula $\omega = 2 \pi n$,where $n$ is the frequency in revolutions per second.
Given,$n = 120 \, \text{revolutions/minute} = \frac{120}{60} \, \text{revolutions/second} = 2 \, \text{rev/sec}$.
Therefore,$\omega = 2 \pi \times 2 \, \text{rad/sec} = 4 \pi \, \text{rad/sec}$.

(B) Let the particle $P$ be at position $B$ at a given instant. The speed of the particle is $u$.
The angular velocity of $P$ about the centre $C$ is given by $\omega_C = \frac{u}{r} = \frac{u}{a}$.
The angular velocity of $P$ about point $A$ is given by $\omega_A = \frac{v_{\perp}}{r_{AP}}$,where $v_{\perp}$ is the component of velocity perpendicular to the line $AP$. At point $B$,the velocity $u$ is perpendicular to the diameter $AB$. The distance $AP = 2a$.
Thus,$\omega_A = \frac{u}{2a}$.
Taking the ratio of angular velocities:
$\frac{\omega_A}{\omega_C} = \frac{u/2a}{u/a} = \frac{1}{2}$.
Therefore,the ratio is $1 : 2$.

(C) The time period $T$ of a particle moving in a circular path is related to its angular velocity $\omega$ by the formula $T = \frac{2\pi}{\omega}$.
Given that the time periods of both particles are the same,let $T_1 = T_2 = T$.
Therefore,$\frac{2\pi}{\omega_1} = \frac{2\pi}{\omega_2}$.
This implies that $\omega_1 = \omega_2$.
Thus,the ratio of their angular velocities is $\frac{\omega_1}{\omega_2} = 1$.

(B) The angular velocity $\omega$ is defined as the rate of change of angular displacement $\theta$ with respect to time $t$,given by $\omega = \frac{\theta}{t}$.
For the seconds hand of a watch,it completes one full revolution,which corresponds to an angular displacement of $\theta = 2\pi \ rad$,in a time period of $t = 60 \ s$.
Substituting these values into the formula:
$\omega = \frac{2\pi}{60} \ rad/s = \frac{\pi}{30} \ rad/s$.

(C) The linear displacement $S$ of a point on the rim of a rotating wheel is given by $S = r \theta$,where $r$ is the radius and $\theta$ is the angular displacement.
Since both wheels are fixed coaxially,they rotate together,meaning they have the same angular displacement $\theta$ in the same time interval.
Let $r$ be the radius of the smaller wheel (supporting $A$) and $2r$ be the radius of the larger wheel (supporting $B$).
The distance travelled by $A$ is $x = r \theta$.
The distance travelled by $B$ is $y = (2r) \theta$.
Dividing the two equations,we get $\frac{x}{y} = \frac{r \theta}{2r \theta} = \frac{1}{2}$.
Therefore,$y = 2x$.

(A) In uniform circular motion,the net force acting on the particle is the centripetal force,which is always directed towards the centre of the circle.
The torque $\vec{\tau}$ about any point is given by $\vec{\tau} = \vec{r} \times \vec{F}$.
Since the centripetal force $\vec{F}$ always passes through the centre of the circle,the position vector $\vec{r}$ relative to the centre is collinear with the force vector $\vec{F}$.
Therefore,the torque about the centre is $\vec{\tau} = 0$.
According to the principle of conservation of angular momentum,if the net external torque about a point is zero,the angular momentum about that point remains conserved.
Thus,the angular momentum of the particle is conserved about the centre of the circle.

(B) In uniform circular motion,the speed of the particle is constant,but the direction of velocity changes continuously.
The velocity vector is always tangent to the circular path at any point.
The acceleration in uniform circular motion is centripetal,meaning it is directed towards the center of the circle.
Since the velocity is tangent and the acceleration is radial (towards the center),they are always perpendicular to each other.
Therefore,the statement that the acceleration vector is tangent to the circle is incorrect.

(C) The angular speed $\omega$ is given by $\omega = \frac{2\pi}{T}$,where $T$ is the time period.
For the hour hand,the time period $T_h = 12 \text{ hours}$.
For the minute hand,the time period $T_m = 1 \text{ hour}$.
The ratio of angular speeds is $\frac{\omega_h}{\omega_m} = \frac{2\pi / T_h}{2\pi / T_m} = \frac{T_m}{T_h}$.
Substituting the values,we get $\frac{\omega_h}{\omega_m} = \frac{1 \text{ hour}}{12 \text{ hours}} = 1 : 12$.

(B) Given: Radius $R = 30 \ cm = 0.3 \ m$.
Angular displacement $\theta = 30^\circ = 30 \times \frac{\pi}{180} \ rad = \frac{\pi}{6} \ rad$.
The linear distance $d$ covered by a point on the circumference of the wheel is given by the arc length formula:
$d = R \theta$
Substituting the values:
$d = 0.3 \times \frac{\pi}{6} \ m$
$d = \frac{3}{10} \times \frac{\pi}{6} \ m = \frac{\pi}{20} \ m$.

(C) Given angular velocity $\omega = 3000 \ rpm$ (revolutions per minute).
To convert $rpm$ to revolutions per second $(rps)$, divide by $60$: $\omega = \frac{3000}{60} \ rps = 50 \ rps$.
Since one full revolution corresponds to $2\pi$ radians, the angular velocity in radians per second is $\omega = 50 \times 2\pi \ rad/s = 100\pi \ rad/s$.
The angle turned in $t = 1 \ s$ is given by $\theta = \omega \times t = 100\pi \times 1 = 100\pi \ radians$.

(B) The angular velocity $\omega$ of a particle moving with velocity $v$ at a distance $r$ from a point,where the velocity vector is perpendicular to the position vector,is given by $\omega = \frac{v}{r}$.
When the particle is at point $B$,its velocity $v$ is perpendicular to the diameter $AB$.
For point $C$ (the center),the distance is $r_C = a$. Thus,$\omega_C = \frac{v}{a}$.
For point $A$,the distance is $r_A = AB = 2a$. Thus,$\omega_A = \frac{v}{2a}$.
The ratio of angular velocity with respect to $A$ and $C$ is $\frac{\omega_A}{\omega_C} = \frac{v/2a}{v/a} = \frac{1}{2}$.
Therefore,the ratio is $1 : 2$.

(C) Given: Radius $r = 4 \, m$,Time period $T = 2 \, s$.
The angular velocity is given by $\omega = \frac{2\pi}{T} = \frac{2\pi}{2} = \pi \, rad/s$.
The centripetal acceleration $a_c$ is given by the formula $a_c = \omega^2 r$.
Substituting the values,$a_c = (\pi)^2 \times 4 = 4\pi^2 \, m/s^2$.

(B) At point $P_1$,the velocity vector $\vec{v}_1$ is directed along the positive $y$-axis,so $\vec{v}_1 = v\hat{j}$.
At point $P_2$,the velocity vector $\vec{v}_2$ is directed along the negative $x$-axis,so $\vec{v}_2 = -v\hat{i}$.
The change in velocity $\Delta\vec{v}$ is given by $\Delta\vec{v} = \vec{v}_2 - \vec{v}_1 = -v\hat{i} - v\hat{j}$.
The magnitude of the change in velocity is $|\Delta\vec{v}| = \sqrt{(-v)^2 + (-v)^2} = \sqrt{v^2 + v^2} = \sqrt{2v^2} = \sqrt{2}v$.

(B) Given the equations of motion:
$x = a \sin \omega t \implies \frac{x}{a} = \sin \omega t$ $(i)$
$y = a \cos \omega t \implies \frac{y}{a} = \cos \omega t$ $(ii)$
Squaring and adding equations $(i)$ and $(ii)$,we get:
$\left(\frac{x}{a}\right)^2 + \left(\frac{y}{a}\right)^2 = \sin^2 \omega t + \cos^2 \omega t$
Since $\sin^2 \theta + \cos^2 \theta = 1$,we have:
$\frac{x^2}{a^2} + \frac{y^2}{a^2} = 1$
$x^2 + y^2 = a^2$
This is the standard equation of a circle with radius $a$ centered at the origin. Therefore,the particle follows a circular path.

(B) Given the position vector: $\vec{r} = \cos(\omega t) \hat{i} + \sin(\omega t) \hat{j}$.
$1$. Velocity $\vec{v}$ is the derivative of position with respect to time:
$\vec{v} = \frac{d\vec{r}}{dt} = -\omega \sin(\omega t) \hat{i} + \omega \cos(\omega t) \hat{j}$.
$2$. Acceleration $\vec{a}$ is the derivative of velocity with respect to time:
$\vec{a} = \frac{d\vec{v}}{dt} = -\omega^2 \cos(\omega t) \hat{i} - \omega^2 \sin(\omega t) \hat{j} = -\omega^2 \vec{r}$.
$3$. Since $\vec{a} = -\omega^2 \vec{r}$,the acceleration is directed towards the origin (centripetal acceleration).
$4$. To check the relationship between $\vec{r}$ and $\vec{v}$,calculate the dot product:
$\vec{r} \cdot \vec{v} = (\cos(\omega t))(-\omega \sin(\omega t)) + (\sin(\omega t))(\omega \cos(\omega t)) = -\omega \sin(\omega t) \cos(\omega t) + \omega \sin(\omega t) \cos(\omega t) = 0$.
Since the dot product is $0$,the velocity is perpendicular to the position vector $\vec{r}$.