Angular Variables and Basic of Uniform Circular Motion Questions in English

(D) In uniform circular motion,the magnitude of centripetal acceleration is $a = \frac{v^2}{R}$.
At point $P_1$,the acceleration vector $\vec{a}_1$ is directed towards the center,i.e.,$\vec{a}_1 = \frac{v^2}{R} \hat{i}$.
At point $P_2$,the acceleration vector $\vec{a}_2$ is directed towards the center,i.e.,$\vec{a}_2 = \frac{v^2}{R} \hat{j}$.
The change in acceleration is $\Delta \vec{a} = \vec{a}_2 - \vec{a}_1 = \frac{v^2}{R} \hat{j} - \frac{v^2}{R} \hat{i}$.
The magnitude of the change in acceleration is $|\Delta \vec{a}| = \sqrt{(\frac{v^2}{R})^2 + (-\frac{v^2}{R})^2} = \sqrt{2(\frac{v^2}{R})^2} = \frac{v^2}{R} \sqrt{2}$.

(B) In uniform circular motion,the speed $v$ remains constant,but the direction of velocity changes.
Let the velocity at point $P$ be $\vec{v}_1$ and at point $Q$ be $\vec{v}_2$.
The magnitude of both velocities is $|\vec{v}_1| = |\vec{v}_2| = v$.
The angle between the velocity vectors $\vec{v}_1$ and $\vec{v}_2$ is equal to the angle subtended by the arc $PQ$ at the center,which is $\theta = 40^\circ$.
The change in velocity is given by $\Delta \vec{v} = \vec{v}_2 - \vec{v}_1$.
The magnitude of the change in velocity is $|\Delta \vec{v}| = \sqrt{v^2 + v^2 - 2v^2 \cos \theta} = \sqrt{2v^2(1 - \cos \theta)} = \sqrt{2v^2(2 \sin^2(\theta/2))} = 2v \sin(\theta/2)$.
Substituting $\theta = 40^\circ$,we get $|\Delta \vec{v}| = 2v \sin(40^\circ/2) = 2v \sin 20^\circ$.

(C) In uniform circular motion,the speed of the particle remains constant,but the direction of motion changes continuously at every point along the circular path.
Since velocity is a vector quantity (having both magnitude and direction),a change in direction implies a change in velocity.
Furthermore,the acceleration in uniform circular motion is the centripetal acceleration,which is directed towards the center of the circle.
As the particle moves,the direction of this centripetal acceleration vector also changes continuously to remain directed towards the center.
Therefore,both velocity and acceleration change in direction,even though their magnitudes remain constant.

(B) The angular velocity $\omega$ is given by the formula $\omega = \frac{2\pi}{T}$,where $T$ is the time period.
For the second hand of a clock,the time taken to complete one full revolution is $T = 60 \text{ seconds}$.
Substituting the value of $T$ into the formula:
$\omega = \frac{2\pi}{60} \text{ rad/s} = \frac{\pi}{30} \text{ rad/s}$.
Therefore,the correct option is $B$.

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

(A) Angular velocity $(\omega)$ is defined as the rate of change of angular displacement with respect to time.
It is a vector quantity used to describe the rotational motion of a rigid body about an axis.
Therefore,it is useful when the body is rotating.
Thus,the correct option is $(A)$.

(B) The angular velocity $\omega$ is given as $3000$ revolutions per minute (rpm).
First,convert this to revolutions per second: $\omega = \frac{3000}{60} = 50$ revolutions per second.
Since $1$ revolution corresponds to $2\pi$ radians,the angular velocity in radians per second is $\omega = 50 \times 2\pi = 100\pi \text{ rad/s}$.
The angle $\theta$ rotated in time $t = 1$ second is given by $\theta = \omega t$.
Substituting the values,$\theta = 100\pi \times 1 = 100\pi \text{ radians}$.

(B) For a particle of mass $m$ rotating in a circle of radius $r$ with a uniform angular speed $\omega$,the centripetal force is given by $F = m \omega^{2} r$.
For the particle on the inner part of the ring at radius $R_{1}$,the centripetal force is $F_{1} = m \omega^{2} R_{1}$.
For the particle on the outer part of the ring at radius $R_{2}$,the centripetal force is $F_{2} = m \omega^{2} R_{2}$.
Taking the ratio of these two forces:
$\frac{F_{1}}{F_{2}} = \frac{m \omega^{2} R_{1}}{m \omega^{2} R_{2}} = \frac{R_{1}}{R_{2}}$.
Thus,the ratio of the forces is $\frac{R_{1}}{R_{2}}$.

(A) The relationship between linear speed $(v)$,angular speed $(\omega)$,and the radius of the circular path $(r)$ is given by the formula: $v = r\omega$.
Given that the linear speed $(v)$ is constant,we can rearrange the formula to solve for angular speed: $\omega = \frac{v}{r}$.
Since $v$ is constant,it follows that $\omega \propto \frac{1}{r}$.
Therefore,the angular speed is inversely proportional to the radius of the circle.

(A) The angular momentum $L$ of a particle is defined as $L = r \times p$. The rate of change of angular momentum is equal to the torque applied,given by $\frac{dL}{dt} = \tau = r \times F$.
In uniform circular motion,the centripetal force $F$ always acts towards the center of the circle.
Since the position vector $r$ is measured from the center,the force $F$ passes through the center,making the lever arm zero.
Therefore,the torque $\tau = r \times F = 0$ about the center of the circle.
Since $\tau = 0$,the angular momentum $L$ remains constant (conserved) about the center of the circle.

(B) Let the particle start from point $A(a, 0)$ at $t = 0$. After time $t$,the particle reaches point $B$ such that the angle subtended at the center is $\theta = \omega t$.
The coordinates of the particle at time $t$ are $(a \cos \omega t, a \sin \omega t)$.
The displacement vector $\vec{d}$ is the vector from $A$ to $B$: $\vec{d} = (a \cos \omega t - a) \hat{i} + (a \sin \omega t) \hat{j}$.
The magnitude of displacement $d$ is given by $d = \sqrt{(a \cos \omega t - a)^2 + (a \sin \omega t)^2}$.
$d = \sqrt{a^2(\cos^2 \omega t - 2 \cos \omega t + 1) + a^2 \sin^2 \omega t}$.
Using $\sin^2 \omega t + \cos^2 \omega t = 1$,we get $d = \sqrt{a^2(2 - 2 \cos \omega t)}$.
Using the identity $1 - \cos \theta = 2 \sin^2(\theta/2)$,we have $d = \sqrt{2a^2(1 - \cos \omega t)} = \sqrt{4a^2 \sin^2(\omega t / 2)}$.
Thus,$d = 2a \sin(\omega t / 2)$.

(D) In uniform circular motion,the acceleration is the centripetal acceleration $\vec{a}_c$,which is always directed towards the center of the circle.
For a point $P$ at an angle $\theta$ with the $x$-axis,the position vector makes an angle $\theta$ with the $x$-axis.
The centripetal acceleration vector $\vec{a}_c$ points from $P$ towards the origin $O$.
The magnitude of centripetal acceleration is $a_c = \frac{V^2}{R}$.
Resolving this vector into components:
The $x$-component is $a_x = -a_c \cos\theta = -\frac{V^2}{R} \cos\theta$.
The $y$-component is $a_y = -a_c \sin\theta = -\frac{V^2}{R} \sin\theta$.
Thus,the acceleration vector is $\vec{a} = -\frac{V^2}{R} \cos\theta \hat{i} - \frac{V^2}{R} \sin\theta \hat{j}$.

(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 taken to complete one full revolution.
Since both cars complete one round in the same time $T$,their time periods are equal $(T_1 = T_2 = T)$.
Therefore,the angular speed of the first car is $\omega_1 = \frac{2\pi}{T}$ and the angular speed of the second car is $\omega_2 = \frac{2\pi}{T}$.
The ratio of their angular speeds is $\frac{\omega_1}{\omega_2} = \frac{2\pi / T}{2\pi / T} = 1$.
Thus,the ratio is $1 : 1$.

(C) The radius of the circle is $r = 5 \; cm = 5 \times 10^{-2} \; m$.
The time period is $T = 0.2 \pi \; sec$.
The speed of the particle is $v = \frac{2 \pi r}{T}$.
Substituting the values: $v = \frac{2 \pi \times 5 \times 10^{-2}}{0.2 \pi} = \frac{10 \times 10^{-2}}{0.2} = 0.5 \; m/s$.
The centripetal acceleration is given by $a = \frac{v^2}{r}$.
Substituting the values: $a = \frac{(0.5)^2}{5 \times 10^{-2}} = \frac{0.25}{0.05} = 5 \; m/s^2$.

(B) Velocity is a vector quantity,meaning it has both magnitude (speed) and direction. For velocity to be constant,both magnitude and direction must remain constant.
If a body has a constant speed,its velocity can still vary if its direction changes. $A$ classic example is uniform circular motion,where the speed is constant,but the direction of motion changes at every point,resulting in a changing velocity and the presence of centripetal acceleration.
Option $A$ is incorrect because constant velocity implies constant speed.
Option $B$ is correct because a body moving in a circle at a constant speed has a varying velocity due to the continuous change in direction.
Option $C$ is incorrect because a body with constant speed in a circular path experiences centripetal acceleration.
Option $D$ is incorrect because if a force acts perpendicular to the displacement (like centripetal force),the work done is zero.

(C) In uniform circular motion,the speed of the object remains constant,but the direction of motion changes at every point.
Since linear velocity is a vector quantity (having both magnitude and direction),it changes continuously.
Acceleration in circular motion is centripetal acceleration,which is directed towards the center of the circle. As the direction of the object changes,the direction of this acceleration also changes continuously.
Since force is related to acceleration by $F = ma$,the direction of the force also changes continuously.
However,the angular velocity $\omega$ remains constant because the rate of change of angular displacement is constant in uniform circular motion.

(B) The man is running with a constant speed along a circular path.
In uniform circular motion,the speed of the object remains constant at all points on the path.
The radius of the path is $r = 2 \sqrt 2 \, m$.
The time taken to complete one round (period) is $T = 10 \, s$.
The distance covered in one round is the circumference of the circle,$C = 2 \pi r$.
$C = 2 \pi (2 \sqrt 2) = 4 \sqrt 2 \pi \, m$.
The constant speed $v$ is given by the ratio of total distance to total time:
$v = \frac{C}{T} = \frac{4 \sqrt 2 \pi}{10} = \frac{2 \sqrt 2 \pi}{5} \, m/s$.
Since the speed is constant,the instantaneous speed at any time $t$,including $t = 2.5 \, s$,is the same as the average speed.
Therefore,the instantaneous speed is $\frac{2 \sqrt 2 \pi}{5} \, m/s$.

(D) The centripetal acceleration $a_c$ of an object moving in a circle is given by $a_c = \omega^2 r$,where $\omega$ is the angular velocity and $r$ is the radius of the circle.
Since both cars complete their circles in the same time $t$,their angular velocities are equal,given by $\omega = \frac{2\pi}{t}$.
Let $a_1$ and $a_2$ be the centripetal accelerations of the two cars respectively.
Then,$a_1 = \omega^2 r_1$ and $a_2 = \omega^2 r_2$.
The ratio of their centripetal acceleration is $\frac{a_1}{a_2} = \frac{\omega^2 r_1}{\omega^2 r_2} = \frac{r_1}{r_2}$.
Thus,the ratio is $r_1 : r_2$.

(A) The object moves in a circular path with the center at the origin $(0,0)$.
Given that at $x = -2 \, m$,the velocity is $\vec{v} = -(4 \, m/s) \hat{j}$.
Since the velocity is tangent to the circular path,at $x = -2 \, m$ (which is on the negative $X$-axis),the velocity vector points in the negative $Y$-direction. This implies the object is moving clockwise.
The radius of the circle is $R = 2 \, m$.
The speed of the object is $v = 4 \, m/s$.
For uniform circular motion,the acceleration is centripetal,directed towards the center.
When the object is at $y = 2 \, m$ (which is on the positive $Y$-axis),the center is at $(0,0)$,so the acceleration vector must point towards the origin,i.e.,in the negative $Y$-direction.
The magnitude of centripetal acceleration is $a_c = \frac{v^2}{R} = \frac{(4)^2}{2} = \frac{16}{2} = 8 \, m/s^2$.
Therefore,the acceleration vector at $y = 2 \, m$ is $\vec{a} = -(8 \, m/s^2) \hat{j}$.

(D) The magnitude of the change in velocity $\Delta \vec{v}$ for a particle moving with constant speed $v$ through an angle $\theta$ is given by the formula:
$\Delta v = 2v \sin\left(\frac{\theta}{2}\right)$
Given:
Speed $v = 10\,m/s$
Angle $\theta = 60^{\circ}$
Substituting the values:
$\Delta v = 2 \times 10 \times \sin\left(\frac{60^{\circ}}{2}\right)$
$\Delta v = 20 \times \sin(30^{\circ})$
Since $\sin(30^{\circ}) = 0.5$:
$\Delta v = 20 \times 0.5 = 10\,m/s$
Therefore,the magnitude of the change in velocity is $10\,m/s$.

(C) Let the initial position of the particle be $A$ and the final position be $B$ on a circular path of radius $r$. The angle subtended at the center $O$ is $\theta = 60^{\circ}$.
The displacement is the straight-line distance between $A$ and $B$,denoted by $x$.
In $\triangle OAB$,$OA = OB = r$ and $\angle AOB = 60^{\circ}$.
Since two sides are equal and the included angle is $60^{\circ}$,the triangle is equilateral.
Therefore,the third side $x = AB = r$.

(C) The velocity vector of a particle moving in a circle is always tangential to the path.
Let the initial velocity at point $P$ be $\vec{v}_i$ and the final velocity at point $Q$ be $\vec{v}_f$.
The magnitude of both velocities is $v$, so $|\vec{v}_i| = |\vec{v}_f| = v$.
The angle between the two velocity vectors is equal to the angle subtended by the arc $PQ$ at the center, which is $\theta = 40^o$.
The magnitude of the change in velocity is given by $|\Delta \vec{v}| = |\vec{v}_f - \vec{v}_i|$.
Using the vector subtraction formula, $|\vec{A} - \vec{B}| = \sqrt{A^2 + B^2 - 2AB \cos \theta}$.
Substituting the values: $|\Delta \vec{v}| = \sqrt{v^2 + v^2 - 2v^2 \cos \theta} = \sqrt{2v^2(1 - \cos \theta)}$.
Using the trigonometric identity $1 - \cos \theta = 2 \sin^2(\theta/2)$:
$|\Delta \vec{v}| = \sqrt{2v^2(2 \sin^2(\theta/2))} = \sqrt{4v^2 \sin^2(\theta/2)} = 2v \sin(\theta/2)$.
Given $\theta = 40^o$, the magnitude of change in velocity is $2v \sin(40^o/2) = 2v \sin 20^o$.

(B) Let $R$ be the radius of the circle. After half rotation,the particle moves from point $A$ to point $B$.
Displacement is the straight-line distance between $A$ and $B$,which is the diameter of the circle,$AB = 2R$.
The distance covered along the circular path is $\pi R$.
Since the speed is constant $v$,the time taken $t$ is given by $t = \frac{\text{distance}}{\text{speed}} = \frac{\pi R}{v}$.
Average velocity is defined as the total displacement divided by the total time taken.
$\text{Average velocity} = \frac{\text{Displacement}}{\text{Time taken}} = \frac{2R}{\pi R / v} = \frac{2v}{\pi}$.

(C) The angular velocity is given as $\omega = 60^\circ/s$.
To convert degrees to radians,we use the relation $180^\circ = \pi\,rad$. Thus,$\omega = 60 \times \frac{\pi}{180} = \frac{\pi}{3}\,rad/s$.
The distance from the centre of rotation is $r = 3.5\,m = \frac{7}{2}\,m$.
The linear velocity $v$ is given by the formula $v = r\omega$.
Substituting the values,$v = \frac{7}{2} \times \frac{\pi}{3} = \frac{7\pi}{6}\,m/s$.

(A) Given:
Radius $R = 1\, m$
Number of revolutions $n = 22$
Time $t = 44\, s$
Frequency $f = \frac{n}{t} = \frac{22}{44} = 0.5\, Hz$
Angular velocity $\omega = 2\pi f = 2\pi \times 0.5 = \pi\, rad/s$
The centripetal acceleration is given by $a_r = \omega^2 R$
Substituting the values: $a_r = (\pi)^2 \times 1 = \pi^2\, m/s^2$
Since the speed is constant,the tangential acceleration $a_t = 0$. Thus,the net acceleration is the centripetal acceleration,which is directed towards the centre of the circle.

(C) Given radius $r = 25\, cm = 0.25\, m$.
Frequency $f = 2\, rev/s$.
The angular velocity is given by $\omega = 2\pi f = 2\pi \times 2 = 4\pi\, rad/s$.
The centripetal acceleration $a_c$ is given by the formula $a_c = r\omega^2$.
Substituting the values: $a_c = 0.25 \times (4\pi)^2$.
$a_c = 0.25 \times 16\pi^2$.
$a_c = 4\pi^2\, m/s^2$.

(D) The angular speed $\omega$ of a particle in uniform circular motion is related to its time period $T$ by the formula $\omega = \frac{2\pi}{T}$.
Given that the time periods of particles $A$ and $B$ are the same,i.e.,$T_A = T_B = T$.
Therefore,the angular speed of particle $A$ is $\omega_A = \frac{2\pi}{T_A} = \frac{2\pi}{T}$.
Similarly,the angular speed of particle $B$ is $\omega_B = \frac{2\pi}{T_B} = \frac{2\pi}{T}$.
Taking the ratio of the angular speeds,we get $\frac{\omega_A}{\omega_B} = \frac{2\pi/T}{2\pi/T} = 1$.
Thus,the ratio of the angular speed of $A$ to that of $B$ is $1: 1$.

(A) Given the position vector: $\overrightarrow{r}(t) = \cos \omega t \hat{i} + \sin \omega t \hat{j}$.
To find the velocity $\overrightarrow{v}(t)$,we differentiate $\overrightarrow{r}(t)$ with respect to time $t$:
$\overrightarrow{v}(t) = \frac{d\overrightarrow{r}}{dt} = -\omega \sin \omega t \hat{i} + \omega \cos \omega t \hat{j}$.
To find the acceleration $\overrightarrow{a}(t)$,we differentiate $\overrightarrow{v}(t)$ with respect to time $t$:
$\overrightarrow{a}(t) = \frac{d\overrightarrow{v}}{dt} = -\omega^2 \cos \omega t \hat{i} - \omega^2 \sin \omega t \hat{j} = -\omega^2 (\cos \omega t \hat{i} + \sin \omega t \hat{j}) = -\omega^2 \overrightarrow{r}$.
Now,check the dot product of $\overrightarrow{v}$ and $\overrightarrow{r}$:
$\overrightarrow{v} \cdot \overrightarrow{r} = (-\omega \sin \omega t)(\cos \omega t) + (\omega \cos \omega t)(\sin \omega t) = 0$.
Since the dot product is $0$,$\overrightarrow{v}$ is perpendicular to $\overrightarrow{r}$.
Since $\overrightarrow{a} = -\omega^2 \overrightarrow{r}$,the acceleration vector is in the opposite direction of the position vector $\overrightarrow{r}$,meaning it is directed towards the origin.

(A) Yes,an object can have a constant speed but a changing velocity if it is moving along a curved path.
Velocity is a vector quantity,meaning it depends on both magnitude (speed) and direction.
If the direction of motion changes,the velocity changes even if the speed remains constant.

(B) In uniform circular motion,the speed of the object remains constant,which is why it is called 'uniform'. However,the direction of the object changes at every point along the circular path. Since acceleration is defined as the rate of change of velocity (which is a vector quantity),the change in direction results in a centripetal acceleration. Therefore,an object in uniform circular motion is accelerating despite having a constant speed.

(C) The instantaneous velocity of a particle moving along a curved or angular path is defined as the limit of the average velocity as the time interval approaches zero.
Mathematically,$\vec{v} = \lim_{\Delta t \to 0} \frac{\Delta \vec{r}}{\Delta t} = \frac{d\vec{r}}{dt}$.
Geometrically,the displacement vector $\Delta \vec{r}$ becomes tangent to the path as $\Delta t$ becomes infinitesimally small.
Therefore,the direction of the instantaneous velocity vector is always tangent to the path of the particle at that specific point.

(N/A) Uniform circular motion is the motion of an object traveling at a constant speed along a circular path.
Consider an object moving with a constant speed $v$ in a circle of radius $r$. Since the direction of velocity changes continuously,the object undergoes acceleration.
Let $\vec{r}$ and $\vec{r}'$ be the position vectors,and $\vec{v}$ and $\vec{v}'$ be the velocity vectors of the object at points $P$ and $P'$ respectively.
The velocity vector at any point is tangent to the path at that point.
From the triangle law of vector addition,the change in velocity $\Delta \vec{v} = \vec{v}' - \vec{v}$ is shown in figure $(a_2)$.
Since the path is circular,$\vec{v}$ is perpendicular to $\vec{r}$,and $\vec{v}'$ is perpendicular to $\vec{r}'$. Thus,$\Delta \vec{v}$ is perpendicular to $\Delta \vec{r}$.
Since the average acceleration $\vec{a} = \frac{\Delta \vec{v}}{\Delta t}$,the direction of $\vec{a}$ is the same as $\Delta \vec{v}$.
As $\Delta t \rightarrow 0$,the triangle formed by $\Delta \vec{v}$ becomes similar to the triangle formed by $\Delta \vec{r}$.
From the similarity of triangles,$\frac{|\Delta \vec{v}|}{v} = \frac{|\Delta \vec{r}|}{r}$.
Dividing by $\Delta t$,we get $\frac{|\Delta \vec{v}|}{\Delta t} = \frac{v}{r} \frac{|\Delta \vec{r}|}{\Delta t}$.
As $\Delta t \rightarrow 0$,$|\Delta \vec{r}| \approx v \Delta t$,so $a_c = \lim_{\Delta t \rightarrow 0} \frac{|\Delta \vec{v}|}{\Delta t} = \frac{v}{r} (v) = \frac{v^2}{r}$.
Thus,the acceleration is directed towards the center.

(N/A) In uniform circular motion,an object moves along a circular path of radius $R$ with constant speed $v$.
As the object moves from point $P$ to $P'$ in time interval $\Delta t$,the position vector turns through an angle $\Delta \theta$.
The angular velocity $\omega$ is defined as the rate of change of angular displacement: $\omega = \frac{\Delta \theta}{\Delta t}$.
Since the arc length $\Delta S = R \Delta \theta$,the linear speed is $v = \frac{\Delta S}{\Delta t} = R \frac{\Delta \theta}{\Delta t} = R \omega$.
The centripetal acceleration $a_c$ is given by $a_c = \frac{v^2}{R}$.
Substituting $v = R \omega$,we get $a_c = \frac{(R \omega)^2}{R} = R \omega^2$.
Since angular velocity $\omega = 2 \pi \nu$,where $\nu$ is the frequency,we can also express centripetal acceleration as:
$a_c = R (2 \pi \nu)^2 = 4 \pi^2 \nu^2 R$.

(N/A) Uniform circular motion is the motion of an object traveling at a constant speed along a circular path.
In this motion,the magnitude of the velocity remains constant,but the direction of the velocity changes continuously at every point.
Since acceleration is defined as the rate of change of velocity,and the direction of velocity is changing,the object experiences an acceleration directed towards the center of the circle.
This acceleration is known as centripetal acceleration $(a_c)$.
The formula for centripetal acceleration is given by: $a_c = \frac{v^2}{r} = \omega^2 r$,where $v$ is the linear speed,$r$ is the radius of the circular path,and $\omega$ is the angular velocity.

(N/A) We are given the centripetal acceleration formula as $a_c = \frac{v^2}{R}$.
We know the relationship between linear velocity $v$ and angular velocity $\omega$ is given by $v = R\omega$,where $R$ is the radius of the circular path.
Substitute $v = R\omega$ into the centripetal acceleration formula:
$a_c = \frac{(R\omega)^2}{R}$
$a_c = \frac{R^2\omega^2}{R}$
By simplifying the expression,we get:
$a_c = R\omega^2$.

(A) In uniform circular motion,the centripetal acceleration is given by $a_c = \omega^2 r$,where $\omega$ is the angular velocity and $r$ is the radius of the circular path.
We know that the relationship between angular velocity $\omega$ and frequency $f$ is $\omega = 2\pi f$.
Substituting this value into the centripetal acceleration formula:
$a_c = (2\pi f)^2 r$
$a_c = 4\pi^2 f^2 r$
Thus,the formula for centripetal acceleration in terms of frequency is $a_c = 4\pi^2 f^2 r$.

(A) The angular speed $\omega$ is defined as the ratio of angular displacement $\Delta \theta$ to the time interval $\Delta t$.
For the hour hand of a clock,it completes one full revolution ($2\pi$ radians) in $12$ hours.
Therefore,$\omega = \frac{\Delta \theta}{\Delta t} = \frac{2\pi \text{ rad}}{12 \text{ h}} = \frac{\pi}{6} \text{ rad h}^{-1}$.

(C) In uniform circular motion,the speed of the particle remains constant,but the direction of velocity changes continuously.
The acceleration in uniform circular motion is centripetal acceleration,which is directed towards the center of the circular path.
The velocity vector is always tangent to the circular path at any given point.
Since the radius (position vector) is perpendicular to the tangent at any point on the circle,the centripetal acceleration (directed along the radius) is always perpendicular to the velocity vector.
Therefore,the angle between the velocity vector $\vec{v}$ and the acceleration vector $\vec{a}$ is $90^{\circ}$.

(B) The formula for centripetal acceleration is $a_{c} = \frac{v^{2}}{r}$.
Given that the new speed is $v' = 2v$ and the new radius is $r' = 2r$.
The new centripetal acceleration $a_{c}'$ is given by:
$a_{c}' = \frac{(v')^{2}}{r'} = \frac{(2v)^{2}}{2r} = \frac{4v^{2}}{2r}$.
Simplifying this,we get $a_{c}' = 2 \left( \frac{v^{2}}{r} \right)$.
Therefore,$a_{c}' = 2a_{c}$.

(N/A) The cyclist is moving along a circular path with a constant speed. Therefore,the motion is uniform circular motion.
In uniform circular motion,the acceleration is purely centripetal,directed towards the centre of the circle.
The magnitude of centripetal acceleration is given by $a = \frac{v^2}{r}$.
Given: speed $v = 10\, m/s$ and radius $r = 1\, km = 1000\, m$.
Substituting the values:
$a = \frac{(10\, m/s)^2}{1000\, m} = \frac{100}{1000}\, m/s^2 = 0.1\, m/s^2$.
The direction of this acceleration is always towards the centre of the circular path,which is point $O$. Thus,the acceleration is $0.1\, m/s^2$ directed along $RO$.

(N/A) When a body of mass $m$ moves on a circular path of radius $R$ with a constant speed $v$,it experiences a centripetal or radial acceleration given by $a = \frac{v^2}{R}$. The direction of this acceleration is always toward the center of the circle.
By Newton's second law of motion,the necessary force to provide this acceleration is $F_c = \frac{mv^2}{R}$. The direction of this force is also toward the center of the circle; hence,it is called centripetal force.
In different situations,centripetal force is provided as follows:
$(1)$ For a planet revolving around the Sun,the necessary centripetal force is provided by the gravitational force.
$(2)$ For an electron revolving around the nucleus in an atom,the necessary centripetal force is provided by the Coulombian force (electric force).
$(3)$ For vehicles moving on a level circular track,the necessary centripetal force is provided by the friction force between the tyres and the road.

(N/A) For an object of mass $m$ moving in a circular path of radius $r$ with a constant speed $v$ (or angular velocity $\omega$),the equations are as follows:
$1$. Centripetal Acceleration $(a_c)$: The acceleration directed towards the center of the circular path is given by:
$a_c = \frac{v^2}{r} = r\omega^2$
$2$. Centripetal Force $(F_c)$: According to Newton's second law $(F = ma)$,the force required to maintain uniform circular motion is:
$F_c = m a_c = \frac{mv^2}{r} = mr\omega^2$
Here,$v$ is the linear speed,$r$ is the radius of the circular path,$m$ is the mass of the object,and $\omega$ is the angular velocity.

(B) The centripetal force is always directed towards the center of the circular path.
In uniform circular motion,the centripetal force is given by $F_c = \frac{mv^2}{r}$,where $m$ is the mass,$v$ is the speed,and $r$ is the radius.
The direction of the centripetal force depends only on the position of the object relative to the center of the circle,not on the direction of its velocity (tangential motion).
Therefore,reversing the direction of motion does not change the direction of the centripetal force; it remains directed towards the center.

(A) The formula for centripetal acceleration $a_c$ is given by $a_c = r \omega^2$,where $r$ is the radius and $\omega$ is the angular speed.
Given: $r = 20 \, cm$ and $a_c = 980 \, cm \, s^{-2}$.
Substituting the values into the formula:
$980 = 20 \times \omega^2$
$\omega^2 = \frac{980}{20} = 49$
$\omega = \sqrt{49} = 7 \, rad \, s^{-1}$.

(A) In uniform circular motion,the centripetal force $\vec{F}$ acts towards the center of the circle,while the displacement $\vec{d}$ (or velocity vector $\vec{v}$) is always tangent to the circular path.
Since the force is always perpendicular to the displacement,the angle $\theta$ between them is $90^{\circ}$.
The work done $W$ is given by the formula $W = F d \cos \theta$.
Substituting $\theta = 90^{\circ}$,we get $W = F d \cos 90^{\circ} = F d (0) = 0$.
Therefore,the work done by the centripetal force is $0$.

(N/A) Angular speed is defined as the rate of change of angular displacement with respect to time. It is a scalar quantity. If an object rotates by an angle $\Delta \theta$ in time $\Delta t$,the average angular speed is $\omega_{avg} = \frac{\Delta \theta}{\Delta t}$. The instantaneous angular speed is $\omega = \lim_{\Delta t \to 0} \frac{\Delta \theta}{\Delta t} = \frac{d\theta}{dt}$.
Angular velocity is defined as the rate of change of angular displacement vector with respect to time. It is a vector quantity,denoted by $\vec{\omega}$. Its direction is given by the right-hand rule,perpendicular to the plane of rotation. The instantaneous angular velocity is $\vec{\omega} = \frac{d\vec{\theta}}{dt}$.

(N/A) The linear speed $v$ of a particle moving in a circular path of radius $r$ is related to its angular speed $\omega$ by the formula:
$v = r\omega$
where:
$v$ is the linear speed (in $m/s$),
$r$ is the radius of the circular path (in $m$),
$\omega$ is the angular speed (in $rad/s$).

(N/A) The relation between linear velocity $\vec{v}$ and angular velocity $\vec{\omega}$ for a particle moving in a circular path of radius $\vec{r}$ is given by the vector product: $\vec{v} = \vec{\omega} \times \vec{r}$.
In terms of magnitude,for a particle moving in a circle of radius $r$,the relation is $v = r\omega$,where $v$ is the linear speed,$\omega$ is the angular speed,and $r$ is the radius of the circular path.

(A) In uniform circular motion,a particle moves along a circular path of radius $A$ with a constant angular velocity $\omega$.
The relationship between linear velocity $v$,angular velocity $\omega$,and the radius of the circular path $A$ is given by the formula:
$v = r\omega$
Substituting the given radius $r = A$,we get:
$v = A\omega$
Therefore,the linear velocity of the particle is $A\omega$.