Angular Variables and Basic of Uniform Circular Motion Questions in English

(A) The angular velocity $\omega$ of a particle moving in a circular path is related to its time period $T$ by the formula: $\omega = \frac{2\pi}{T}$.
Since the time period $T$ is the same for both particles $(T_1 = T_2 = T)$,their angular velocities are:
$\omega_1 = \frac{2\pi}{T_1}$ and $\omega_2 = \frac{2\pi}{T_2}$.
Therefore,$\omega_1 = \omega_2$.
The ratio of the angular velocity of the first particle to the second particle is $\frac{\omega_1}{\omega_2} = \frac{1}{1}$.

(D) particle moves along a circular path of radius $r$ with uniform speed $v$.
The angular velocity $\omega$ of the particle is given by the relation $\omega = \frac{v}{r}$.
The angle $\theta$ described by the particle in time $t$ is given by $\theta = \omega t$.
For $t = 1 \ s$,the angle described is $\theta = \omega \times 1 = \frac{v}{r}$ radians.

(B) The angular velocity $\omega$ is defined as the angle described per unit time.
For the minute hand of a clock,it completes one full revolution $(360^{\circ})$ in $60$ minutes.
Since $1$ minute $= 60$ seconds,the time taken for one full revolution is $60 \times 60 = 3600 \ s$.
Therefore,the angular velocity $\omega = \frac{360^{\circ}}{3600 \ s} = 0.1^{\circ}/s$.

(B) The body moves from one end of the diameter to the other end along the circular path. This corresponds to covering an angular displacement of $\Delta \theta = \pi \ rad$ (half of the circle).
Given time taken,$\Delta t = 3 \ s$.
The angular speed $\omega$ is defined as the rate of change of angular displacement:
$\omega = \frac{\Delta \theta}{\Delta t} = \frac{\pi \ rad}{3 \ s} = \frac{\pi}{3} \ rad/s$.

(C) The kinetic energy of a particle is given by the formula $K = \frac{1}{2} mv^2$.
Since the particle is moving in a circle with uniform speed,the magnitude of the velocity $(v)$ remains constant.
Consequently,the kinetic energy $(K)$ remains constant throughout the motion.
Velocity and displacement are vectors that change direction continuously in circular motion,and acceleration (centripetal) changes direction as well.

(A) Given,diameter,$d = 80 \ m$.
Therefore,radius,$r = d/2 = 40 \ m$.
Distance travelled after completion of $3/4$ revolution is given by:
Distance $= (3/4) \times (2 \pi r) = (3/2) \times \pi \times 40 = 60 \pi \ m$.
Displacement is the shortest distance between the initial point $A$ and the final point $B$. Since the athlete covers $3/4$ of the circle,the angle between the initial and final position vectors is $90^{\circ}$.
Using the Pythagorean theorem for the right-angled triangle formed by the two radii:
Displacement $= \sqrt{r^2 + r^2} = r \sqrt{2} = 40 \sqrt{2} \ m$.

(B) motion is said to be periodic if it repeats its path at regular intervals of time. In uniform circular motion,the particle returns to the same position after every time interval $T = \frac{2\pi r}{v}$,so it is periodic.
Simple Harmonic Motion $(SHM)$ is a specific type of oscillatory motion where the restoring force is directly proportional to the displacement from the mean position $(F = -kx)$.
In uniform circular motion,the acceleration is always directed towards the center (centripetal acceleration),which does not satisfy the condition for $SHM$ along a straight line.
Therefore,uniform circular motion is periodic but not $SHM$.

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

(B) The given situation is shown in the figure. When the body covers a distance of $C / 4$,where $C$ is the circumference,it moves from point $A$ to point $B$ along the circular path.
Since the distance covered is one-fourth of the circumference,the angle subtended at the center $O$ is $90^{\circ}$.
Thus,$\triangle OAB$ is a right-angled triangle with $OA = OB = r$.
The displacement is the straight-line distance between the initial position $A$ and the final position $B$.
Using the Pythagorean theorem:
$\text{Displacement} = AB = \sqrt{OA^2 + OB^2}$
$\text{Displacement} = \sqrt{r^2 + r^2} = \sqrt{2r^2} = r \sqrt{2}$

(D) Average angular velocity is defined as the total angular displacement divided by the total time taken.
Total angular displacement for one full revolution is $\theta_{\text{total}} = 2\pi \ rad$.
Total time taken is $t_{\text{total}} = 4 \ s + 2 \ s = 6 \ s$.
Therefore,the average angular velocity $\omega_{\text{avg}} = \frac{\theta_{\text{total}}}{t_{\text{total}}} = \frac{2\pi}{6} = \frac{\pi}{3} \ rad/s$.

(C) The speed of the body is $v = 20 \,m/s$.
In a uniform circular motion, the velocity vector is always tangent to the circle.
Let the initial velocity be $\vec{v}_1 = v \hat{j}$ (directed towards North).
After half a revolution, the body moves in the opposite direction, so the final velocity is $\vec{v}_2 = -v \hat{j}$ (directed towards South).
The change in velocity $\Delta \vec{v}$ is given by:
$\Delta \vec{v} = \vec{v}_2 - \vec{v}_1$
$\Delta \vec{v} = (-v \hat{j}) - (v \hat{j}) = -2v \hat{j}$
The magnitude of the change in velocity is:
$|\Delta \vec{v}| = |-2v| = 2v$
Substituting the value of $v = 20 \,m/s$:
$|\Delta \vec{v}| = 2 \times 20 = 40 \,m/s$.

(A) Let the velocity at point $A$ be $\vec{v}_1$ and at point $B$ be $\vec{v}_2$. Since the speed is constant,$|\vec{v}_1| = |\vec{v}_2| = v$.
The change in velocity is given by $\Delta \vec{v} = \vec{v}_2 - \vec{v}_1$.
The magnitude of the change in velocity for a particle moving in a circle with constant speed $v$ through an angle $\theta$ is given by the vector subtraction formula:
$|\Delta \vec{v}| = \sqrt{v^2 + v^2 - 2v^2 \cos \theta}$
$|\Delta \vec{v}| = \sqrt{2v^2(1 - \cos \theta)}$
Using the trigonometric identity $1 - \cos \theta = 2 \sin^2 \frac{\theta}{2}$,we get:
$|\Delta \vec{v}| = \sqrt{2v^2 \cdot 2 \sin^2 \frac{\theta}{2}}$
$|\Delta \vec{v}| = \sqrt{4v^2 \sin^2 \frac{\theta}{2}}$
$|\Delta \vec{v}| = 2v \sin \frac{\theta}{2}$

(A) 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 $A$ and $B$ complete their circular paths in the same time,their time periods are equal,i.e.,$T_A = T_B$.
Therefore,the ratio of their angular speeds is: $\frac{\omega_A}{\omega_B} = \frac{2\pi / T_A}{2\pi / T_B} = \frac{T_B}{T_A}$.
Substituting $T_A = T_B$,we get: $\frac{\omega_A}{\omega_B} = \frac{T_A}{T_A} = 1$.
Thus,the ratio of the angular speed of $A$ to $B$ is $1: 1$.

(B) In uniform circular motion,the speed of the particle remains constant,and the particle moves along a circular path.
$1$. Velocity: The velocity vector is always tangent to the circular path at any point. This component is known as the transverse velocity $(v_t = r\omega)$. There is no radial component of velocity $(v_r = 0)$ because the distance from the center remains constant.
$2$. Acceleration: Since the direction of the velocity changes continuously,there is an acceleration directed towards the center of the circle,known as centripetal or radial acceleration $(a_r = v^2/r = r\omega^2)$. There is no tangential (transverse) acceleration $(a_t = 0)$ because the speed is constant.
Therefore,the particle has transverse velocity and radial acceleration.

(C) $(i)$ Assertion is correct: In uniform circular motion,a body moves with constant speed. It is the magnitude of velocity that remains constant.
$(ii)$ Reason is incorrect: In uniform circular motion,the acceleration is centripetal acceleration,which is directed towards the center of the circular path. Since the direction of the particle changes at every point,the direction of the centripetal acceleration also changes continuously. Because acceleration is a vector quantity,a change in direction implies that the acceleration is not constant.

(B) The angular velocity $\omega$ is given by $\omega = \frac{2\pi}{T}$,where $T$ is the time period.
For the hour hand of a watch,the time period $T_h = 12 \text{ hours}$.
Thus,$\omega_h = \frac{2\pi}{12} \text{ rad/hour}$.
For the rotation of the Earth,the time period $T_e = 24 \text{ hours}$.
Thus,$\omega_e = \frac{2\pi}{24} \text{ rad/hour}$.
The ratio of the angular velocity of the hour hand to the angular velocity of the Earth is $\frac{\omega_h}{\omega_e} = \frac{2\pi/12}{2\pi/24} = \frac{24}{12} = 2:1$.

(B) The power delivered by a force is given by $P = \vec{F} \cdot \vec{v}$.
Since the force $\vec{F}$ is proportional to acceleration $\vec{a}$ (by Newton's second law,$\vec{F} = m\vec{a}$),the condition that velocity $\vec{v}$ and acceleration $\vec{a}$ are perpendicular implies that $\vec{F} \cdot \vec{v} = 0$.
Therefore,the power delivered to the particle is zero $(P = 0)$.
Since power is the rate of change of kinetic energy $(P = \frac{dK}{dt})$,$P = 0$ implies that the kinetic energy $K$ is constant.
Thus,the kinetic energy of the particle remains constant throughout the motion.

(B) The total time given is $2 \,min \,20 \,s$.
Converting this into seconds: $2 \times 60 \,s + 20 \,s = 120 \,s + 20 \,s = 140 \,s$.
The athlete completes one round in $40 \,s$.
The number of rounds completed in $140 \,s$ is $\frac{140}{40} = 3.5$ rounds.
After $3$ complete rounds, the athlete returns to the starting point, so the displacement for these $3$ rounds is $0$.
In the remaining $0.5$ round, the athlete moves from the starting point $A$ to the diametrically opposite point $B$.
The displacement is the shortest distance between the initial and final positions, which is the diameter of the circular track.
Therefore, the displacement $= 2R$.

(C) For a body moving in a circular path with a constant speed $v$, the centripetal acceleration $a$ is given by the formula:
$a = \frac{v^2}{r}$
Given that the speed $v = 10 \,ms^{-1}$, we substitute this value into the equation:
$a = \frac{(10)^2}{r} = \frac{100}{r}$
This shows that the acceleration $a$ is inversely proportional to the radius $r$ $(a \propto \frac{1}{r})$.
Therefore, the graph representing this relationship is a rectangular hyperbola, which corresponds to the graph shown in option $C$.

(B) The angular displacement for one complete rotation is $2\pi \text{ radians}$.
Since the merry-go-round completes $9$ rotations,the total angular displacement $\Delta\theta = 9 \times 2\pi = 18\pi \text{ radians}$.
The time taken $\Delta t = 18 \text{ seconds}$.
The angular speed $\omega$ is given by the formula $\omega = \frac{\Delta\theta}{\Delta t}$.
Substituting the values,$\omega = \frac{18\pi}{18} = \pi \text{ rad/s}$.

(A) Radius of the circle,$r = 5 \,cm$.
Time period,$T = 5 \,s$.
Angular velocity is given by $\omega = \frac{2 \pi}{T} = \frac{2 \pi}{5} \,rad/s$.
Linear speed is $v = \omega r = (\frac{2 \pi}{5}) \times 5 = 2 \pi \,cm/s$.
Since the particle is in uniform circular motion,the acceleration is centripetal acceleration,given by $a = \frac{v^2}{r}$.
Substituting the values,$a = \frac{(2 \pi)^2}{5} = \frac{4 \pi^2}{5} = 0.8 \pi^2 \,cm/s^2$.

(D) In uniform circular motion,the following properties hold true:
$(a)$ Since the centripetal force is always perpendicular to the displacement,the work done during one complete cycle is zero.
$(b)$ The centripetal force is a radial force that acts towards the centre of the circle.
$(c)$ The angular velocity $\omega$ remains constant in magnitude and direction.
$(d)$ While the speed (magnitude of velocity) is constant,the tangential velocity is a vector quantity. Since the direction of motion changes continuously at every point on the circular path,the tangential velocity is not constant.
Therefore,the statement that tangential velocity is constant is wrong.

(A) In uniform circular motion,the particle moves along a circular path with a constant speed.
At any point $P$ on the circle,the velocity vector $\vec{v}$ is directed along the tangent to the circle at that point. This direction is perpendicular to the radius,which is referred to as the transverse direction.
The acceleration in uniform circular motion is the centripetal acceleration $\vec{a}_c$,which is always directed towards the center $O$ of the circle. This direction is along the radius,which is referred to as the radial direction.
Therefore,the velocity is transverse and the acceleration is radial.

(C) In uniform circular motion,the speed of the particle is constant. Therefore,the acceleration of the particle is purely centripetal,meaning it is directed towards the center of the circular path.
Since the centripetal acceleration is always directed towards the center,it is always parallel to the position vector $r$ (measured from the center) and perpendicular to the velocity vector $v$ (which is tangential to the path).
$1$. Because the acceleration $a$ is perpendicular to the velocity $v$,their dot product must be zero: $v \cdot a = 0$.
$2$. Because the acceleration $a$ is directed towards the center,it is collinear with the position vector $r$. Two vectors that are collinear (parallel or anti-parallel) have a cross product equal to zero: $r \times a = 0$.
Thus,the correct statement is $v \cdot a = 0$ and $r \times a = 0$.

(C) Given: Radius $R = 5 \ m$,tangential velocity $v = 2 \ ms^{-1}$.
For uniform circular motion,the angular velocity $\omega$ is given by $\omega = \frac{v}{R} = \frac{2}{5} = 0.4 \ rad \ s^{-1}$.
The time period $T$ for one revolution is $T = \frac{2 \pi}{\omega} = \frac{2 \pi}{0.4} = 5 \pi \ s$.
The time taken to complete $2$ revolutions is $t = 2 \times T = 2 \times 5 \pi = 10 \pi \ s$.
The magnitude of centripetal acceleration $a_c$ is $a_c = \frac{v^2}{R} = \frac{2^2}{5} = \frac{4}{5} = 0.8 \ ms^{-2}$.
Thus,the time taken is $10 \pi \ s$ and the acceleration is $0.8 \ ms^{-2}$.