Kinematics Circular Motion (Uniform Angular Accelaration) Questions in English

(B) Given: Initial angular velocity $\omega_{0} = 20\pi \, rad/s$,final angular velocity $\omega = 40\pi \, rad/s$,and time $t = 10 \, s$.
Using the equation of motion $\omega = \omega_{0} + \alpha t$,we find the angular acceleration $\alpha$:
$40\pi = 20\pi + \alpha(10) \Rightarrow 20\pi = 10\alpha \Rightarrow \alpha = 2\pi \, rad/s^{2}$.
Now,calculate the total angular displacement $\theta$ using $\theta = \omega_{0}t + \frac{1}{2}\alpha t^{2}$:
$\theta = (20\pi)(10) + \frac{1}{2}(2\pi)(10)^{2} = 200\pi + 100\pi = 300\pi \, rad$.
Since one rotation is equal to $2\pi \, rad$,the number of rotations $n$ is given by $n = \frac{\theta}{2\pi}$:
$n = \frac{300\pi}{2\pi} = 150$ rotations.

(A) The speed of the particle is given by the equation $v = 2t + 2$.
At time $t = 1 \ s$,the speed $v$ is $v = 2(1) + 2 = 4 \ m/s$.
The centripetal acceleration $a_{cp}$ is given by the formula $a_{cp} = \frac{v^2}{r}$.
Given the radius $r = 2 \ m$,we substitute the values: $a_{cp} = \frac{4^2}{2} = \frac{16}{2} = 8 \ m/s^2$.

(C) For an object starting from rest with uniform angular acceleration $\alpha$,the angular displacement $\theta$ in time $t$ is given by $\theta = \frac{1}{2}\alpha t^2$.
Since the number of rotations $n$ is proportional to the angular displacement,$n \propto t^2$.
Let $n_1$ be the rotations in the first interval $t_1 = 3\,s$,and $n_2$ be the rotations in the next interval $t_2 = 3\,s$.
The total rotations in time $2t$ is $n_{total} = n_1 + n_2$.
Using the property of uniform acceleration starting from rest,the ratio of rotations in successive equal time intervals is $1 : 3 : 5 : \dots$.
Therefore,$n_1 : n_2 = 1 : 3$.
Given $n_1 = 10$,we have $10 : n_2 = 1 : 3$.
Thus,$n_2 = 10 \times 3 = 30$ rotations.

(A) Given: Initial angular velocity $\omega_{0} = 0 \, rad \, s^{-1}$,final angular velocity $\omega = 15 \, rad \, s^{-1}$,time $t = 0.270 \, s$,and radius $r = 0.810 \, m$.
First,we calculate the angular acceleration $\alpha$ using the formula $\omega = \omega_{0} + \alpha t$.
Since $\omega_{0} = 0$,we have $\alpha = \frac{\omega}{t} = \frac{15}{0.270} \, rad \, s^{-2}$.
The tangential acceleration $a_{t}$ is given by $a_{t} = r \alpha$.
Substituting the values: $a_{t} = 0.810 \times \frac{15}{0.270} = 0.810 \times 55.55 = 45 \, m \, s^{-2}$.
Thus,the acceleration of the discus is $45 \, m \, s^{-2}$.

(C) The seconds hand of a clock completes one full rotation in $T = 60 \, s$. The distance traveled by the tip in one rotation is the circumference of the circle,$2 \pi \ell$,where $\ell$ is the length of the seconds hand.
Linear speed $v$ is given by $v = \frac{2 \pi \ell}{T}$.
Given $v = 1.05 \, cm \, s^{-1}$ and $T = 60 \, s$,we have:
$1.05 = \frac{2 \times 3.14159 \times \ell}{60}$.
Rearranging for $\ell$:
$\ell = \frac{1.05 \times 60}{2 \times 3.14159} \approx \frac{63}{6.283} \approx 10.027 \, cm$.
Thus,the length of the seconds hand is nearly $10 \, cm$.

(C) Given: Initial angular velocity $\omega_i = 0 \ rad/s$,Final angular velocity $\omega_f = 20 \ rad/s$,Time $t = 5 \ s$.
The angular displacement $\theta$ is given by the formula: $\theta = \left( \frac{\omega_i + \omega_f}{2} \right) \times t$.
Substituting the values: $\theta = \left( \frac{0 + 20}{2} \right) \times 5 = 10 \times 5 = 50 \ rad$.
The number of revolutions $n$ is given by $n = \frac{\theta}{2\pi}$.
Therefore,$n = \frac{50}{2\pi} = \frac{25}{\pi} \text{ revolutions}$.

(C) The angular displacement is given by $\theta = 2t^3 + 0.5$.
The angular velocity $\omega$ is defined as the rate of change of angular displacement with respect to time:
$\omega = \frac{d\theta}{dt} = \frac{d}{dt}(2t^3 + 0.5)$.
Applying the power rule for differentiation:
$\omega = 6t^2$.
To find the angular velocity at $t = 2 \, s$,substitute the value of $t$ into the expression:
$\omega = 6(2)^2 = 6 \times 4 = 24 \, rad/s$.

(D) Given: Angular acceleration $\alpha = \pi \, rad/s^2$,initial angular velocity $\omega_0 = 0$,and time $t = 10 \, s$.
Using the kinematic equation for angular displacement: $\theta = \omega_0 t + \frac{1}{2} \alpha t^2$.
Substituting the values: $\theta = 0 \cdot (10) + \frac{1}{2} \cdot \pi \cdot (10)^2$.
$\theta = \frac{1}{2} \cdot \pi \cdot 100 = 50\pi \, rad$.
Since one revolution corresponds to $2\pi \, rad$,the number of revolutions $n$ is given by $n = \frac{\theta}{2\pi}$.
$n = \frac{50\pi}{2\pi} = 25$.
Therefore,the particle makes $25$ revolutions.

(C) The angular displacement $\theta$ for a body starting from rest with uniform angular acceleration $\alpha$ is given by $\theta = \frac{1}{2} \alpha t^2$.
For the first $t_1 = 2 \ s$,the angle rotated is $\theta = \frac{1}{2} \alpha (2)^2 = 2 \alpha$.
This implies $\alpha = \frac{\theta}{2}$.
For the total time $t_2 = 2 \ s + 2 \ s = 4 \ s$,the total angle rotated is $\theta_{total} = \frac{1}{2} \alpha (4)^2 = 8 \alpha$.
The angle rotated in the next $2 \ s$ is $\Delta \theta = \theta_{total} - \theta = 8 \alpha - 2 \alpha = 6 \alpha$.
Substituting $\alpha = \frac{\theta}{2}$ into the expression,we get $\Delta \theta = 6 \left( \frac{\theta}{2} \right) = 3 \theta$.

(C) First,convert the initial and final velocities from $km/h$ to $m/s$:
$u = 18 \times \frac{5}{18} = 5\,m/s$
$v = 36 \times \frac{5}{18} = 10\,m/s$
Calculate the linear acceleration $a$ using the formula $a = \frac{v-u}{t}$:
$a = \frac{10 - 5}{5} = \frac{5}{5} = 1\,m/s^2$
The radius $r$ of the wheel is half the diameter:
$r = \frac{0.8}{2} = 0.4\,m$
The angular acceleration $\alpha$ is related to linear acceleration $a$ by the formula $\alpha = \frac{a}{r}$:
$\alpha = \frac{1}{0.4} = 2.5\,rad/s^2$

(B) The centripetal acceleration is given by $a_c = k^2rt^2$.
Since $a_c = \frac{v^2}{r}$,we have $\frac{v^2}{r} = k^2rt^2$.
Solving for $v$,we get $v^2 = k^2r^2t^2$,which implies $v = krt$.
The tangential acceleration $a_t$ is the rate of change of speed: $a_t = \frac{dv}{dt} = \frac{d}{dt}(krt) = kr$.
The tangential force $F_t$ responsible for the change in speed is $F_t = m a_t = mkr$.
The power $P$ delivered to the particle is given by $P = F_t \cdot v$.
Substituting the values,$P = (mkr) \cdot (krt) = mk^2r^2t$.

(B) Let the angular positions of $A, B$ and $C$ at $t=0$ be $\theta_A(0) = 0$,$\theta_B(0) = \pi/2$,and $\theta_C(0) = \pi$ respectively.
The angular velocities are $\omega_A = v_A/r = \pi \, rad/s$,$\omega_B = v_B/r = 2.5\pi \, rad/s$,and $\omega_C = v_C/r = 2\pi \, rad/s$.
At time $t$,the angular positions are $\theta_A(t) = \pi t$,$\theta_B(t) = \pi/2 + 2.5\pi t$,and $\theta_C(t) = \pi + 2\pi t$.
For the particles to meet,their angular positions must be equal modulo $2\pi$. Let $\theta_A(t) = \theta_B(t) + 2n\pi$ and $\theta_B(t) = \theta_C(t) + 2m\pi$.
$\pi t = \pi/2 + 2.5\pi t + 2n\pi \implies -1.5t = 0.5 + 2n \implies t = -(2n + 0.5)/1.5$.
$\pi/2 + 2.5\pi t = \pi + 2\pi t + 2m\pi \implies 0.5t = 0.5 + 2m \implies t = 1 + 4m$.
Equating $t$: $1 + 4m = -(2n + 0.5)/1.5 \implies 1.5 + 6m = -2n - 0.5 \implies 2n + 6m = -2 \implies n + 3m = -1$.
For the first meeting,let $m = -1$,then $n = 2$. Thus $t = 1 + 4(-1) = -3$ (taking magnitude $t=3s$).
Distance $s_B = v_B t = 2.5 \times 3 = 7.5 \, m$.
Distance $s_C = v_C t = 2 \times 3 = 6 \, m$.
Ratio $s_B : s_C = 7.5 : 6 = 5 : 4$.

(A) Given: Radius $r = \frac{20}{\pi} \, m$,initial angular velocity $\omega_i = 0$,and total angular displacement $\theta = 2 \times (2 \pi) = 4 \pi \, rad$ (since it completes two revolutions).
Final linear velocity $v = 80 \, m/s$.
The final angular velocity is $\omega_f = \frac{v}{r} = \frac{80}{20/\pi} = 4 \pi \, rad/s$.
Using the rotational kinematic equation $\omega_f^2 = \omega_i^2 + 2 \alpha \theta$,we get:
$(4 \pi)^2 = 0^2 + 2 \alpha (4 \pi)$
$16 \pi^2 = 8 \pi \alpha$
$\alpha = 2 \pi \, rad/s^2$.
The tangential acceleration is $a_t = r \alpha = \left( \frac{20}{\pi} \right) \times (2 \pi) = 40 \, m/s^2$.

(D) The angular displacement is given by $\theta = 2t^3 + 0.5$.
Angular velocity $\omega$ is defined as the rate of change of angular displacement with respect to time: $\omega = \frac{d\theta}{dt}$.
Differentiating $\theta$ with respect to $t$: $\omega = \frac{d}{dt}(2t^3 + 0.5) = 6t^2$.
To find the angular velocity after $t = 2$ seconds,substitute $t = 2$ into the expression for $\omega$:
$\omega = 6(2)^2 = 6 \times 4 = 24 \ rad/sec$.

(B) Given:
Initial angular speed,$\omega_0 = 2.00\, rad/s$
Angular acceleration,$\alpha = 3.0\, rad/s^2$
Time,$t = 2\, s$
Using the kinematic equation for angular displacement:
$\theta = \omega_0 t + \frac{1}{2} \alpha t^2$
Substituting the given values:
$\theta = (2.00)(2) + \frac{1}{2}(3.0)(2)^2$
$\theta = 4 + \frac{1}{2}(3.0)(4)$
$\theta = 4 + 6 = 10\, radians$
Thus,the wheel has rotated through an angle of $10\, radians$.

(C) The particle starts from rest,so initial angular velocity $\omega_{0} = 0$.
After $n$ rounds,the total angular displacement is $\theta = 2 \pi n$.
The final linear velocity is $V_{0}$,so the final angular velocity is $\omega = \frac{V_{0}}{r}$.
Using the kinematic equation $\omega^{2} = \omega_{0}^{2} + 2 \alpha \theta$,we get:
$\alpha = \frac{\omega^{2} - \omega_{0}^{2}}{2 \theta} = \frac{(V_{0}/r)^{2} - 0}{2(2 \pi n)} = \frac{V_{0}^{2}/r^{2}}{4 \pi n} = \frac{V_{0}^{2}}{4 \pi n r^{2}} \; rad/s^{2}$.

(N/A) The angular acceleration $\alpha$ is uniform,hence the rate of change of angular velocity $\omega$ with respect to time $t$ is constant:
$\frac{d\omega}{dt} = \alpha$
Rearranging the terms for integration:
$d\omega = \alpha dt$
Integrating both sides:
$\int d\omega = \int \alpha dt$
$\omega = \alpha t + c$,where $c$ is the constant of integration.
Applying the initial condition: at $t = 0$,$\omega = \omega_{0}$.
Substituting these values into the equation:
$\omega_{0} = \alpha(0) + c \implies c = \omega_{0}$
Substituting the value of $c$ back into the integrated equation:
$\omega = \omega_{0} + \alpha t$.

(N/A) $(i)$ We use the equation $\omega = \omega_{0} + \alpha t$.
Initial angular speed $\omega_{0} = \frac{2 \pi \times 1200}{60} \; rad/s = 40 \pi \; rad/s$.
Final angular speed $\omega = \frac{2 \pi \times 3120}{60} \; rad/s = 104 \pi \; rad/s$.
Angular acceleration $\alpha = \frac{\omega - \omega_{0}}{t} = \frac{104 \pi - 40 \pi}{16} = \frac{64 \pi}{16} = 4 \pi \; rad/s^{2}$.
$(ii)$ The angular displacement $\theta$ is given by $\theta = \omega_{0} t + \frac{1}{2} \alpha t^{2}$.
$\theta = (40 \pi \times 16) + \frac{1}{2} \times (4 \pi) \times (16)^{2} = 640 \pi + 512 \pi = 1152 \pi \; rad$.
Number of revolutions $n = \frac{\theta}{2 \pi} = \frac{1152 \pi}{2 \pi} = 576$.

(A) The tangential component of the linear acceleration of a particle in rotational motion is given by $a_{T} = r\alpha$,where $r$ is the radius and $\alpha$ is the angular acceleration.
Since the rigid body is rotating with a constant angular velocity,the angular acceleration $\alpha = \frac{d\omega}{dt} = 0$.
Therefore,the tangential component of the linear acceleration $a_{T} = r \times 0 = 0$.

(B) The tangential acceleration $a_t$ is defined as the rate of change of the magnitude of velocity,given by $a_t = \frac{dv}{dt} = r \alpha$,where $r$ is the radius and $\alpha$ is the angular acceleration.
If the angular speed $\omega$ is constant,then the angular acceleration $\alpha = \frac{d\omega}{dt} = 0$.
Therefore,the tangential acceleration $a_t$ is zero only when the particle undergoes uniform circular motion,meaning the angular speed remains constant.

(A) The length of the second's hand is the radius of the circular path,$R = 0.1\, m$.
The time period for the second's hand is $T = 60\, s$.
The angular velocity $\omega$ is given by $\omega = \frac{2\pi}{T} = \frac{2\pi}{60} \approx 0.105\, rad/s$.
The centripetal acceleration $a$ of the tip of the hand is given by $a = \omega^2 R$.
Substituting the values: $a = (0.105)^2 \times 0.1$.
$a \approx 0.011025 \times 0.1 = 0.0011025\, m/s^2$.
This can be written as $1.1 \times 10^{-3}\, m/s^2$.
Therefore,the average acceleration is of the order of $10^{-3}$.

(B) Initial angular speed $\omega_0 = 360 \; rpm = 360 \times \frac{2\pi}{60} \; rad/s = 12\pi \; rad/s$.
Final angular speed $\omega = 1200 \; rpm = 1200 \times \frac{2\pi}{60} \; rad/s = 40\pi \; rad/s$.
Time taken $t = 14 \; s$.
Using the equation of rotational motion $\omega = \omega_0 + \alpha t$,where $\alpha$ is the angular acceleration:
$40\pi = 12\pi + \alpha(14)$.
$28\pi = 14\alpha$.
$\alpha = \frac{28\pi}{14} = 2\pi \; rad/s^2$.

(B) Given:
Initial angular speed,$\omega_1 = 900 \, rpm = \frac{900}{60} \, rev/s = 15 \, rev/s$.
Final angular speed,$\omega_2 = 2460 \, rpm = \frac{2460}{60} \, rev/s = 41 \, rev/s$.
Time interval,$t = 26 \, s$.
Since the angular acceleration is uniform,the average angular speed is $\omega_{avg} = \frac{\omega_1 + \omega_2}{2}$.
$\omega_{avg} = \frac{15 + 41}{2} = \frac{56}{2} = 28 \, rev/s$.
The total number of revolutions $N$ is given by the product of average angular speed and time:
$N = \omega_{avg} \times t = 28 \, rev/s \times 26 \, s = 728 \, revolutions$.

(D) Initial angular speed,$\omega_0 = 600 \, rpm = 10 \, rev/s$.
Final angular speed,$\omega_f = 1800 \, rpm = 30 \, rev/s$.
Time taken,$t = 10 \, s$.
Since the acceleration is uniform,the average angular speed is $\omega_{avg} = \frac{\omega_0 + \omega_f}{2} = \frac{10 + 30}{2} = 20 \, rev/s$.
The total number of rotations (revolutions) made is $\theta = \omega_{avg} \times t$.
$\theta = 20 \, rev/s \times 10 \, s = 200 \, rev$.

(A) The formula for angular velocity under uniform angular acceleration is $\omega = \omega_{0} + \alpha t$.
Here, the initial angular speed $\omega_{0} = 1200\,rpm$ and the final angular speed $\omega = 3120\,rpm$.
The time interval is $t = 16\,s$.
First, convert the angular speeds from $rpm$ to $rad/s$ using the relation $1\,rpm = \frac{2\pi}{60}\,rad/s$.
The change in angular speed is $\Delta\omega = 3120 - 1200 = 1920\,rpm$.
Converting this to $rad/s$: $\Delta\omega = 1920 \times \frac{2\pi}{60} = 32 \times 2\pi = 64\pi\,rad/s$.
The angular acceleration $\alpha$ is given by $\alpha = \frac{\Delta\omega}{t} = \frac{64\pi}{16} = 4\pi\,rad/s^{2}$.

(B) Given angular acceleration $\alpha = \frac{d\omega}{dt} = 6t^2 - 2t$.
Integrating with respect to $t$ to find angular velocity $\omega$:
$\int_{10}^{\omega} d\omega = \int_{0}^{t} (6t^2 - 2t) dt$
$\omega - 10 = [2t^3 - t^2]_0^t$
$\omega = 2t^3 - t^2 + 10$.
Now,$\omega = \frac{d\theta}{dt} = 2t^3 - t^2 + 10$.
Integrating with respect to $t$ to find angular position $\theta$:
$\int_{4}^{\theta} d\theta = \int_{0}^{t} (2t^3 - t^2 + 10) dt$
$\theta - 4 = [\frac{2t^4}{4} - \frac{t^3}{3} + 10t]_0^t$
$\theta - 4 = \frac{t^4}{2} - \frac{t^3}{3} + 10t$
$\theta = \frac{t^4}{2} - \frac{t^3}{3} + 10t + 4$.

(B) In uniform circular motion,the centripetal acceleration $a_c$ is given by the formula $a_c = \omega^2 r$,where $\omega$ is the angular velocity and $r$ is the radius of the circular path.
Since angular velocity $\omega = 2 \pi f$,where $f$ is the frequency,we can substitute this into the acceleration formula:
$a_c = (2 \pi f)^2 r = 4 \pi^2 f^2 r$.
From this expression,we can see that the acceleration is directly proportional to the square of the frequency: $a_c \propto f^2$.
If the frequency $f$ is doubled $(f' = 2f)$,the new acceleration $a_c'$ becomes:
$a_c' \propto (2f)^2 = 4f^2$.
Therefore,the acceleration becomes $4$ times the original value.

(D) Angular acceleration $(\alpha)$ is defined as the rate of change of angular velocity $(\omega)$ with respect to time,given by $\alpha = \frac{d\omega}{dt}$.
Since the particle is revolving with a constant angular speed,the angular velocity $\omega$ does not change over time.
Therefore,the rate of change of angular velocity is zero,which implies $\alpha = 0 \, rad/s^2$.

(A) Given: Initial angular velocity $\omega_0 = 0 \text{ rad/s}$ (since it starts from rest).
Angular displacement $\theta = 100 \pi \text{ rad}$.
Time $t = 5 \text{ s}$.
Using the second equation of rotational motion: $\theta = \omega_0 t + \frac{1}{2} \alpha t^2$.
Substituting the values: $100 \pi = 0 + \frac{1}{2} \alpha (5)^2$.
$100 \pi = \frac{25}{2} \alpha$.
$\alpha = \frac{200 \pi}{25} = 8 \pi \text{ rad/s}^2$.
Now,using the first equation of rotational motion: $\omega = \omega_0 + \alpha t$.
$\omega = 0 + (8 \pi) \times 5 = 40 \pi \text{ rad/s}$.
Thus,the angular speed at the end of $5 \text{ s}$ is $40 \pi \text{ rad/s}$.

(A) Angular acceleration $(\vec{\alpha})$ is defined as the rate of change of angular velocity $(\vec{\omega})$ with respect to time, given by $\vec{\alpha} = \frac{d\vec{\omega}}{dt}$.
Since the angular velocity vector $(\vec{\omega})$ for a body moving in a circular path is directed along the axis of rotation, the rate of change of this vector, which is the angular acceleration $(\vec{\alpha})$, also acts along the same axis of rotation.

(C) The length of the second hand is $r_s = 75 \ cm = 0.75 \ m$. The length of the minute hand is $r_m = 60 \ cm = 0.60 \ m$.
In $30$ minutes,the second hand completes $30$ full revolutions. The distance traveled by the tip of the second hand is $d_s = 30 \times (2 \pi r_s) = 30 \times 2 \times 3.14 \times 0.75 = 141.3 \ m$.
In $30$ minutes,the minute hand completes $0.5$ revolutions (half a circle). The distance traveled by the tip of the minute hand is $d_m = 0.5 \times (2 \pi r_m) = \pi r_m = 3.14 \times 0.60 = 1.884 \ m$.
The difference in distance is $x = d_s - d_m = 141.3 - 1.884 = 139.416 \ m$.
Rounding to one decimal place,$x \approx 139.4 \ m$.

(A) The particle starts from rest,so the initial velocity $u = 0$.
The distance covered in one revolution is the circumference of the circle,which is $2 \pi r$.
The distance covered in two revolutions is $s = 2 \times 2 \pi r = 4 \pi r$.
Using the equation of motion $v^2 - u^2 = 2as$,where $a$ is the tangential acceleration:
$v^2 - 0^2 = 2 \times a \times (4 \pi r)$
$v^2 = 8 \pi r a$
Therefore,the tangential acceleration is $a = \frac{v^2}{8 \pi r}$.

(D) Using the rotational kinematic equation for angular displacement:
$\theta = \omega_0 t + \frac{1}{2} \alpha t^2$
Since the particle starts from rest,$\omega_0 = 0$,so:
$\theta = \frac{1}{2} \alpha t^2$
Solving for time $t$:
$t = \sqrt{\frac{2 \theta}{\alpha}} \quad ...(i)$
The linear distance (arc length) covered by the particle for an angular displacement $\theta$ is:
$s = r \theta \quad ...(ii)$
The average velocity is defined as the total displacement divided by the total time. For a small displacement $\theta$,the magnitude of displacement is approximately the arc length $s$:
$V_{\text{average}} = \frac{s}{t} = \frac{r \theta}{\sqrt{\frac{2 \theta}{\alpha}}}$
$V_{\text{average}} = r \theta \cdot \sqrt{\frac{\alpha}{2 \theta}} = r \sqrt{\frac{\alpha \theta^2}{2 \theta}} = \frac{r}{\sqrt{2}} \sqrt{\alpha \theta}$

(B) Using the $3^{rd}$ equation of motion: $v^2 = u^2 + 2as$.
Since the motion starts from rest,$u = 0$,so $v^2 = 2 a_t s$.
Therefore,$a_t = \frac{v^2}{2s}$.
By the end of the $3^{rd}$ revolution,the total distance covered is $s = 3 \times (2 \pi r) = 6 \pi r$.
Substituting $s$ into the acceleration formula: $a_t = \frac{v^2}{2 \times 6 \pi r} = \frac{v^2}{12 \pi r}$.
Given the kinetic energy $E = \frac{1}{2} m v^2$,we have $v^2 = \frac{2E}{m}$.
Substituting $v^2$ into the expression for $a_t$: $a_t = \frac{2E}{m \times 12 \pi r} = \frac{E}{6 \pi rm}$.

(D) The radius of the circular path is $r = \frac{\pi}{2} \ m$.
In $x$ revolutions,the total distance traveled by the particle is $d = x \times (2\pi r)$.
Substituting the value of $r$: $d = x \times (2\pi \times \frac{\pi}{2}) = x \times \pi^2 = \pi^2 x \ m$.
Tangential velocity $v$ is defined as the total distance traveled divided by the time taken: $v = \frac{d}{t}$.
Therefore,$v = \frac{\pi^2 x}{t} \ m/s$.

(C) Given: Mass $m = 10 \text{ g} = 10^{-2} \text{ kg}$, Radius $r = 6.4 \text{ cm} = 6.4 \times 10^{-2} \text{ m}$, Kinetic Energy $K = 8 \times 10^{-4} \text{ J}$.
Since the particle starts from rest, its initial velocity $u = 0$.
The distance covered in two revolutions is $s = 2 \times (2 \pi r) = 4 \pi r$.
Using the work-energy theorem, the work done by the tangential force $F_t$ equals the change in kinetic energy:
$W = F_t \cdot s = \Delta K$
Since $F_t = m a_t$, we have:
$m a_t \cdot (4 \pi r) = K - 0$
$a_t = \frac{K}{m \cdot 4 \pi r}$
Substituting the values:
$a_t = \frac{8 \times 10^{-4} \text{ J}}{10^{-2} \text{ kg} \times 4 \times 3.14 \times 6.4 \times 10^{-2} \text{ m}}$
$a_t = \frac{8 \times 10^{-4}}{25.6 \pi \times 10^{-4}} = \frac{8}{25.6 \times 3.14} \approx 0.1 \text{ m/s}^2$.

(B) The particle is performing uniform circular motion $(U.C.M.)$.
The total angle covered in $x$ revolutions is $\theta = 2 \pi x$ radians.
The angular velocity $\omega$ is given by $\omega = \frac{\theta}{t} = \frac{2 \pi x}{t}$.
The tangential velocity $v$ is related to angular velocity by $v = \omega R$.
Given the radius $R = \frac{\pi}{2} \ m$.
Substituting the values,$v = \left( \frac{2 \pi x}{t} \right) \times \left( \frac{\pi}{2} \right) = \frac{\pi^2 x}{t} \ m/s$.

(D) The angular displacement is given by $\theta = 5 \sin \frac{\pi t}{6}$.
Angular velocity $\omega$ is the rate of change of angular displacement,defined as $\omega = \frac{d\theta}{dt}$.
Differentiating $\theta$ with respect to $t$:
$\omega = \frac{d}{dt} \left( 5 \sin \frac{\pi t}{6} \right) = 5 \cdot \cos \left( \frac{\pi t}{6} \right) \cdot \frac{\pi}{6} = \frac{5\pi}{6} \cos \left( \frac{\pi t}{6} \right)$.
At $t = 3 \ s$:
$\omega = \frac{5\pi}{6} \cos \left( \frac{\pi \times 3}{6} \right) = \frac{5\pi}{6} \cos \left( \frac{\pi}{2} \right)$.
Since $\cos \frac{\pi}{2} = 0$,we get $\omega = \frac{5\pi}{6} \times 0 = 0 \ rad/s$.

(D) The frequency of revolution is given by $f = \frac{x}{t}$.
Angular velocity $\omega$ is related to frequency by $\omega = 2 \pi f = \frac{2 \pi x}{t}$.
Tangential velocity $V$ is related to angular velocity by $V = \omega r$.
Given radius $r = \frac{\pi}{2} \ m$.
Substituting the values: $V = \left( \frac{2 \pi x}{t} \right) \cdot \left( \frac{\pi}{2} \right) = \frac{\pi^{2} x}{t}$.

(A) The particle starts from rest,so initial angular velocity $\omega_0 = 0$.
Using the equation of rotational motion: $\theta = \omega_0 t + \frac{1}{2} \alpha t^2$.
Since $\omega_0 = 0$,we have $\theta = \frac{1}{2} \alpha t^2$,which gives $t = \sqrt{\frac{2 \theta}{\alpha}}$.
The displacement of the particle for an angular displacement $\theta$ is the chord length $d = 2r \sin(\theta/2)$. For small $\theta$,$\sin(\theta/2) \approx \theta/2$,so $d \approx r \theta$.
The average velocity is defined as $v_{avg} = \frac{\text{displacement}}{\text{time}} = \frac{r \theta}{t}$.
Substituting $t$: $v_{avg} = \frac{r \theta}{\sqrt{2 \theta / \alpha}} = r \theta \sqrt{\frac{\alpha}{2 \theta}} = r \sqrt{\frac{\alpha \theta^2}{2 \theta}} = r \sqrt{\frac{\alpha \theta}{2}}$.
Thus,the correct option is $r \sqrt{\frac{\alpha \theta}{2}}$,which matches option $A$ if written as $r \left(\frac{\alpha \theta}{2}\right)^{1/2}$.

(B) Let the initial kinetic energy be $E_1 = E$ and the final kinetic energy be $E_2 = 3E$.
Since $E = \frac{1}{2}mv^2 = \frac{1}{2}mr^2\omega^2$,we have $E \propto \omega^2$.
Thus,$\frac{E_2}{E_1} = \frac{\omega_2^2}{\omega_1^2} = 3$,which implies $\omega_2^2 = 3\omega_1^2$.
Assuming the particle starts from rest,$\omega_1 = 0$. However,the problem implies motion starts from an initial state. Let initial angular velocity be $\omega_0$. Then $\omega_f^2 = 3\omega_0^2$.
Using the rotational kinematic equation $\omega_f^2 = \omega_0^2 + 2\alpha\theta$,where $\theta = 3 \times 2\pi = 6\pi$ radians.
$3\omega_0^2 = \omega_0^2 + 2\alpha(6\pi) \implies 2\omega_0^2 = 12\alpha\pi \implies \alpha = \frac{\omega_0^2}{6\pi}$.
Given $E = \frac{1}{2}mr^2\omega_0^2$,we have $\omega_0^2 = \frac{2E}{mr^2}$.
Substituting $\omega_0^2$ into the expression for $\alpha$: $\alpha = \frac{2E}{mr^2} \cdot \frac{1}{6\pi} = \frac{E}{3\pi mr^2}$.
The tangential acceleration is $a_t = r\alpha = r \cdot \frac{E}{3\pi mr^2} = \frac{E}{3\pi mr}$.

(C) Initial angular velocity $\omega_{0} = 1200 \text{ rpm} = \frac{1200 \times 2\pi}{60} \text{ rad/s} = 40\pi \text{ rad/s}$.
Final angular velocity $\omega = 0 \text{ rad/s}$.
Angular deceleration $\alpha = 4 \text{ rad/s}^{2}$.
Using the kinematic equation $\omega^{2} = \omega_{0}^{2} - 2\alpha\theta$:
$0 = (40\pi)^{2} - 2(4)\theta$
$8\theta = 1600\pi^{2}$
$\theta = 200\pi^{2} \text{ rad}$.
Since the total angle $\theta = 2\pi n$,where $n$ is the number of revolutions:
$2\pi n = 200\pi^{2}$
$n = 100\pi = 100 \times 3.14159 \approx 314 \text{ revolutions}$.

(C) The angular speed $\omega$ is defined as $\omega = \frac{2\pi}{T}$,where $T$ is the time period.
For the minute hand,the time period $T_{\min} = 60 \text{ minutes}$. Thus,$\omega_{\min} = \frac{2\pi}{60} \text{ rad/min}$.
For the hour hand,the time period $T_{hr} = 12 \text{ hours} = 12 \times 60 \text{ minutes}$. Thus,$\omega_{hr} = \frac{2\pi}{12 \times 60} \text{ rad/min}$.
The ratio of angular speeds is $\frac{\omega_{\min}}{\omega_{hr}} = \frac{2\pi / 60}{2\pi / (12 \times 60)} = \frac{12 \times 60}{60} = 12$.
Therefore,the ratio is $12: 1$.

(D) The angular displacement $\theta$ for an object starting from rest with constant angular acceleration $\alpha$ is given by $\theta = \frac{1}{2} \alpha t^2$.
For the first $2 \ s$ $(t_1 = 2 \ s)$: $\theta_1 = \frac{1}{2} \alpha (2)^2 = 2\alpha$.
For the total time of $4 \ s$ $(t_2 = 2 + 2 = 4 \ s)$: $\theta_{total} = \frac{1}{2} \alpha (4)^2 = 8\alpha$.
The angle rotated in the next $2 \ s$ is $\theta_2 = \theta_{total} - \theta_1 = 8\alpha - 2\alpha = 6\alpha$.
Therefore,the ratio $\theta_1 : \theta_2 = 2\alpha : 6\alpha = 1 : 3$.

(C) The angular displacement $\theta_n$ covered in the $n^{\text{th}}$ second for rotational motion is given by the formula: $\theta_n = \omega_0 + \frac{\alpha}{2}(2n - 1)$.
Given that the particle starts from rest,the initial angular velocity $\omega_0 = 0$.
The angular displacement in the $P^{\text{th}}$ second is given as $\beta$.
Substituting these values into the formula: $\beta = 0 + \frac{\alpha}{2}(2P - 1)$.
Rearranging the equation to solve for the angular acceleration $\alpha$: $\alpha = \frac{2 \beta}{(2P - 1)}$.

(C) The correct option is $C$.
Concept: The angular speed is given by $\omega = \frac{\Delta \theta}{\Delta t}$.
Relative angular speed is given by $\omega_{rel} = \omega_{s} - \omega_{h}$.
The angular speed of the second hand is $\omega_{s} = \frac{2 \pi}{60} = \frac{\pi}{30} \ rad/s$.
The angular speed of the hour hand is $\omega_{h} = \frac{2 \pi}{12 \times 3600} = \frac{2 \pi}{43200} = \frac{\pi}{21600} \ rad/s$.
Therefore,the relative angular speed is $\omega_{rel} = \omega_{s} - \omega_{h} = \frac{\pi}{30} - \frac{\pi}{21600}$.
Taking the common denominator $21600$: $\omega_{rel} = \frac{720 \pi - \pi}{21600} = \frac{719 \pi}{21600} \ rad/s$.

(B) The angular speed of the hour hand $(\omega_h)$ is given by the angle covered in one full rotation ($2 \pi$ radians) divided by the time taken $(12 \text{ hours} = 12 \times 3600 \text{ seconds})$: $\omega_h = \frac{2 \pi}{12 \times 3600} \text{ rad/s}$.
The angular speed of the minute hand $(\omega_m)$ is given by the angle covered in one full rotation ($2 \pi$ radians) divided by the time taken $(1 \text{ hour} = 3600 \text{ seconds})$: $\omega_m = \frac{2 \pi}{3600} \text{ rad/s}$.
The relative angular speed is the difference between the angular speeds of the two hands: $\Delta \omega = \omega_m - \omega_h$.
Substituting the values: $\Delta \omega = \frac{2 \pi}{3600} - \frac{2 \pi}{12 \times 3600} = \frac{2 \pi}{3600} \left(1 - \frac{1}{12}\right) = \frac{2 \pi}{3600} \left(\frac{11}{12}\right) = \frac{22 \pi}{43200} = \frac{11 \pi}{21600} \text{ rad/s}$.

(C) Given: Angular acceleration $\alpha = 4 \ rad/s^2$,initial angular velocity $\omega_0 = 0$.
Centripetal acceleration is given by $a_c = r\omega^2$.
Tangential acceleration is given by $a_t = r\alpha$.
We are given that $a_c = a_t$,so $r\omega^2 = r\alpha$.
This simplifies to $\omega^2 = \alpha$,which means $\omega = \sqrt{\alpha} = \sqrt{4} = 2 \ rad/s$.
Using the kinematic equation for rotational motion,$\omega = \omega_0 + \alpha t$.
Substituting the known values: $2 = 0 + 4t$.
Solving for $t$: $t = 2/4 = 1/2 \ s$.