Kinematics Circular Motion (Uniform Angular Accelaration) Questions in English

(B) The angular speed of the hour hand is $\omega_{h} = \frac{2 \pi}{T_{h}} = \frac{2 \pi}{12 \times 3600} \text{ rad/s}$.
The angular speed of the second hand is $\omega_{s} = \frac{2 \pi}{T_{s}} = \frac{2 \pi}{60} \text{ rad/s}$.
The relative angular speed is given by $\omega_{rel} = |\omega_{s} - \omega_{h}|$.
$\omega_{rel} = \left| \frac{2 \pi}{60} - \frac{2 \pi}{43200} \right| = 2 \pi \left( \frac{720 - 1}{43200} \right)$.
$\omega_{rel} = 2 \pi \left( \frac{719}{43200} \right) = \frac{719 \pi}{21600} \text{ rad/s}$.

(D) Given: Initial angular velocity $\omega_0 = 0 \ rad \ s^{-1}$,final angular velocity $\omega = 24 \ rad \ s^{-1}$,and time $t = 8 \ s$.
First,calculate the constant angular acceleration $\alpha$ using the formula $\alpha = \frac{\omega - \omega_0}{t}$.
$\alpha = \frac{24 - 0}{8} = 3 \ rad \ s^{-2}$.
Now,calculate the total angle $\theta$ turned through using the kinematic equation $\theta = \omega_0 t + \frac{1}{2} \alpha t^2$.
$\theta = 0 \times 8 + \frac{1}{2} \times 3 \times (8)^2$.
$\theta = \frac{1}{2} \times 3 \times 64 = 3 \times 32 = 96 \ rad$.

(C) Given that,angular acceleration $\alpha = 4 \ rad/s^2$ and initial angular velocity $\omega_0 = 0$.
Centripetal (radial) acceleration is given by $a_r = r \omega^2$.
Tangential acceleration is given by $a_t = r \alpha$.
We are given that the magnitudes of centripetal and tangential acceleration are equal,so $a_r = a_t$.
Substituting the expressions,we get $r \omega^2 = r \alpha$.
Dividing both sides by $r$,we have $\omega^2 = \alpha = 4$.
Taking the square root,$\omega = 2 \ rad/s$.
Using the kinematic equation for rotational motion,$\omega = \omega_0 + \alpha t$.
Since $\omega_0 = 0$,we have $\omega = \alpha t$.
Substituting the known values,$2 = 4t$.
Therefore,$t = 2/4 = 1/2 \ s$.

(C) Initial angular speed $\omega_{1} = 1800 \ rpm = \frac{1800 \times 2\pi}{60} \ rad \ s^{-1} = 60\pi \ rad \ s^{-1}$.
Final angular speed $\omega_{2} = 3000 \ rpm = \frac{3000 \times 2\pi}{60} \ rad \ s^{-1} = 100\pi \ rad \ s^{-1}$.
Time interval $t = 20 \ s$.
Angular acceleration $\alpha = \frac{\omega_{2} - \omega_{1}}{t}$.
$\alpha = \frac{100\pi - 60\pi}{20} = \frac{40\pi}{20} = 2\pi \ rad \ s^{-2}$.

(A) For the second-hand,the time period $T_s = 60 \text{ s}$.
So,the angular speed $\omega_s = \frac{2\pi}{T_s} = \frac{2\pi}{60} \text{ rad/s}$.
For the hour-hand,the time period $T_h = 12 \text{ hours} = 12 \times 60 \times 60 \text{ s} = 43200 \text{ s}$.
So,the angular speed $\omega_h = \frac{2\pi}{T_h} = \frac{2\pi}{43200} \text{ rad/s}$.
The ratio of the angular speed of the second-hand to the hour-hand is $\frac{\omega_s}{\omega_h} = \frac{2\pi / 60}{2\pi / 43200} = \frac{43200}{60} = 720$.
Thus,the ratio is $720: 1$.

(A) The velocity vector is given by $\vec{v}(t) = A \cos(kt) \hat{i} - A \sin(kt) \hat{j}$.
To find the force $\vec{F}$,we differentiate the velocity with respect to time $t$:
$\vec{F} = m \frac{d\vec{v}}{dt} = m \frac{d}{dt} [A \cos(kt) \hat{i} - A \sin(kt) \hat{j}]$
$\vec{F} = m A [-k \sin(kt) \hat{i} - k \cos(kt) \hat{j}] = -mkA [\sin(kt) \hat{i} + \cos(kt) \hat{j}]$.
Now,calculate the dot product of $\vec{F}$ and $\vec{v}$:
$\vec{F} \cdot \vec{v} = (-mkA [\sin(kt) \hat{i} + \cos(kt) \hat{j}]) \cdot (A \cos(kt) \hat{i} - A \sin(kt) \hat{j})$
$\vec{F} \cdot \vec{v} = -mkA^2 [\sin(kt)\cos(kt) - \cos(kt)\sin(kt)] = 0$.
Since the dot product is zero,the angle between the force $\vec{F}$ and the velocity $\vec{v}$ is $90^{\circ}$.

(B) Initial angular speed $\omega_0 = 300 \text{ rpm} = \frac{300 \times 2\pi}{60} \text{ rad/s} = 10\pi \text{ rad/s}$.
Final angular speed $\omega = 0 \text{ rad/s}$.
Time taken $t = 80 \text{ s}$.
Using the equation of motion $\omega = \omega_0 + \alpha t$,we find the angular deceleration $\alpha$:
$0 = 10\pi + \alpha(80) \Rightarrow \alpha = -\frac{10\pi}{80} = -\frac{\pi}{8} \text{ rad/s}^2$.
The total angular displacement $\theta$ is given by $\theta = \omega_0 t + \frac{1}{2} \alpha t^2$:
$\theta = (10\pi)(80) + \frac{1}{2} \left(-\frac{\pi}{8}\right) (80)^2 = 800\pi - \frac{\pi}{16} (6400) = 800\pi - 400\pi = 400\pi \text{ rad}$.
The number of revolutions $n$ is given by $n = \frac{\theta}{2\pi}$:
$n = \frac{400\pi}{2\pi} = 200$.

(A) The angular velocity $\omega$ is given by $\omega = \frac{2\pi}{T}$ or $\omega = 2\pi n$.
$1$. For Earth rotating about its own axis,$T = 24 \ h = 86400 \ s$.
$\omega_1 = \frac{2\pi}{86400} \ rad/s \approx 7.27 \times 10^{-5} \ rad/s$.
$2$. For the hour hand of a clock,$T = 12 \ h = 43200 \ s$.
$\omega_2 = \frac{2\pi}{43200} \ rad/s \approx 1.45 \times 10^{-4} \ rad/s$.
$3$. For the second hand of a clock,$T = 60 \ s$.
$\omega_3 = \frac{2\pi}{60} \ rad/s \approx 0.105 \ rad/s$.
$4$. For the flywheel,$n = 300 \ rpm = 5 \ rev/s$.
$\omega_4 = 2\pi \times 5 = 10\pi \ rad/s \approx 31.4 \ rad/s$.
Comparing the values: $\omega_1 < \omega_2 < \omega_3 < \omega_4$.
Thus,the increasing order is $1, 2, 3, 4$.

(B) Given: Initial angular velocity $\omega_0 = 0 \ rad/s$,angular acceleration $\alpha = \pi \ rad/s^2$,and time $t = 6 \ s$.
Using the kinematic equation for angular displacement: $\theta = \omega_0 t + \frac{1}{2} \alpha t^2$.
Substituting the values: $\theta = 0 \times 6 + \frac{1}{2} \times \pi \times (6)^2$.
$\theta = \frac{1}{2} \times \pi \times 36 = 18\pi \ rad$.
The number of rotations $n$ is given by $n = \frac{\theta}{2\pi}$.
$n = \frac{18\pi}{2\pi} = 9$ rotations.

(A) The normal acceleration is given by $a_n = \frac{v^2}{R} = k t^\alpha$,where $k$ is a constant.
From this,the square of the velocity is $v^2 = R k t^\alpha$.
The kinetic energy of the object is $K = \frac{1}{2} m v^2 = \frac{1}{2} m R k t^\alpha$.
The power $P$ developed by all forces is equal to the rate of change of kinetic energy,$P = \frac{dK}{dt}$.
$P = \frac{d}{dt} \left( \frac{1}{2} m R k t^\alpha \right) = \frac{1}{2} m R k \alpha t^{\alpha-1}$.
Since $m$,$R$,$k$,and $\alpha$ are constants,we have $P \propto t^{\alpha-1}$.

(A) The angular velocity $\omega$ is given by $\omega = \frac{2\pi}{T}$ or $\omega = 2\pi n$.
$1$. For Earth rotating about its own axis,$T = 24 \ h = 86400 \ s$:
$\omega_1 = \frac{2\pi}{86400} \ rad/s \approx 7.27 \times 10^{-5} \ rad/s$.
$2$. For the hour hand of a clock,$T = 12 \ h = 43200 \ s$:
$\omega_2 = \frac{2\pi}{43200} \ rad/s \approx 1.45 \times 10^{-4} \ rad/s$.
$3$. For the second hand of a clock,$T = 60 \ s$:
$\omega_3 = \frac{2\pi}{60} \ rad/s \approx 0.105 \ rad/s$.
$4$. For the flywheel,$n = 300 \ rpm = 5 \ rev/s$:
$\omega_4 = 2\pi \times 5 = 10\pi \ rad/s \approx 31.4 \ rad/s$.
Comparing the values: $\omega_1 < \omega_2 < \omega_3 < \omega_4$.
Thus,the increasing order is $1, 2, 3, 4$.

(A) From the kinematic equations of circular motion,we have $\omega = \omega_0 + \alpha t$.
Given $\omega_0 = 0$ at $t=0$,at $t=2 \ s$,$\omega = 5 \ rad/s$.
$5 = 0 + \alpha \times 2 \Rightarrow \alpha = 2.5 \ rad/s^2$.
For the interval $t=0 \ s$ to $t=20 \ s$,the angular displacement $\theta_1$ is:
$\theta_1 = \omega_0 t + \frac{1}{2} \alpha t^2 = 0 + \frac{1}{2} \times 2.5 \times (20)^2 = 500 \ rad$.
The angular velocity at $t=20 \ s$ is $\omega_{20} = \omega_0 + \alpha \times 20 = 0 + 2.5 \times 20 = 50 \ rad/s$.
For the interval $t=20 \ s$ to $t=50 \ s$,the acceleration is zero,so the angular velocity remains constant at $50 \ rad/s$.
The angular displacement $\theta_2$ is:
$\theta_2 = \omega_{20} \times \Delta t = 50 \times (50 - 20) = 50 \times 30 = 1500 \ rad$.
Total angular displacement $\theta = \theta_1 + \theta_2 = 500 + 1500 = 2000 \ rad$.
Number of revolutions $n = \frac{\theta}{2\pi} = \frac{2000}{2\pi} = \frac{1000}{\pi}$.

(B) Since the angular acceleration $\alpha$ is constant and the initial angular velocity $\omega_0 = 0$,the angle $\theta$ rotated in time $t$ is given by $\theta = \frac{1}{2} \alpha t^2$.
Given $\theta = 15^{\circ}$ for time $t$,we have $15^{\circ} = \frac{1}{2} \alpha t^2$ (Equation $1$).
Now,we need to find the angle rotated in the next $2t \ s$. The total time elapsed is $t + 2t = 3t \ s$.
The total angle $\theta_{total}$ rotated in time $3t$ is $\theta_{total} = \frac{1}{2} \alpha (3t)^2 = 9 \times (\frac{1}{2} \alpha t^2)$.
Substituting Equation $1$ into this,we get $\theta_{total} = 9 \times 15^{\circ} = 135^{\circ}$.
The increase in angle in the next $2t \ s$ is $\Delta \theta = \theta_{total} - \theta = 135^{\circ} - 15^{\circ} = 120^{\circ}$.

(C) The angular position is given by $\theta = \frac{5t^4}{40} - \frac{t^3}{3} = \frac{t^4}{8} - \frac{t^3}{3}$.
Angular velocity $\omega$ is the rate of change of angular position: $\omega = \frac{d\theta}{dt} = \frac{d}{dt}(\frac{t^4}{8} - \frac{t^3}{3}) = \frac{4t^3}{8} - \frac{3t^2}{3} = 0.5t^3 - t^2$.
Angular acceleration $\alpha$ is the rate of change of angular velocity: $\alpha = \frac{d\omega}{dt} = \frac{d}{dt}(0.5t^3 - t^2) = 1.5t^2 - 2t$.
At $t = 10 \text{ s}$,substituting the value of $t$ in the expression for $\alpha$:
$\alpha = 1.5(10)^2 - 2(10) = 1.5(100) - 20 = 150 - 20 = 130 \text{ rad/s}^2$.

(B) The angular displacement is given by the formula $\theta = \frac{1}{2}\alpha t^2$,where $\alpha$ is the uniform angular acceleration and $t$ is the time.
For the first $2 \text{ s}$ $(t = 2 \text{ s})$,the angular displacement is $\theta_1 = \frac{1}{2}\alpha (2)^2 = 2\alpha$.
For the total time of $4 \text{ s}$ $(t = 4 \text{ s})$,the total angular displacement is $\theta_{\text{total}} = \frac{1}{2}\alpha (4)^2 = 8\alpha$.
The angular displacement in the next $2 \text{ s}$ is $\theta_2 = \theta_{\text{total}} - \theta_1 = 8\alpha - 2\alpha = 6\alpha$.
Therefore,the ratio $\frac{\theta_2}{\theta_1} = \frac{6\alpha}{2\alpha} = 3$.

(D) Initial angular speed $\omega_i = 600 \text{ rpm} = \frac{600 \times 2\pi}{60} = 20\pi \text{ rad/s}$.
Final angular speed $\omega_f = 1200 \text{ rpm} = \frac{1200 \times 2\pi}{60} = 40\pi \text{ rad/s}$.
Angular acceleration $\alpha = \frac{\omega_f - \omega_i}{t} = \frac{40\pi - 20\pi}{10} = 2\pi \text{ rad/s}^2$.
Total angle rotated $\theta = \omega_i t + \frac{1}{2} \alpha t^2 = (20\pi)(10) + \frac{1}{2}(2\pi)(10^2) = 200\pi + 100\pi = 300\pi \text{ rad}$.
Number of revolutions $n = \frac{\theta}{2\pi} = \frac{300\pi}{2\pi} = 150$.