Relation between Torque and Angular acceleration and it's Application Questions in English

(B) Given: Moment of inertia $I = 2 \, kg \cdot m^2$,initial frequency $n_1 = 60 \, rpm = 1 \, rps$,final frequency $n_2 = 0 \, rps$,and time $t = 1 \, min = 60 \, s$.
The angular velocity is given by $\omega = 2\pi n$.
The torque $\tau$ is given by the rate of change of angular momentum: $\tau = I \alpha = I \frac{\Delta \omega}{t}$.
Substituting the values: $\tau = I \frac{2\pi(n_2 - n_1)}{t}$.
$\tau = 2 \times \frac{2\pi(0 - 1)}{60} = \frac{-4\pi}{60} = -\frac{\pi}{15} \, N \cdot m$.
The magnitude of the torque required to stop the wheel is $\frac{\pi}{15} \, N \cdot m$.

(A) The power $P$ delivered by an engine to a rotating body is given by the product of torque $\tau$ and angular velocity $\omega$.
Given:
Torque $\tau = 100\, Nm$
Angular speed $\omega = 100\, rad\, s^{-1}$
Using the formula $P = \tau \omega$:
$P = (100\, Nm) \times (100\, rad\, s^{-1})$
$P = 10,000\, W$
Since $1\, kW = 1,000\, W$,we have:
$P = 10\, kW$.

(D) The torque $\tau$ applied to the cylinder is given by $\tau = F \times r$,where $F = 20\,N$ and $r = 0.5\,m$.
$\tau = 20 \times 0.5 = 10\,N\cdot m$.
For a hollow cylinder,the moment of inertia $I$ about its central axis is $I = M r^2$,where $M = 5\,kg$ and $r = 0.5\,m$.
$I = 5 \times (0.5)^2 = 5 \times 0.25 = 1.25\,kg\cdot m^2$.
Using the relation $\tau = I \alpha$,where $\alpha$ is the angular acceleration:
$10 = 1.25 \times \alpha$.
$\alpha = \frac{10}{1.25} = 8\,rad/s^2$.

(B) Given: $r_1 = 0.05\,m$,$r_2 = 0.30\,m$,$I = 1500\,kg\cdot m^2$.
The torque $\tau$ is calculated as the sum of torques produced by each force:
$\tau = \tau_{10N} + \tau_{9N} - \tau_{12N}$
$\tau = (10 \times r_2) + (9 \times r_2) - (12 \times r_1 \times \cos(30^\circ))$
$\tau = (10 \times 0.3) + (9 \times 0.3) - (12 \times 0.05 \times \frac{\sqrt{3}}{2})$
$\tau = 3 + 2.7 - (0.6 \times 0.866) = 5.7 - 0.5196 = 5.1804\,N\cdot m$.
Using $\tau = I \alpha$:
$\alpha = \frac{\tau}{I} = \frac{5.1804}{1500} \approx 0.00345\,rad/s^2 \approx 3.45 \times 10^{-3}\,rad/s^2$.
Comparing with the given options,the closest value is $3 \times 10^{-3}\,rad/s^2$.

(A) The net force acting on the pulley due to the belt tension difference is $F_{net} = T_1 - T_2 = 135\,N - 45\,N = 90\,N$.
The radius of the motor pulley is $r = 8\,cm = 0.08\,m$.
The torque exerted by the motor on the belt is $\tau = F_{net} \times r = 90\,N \times 0.08\,m = 7.2\,Nm$.
The power of the motor is given by $P = \tau \omega$,where $\omega = 200\,rad\,s^{-1}$.
$P = 7.2\,Nm \times 200\,rad\,s^{-1} = 1440\,W$.
Converting to kilowatts,$P = 1440 / 1000 = 1.44\,kW$.

(A) The torque $\tau$ about the hinged end is given by $\tau = Mg \frac{L}{2}$.
The moment of inertia $I$ of a uniform rod of mass $M$ and length $L$ about an axis passing through one end is $I = \frac{ML^2}{3}$.
Using the rotational analogue of Newton's second law,$\tau = I \alpha$,where $\alpha$ is the angular acceleration:
$Mg \frac{L}{2} = \left( \frac{ML^2}{3} \right) \alpha$.
Solving for $\alpha$:
$\alpha = \frac{MgL/2}{ML^2/3} = \frac{3g}{2L}$.
Given the length of the metre stick is $L = 1 \ m$,we substitute $L = 1$:
$\alpha = \frac{3g}{2(1)} = \frac{3}{2} g \ rad/s^2$.

(B) The torque $\tau$ about the hinged end is given by $\tau = F \times r$,where $r = \frac{5L}{6}$.
$\tau = \left( \frac{Mg}{2} \right) \times \left( \frac{5L}{6} \right) = \frac{5MgL}{12}$.
The moment of inertia $I$ of a rod about an axis passing through one end is $I = \frac{ML^2}{3}$.
Using the relation $\tau = I \alpha$,where $\alpha$ is the angular acceleration:
$\alpha = \frac{\tau}{I} = \frac{5MgL}{12} \times \frac{3}{ML^2}$.
$\alpha = \frac{5g}{4L}$.

(B) When the string is cut,the only torque acting on the rod about the hinge $P$ is due to the gravitational force acting at the center of mass of the rod.
The gravitational force $Mg$ acts at a distance of $L/2$ from the hinge $P$.
The torque $\tau$ about the hinge $P$ is given by:
$\tau = Mg \times \frac{L}{2}$
The moment of inertia $I$ of the rod about an axis passing through its end $P$ is:
$I = \frac{ML^2}{3}$
Using the rotational analog of Newton's second law,$\tau = I\alpha$,where $\alpha$ is the angular acceleration:
$Mg \times \frac{L}{2} = \left(\frac{ML^2}{3}\right) \alpha$
Solving for $\alpha$:
$\alpha = \frac{MgL/2}{ML^2/3} = \frac{3g}{2L}$

(A) The relationship between torque $\tau$ and angular momentum $L$ is given by Newton's second law for rotation: $\tau = \frac{dL}{dt}$.
Given the initial angular momentum $L_i = L_0$ and the final angular momentum $L_f = 4L_0$.
The time interval is $\Delta t = 4 \ s$.
The change in angular momentum is $\Delta L = L_f - L_i = 4L_0 - L_0 = 3L_0$.
Therefore,the magnitude of the constant torque is $\tau = \frac{\Delta L}{\Delta t} = \frac{3L_0}{4}$.

(B) Initial angular velocity $\omega_1 = 720 \, rpm = \frac{720 \times 2\pi}{60} \, rad/s = 24\pi \, rad/s$.
Final angular velocity $\omega_2 = 0 \, rad/s$.
Time taken $t = 8 \, s$.
Using the equation of rotational motion $\omega_2 = \omega_1 - \alpha t$,where $\alpha$ is the angular retardation:
$0 = 24\pi - \alpha(8) \implies \alpha = \frac{24\pi}{8} = 3\pi \, rad/s^2$.
The magnitude of torque $\tau$ is given by $\tau = I\alpha$.
Given $I = \frac{24}{\pi} \, kg \cdot m^2$,we have:
$\tau = \left(\frac{24}{\pi}\right) \times (3\pi) = 72 \, N \cdot m$.

(A) The torque $\vec{\tau}$ acting on a particle is given by $\vec{\tau} = \vec{r} \times \vec{F} = \vec{r} \times \frac{d\vec{p}}{dt} = \frac{d}{dt}(\vec{r} \times \vec{p}) = \frac{d\vec{L}}{dt}$.
Given mass $m = 2\, kg$,position $\vec{r}(t) = \vec{r}_0 + \int_0^t \vec{v} dt = (4\hat{i} - 2\hat{j}) + \int_0^t (2t^2\hat{i}) dt = (4 + \frac{2}{3}t^3)\hat{i} - 2\hat{j}$.
Velocity $\vec{v}(t) = 2t^2\hat{i}$.
Angular momentum $\vec{L} = \vec{r} \times m\vec{v} = m [((4 + \frac{2}{3}t^3)\hat{i} - 2\hat{j}) \times (2t^2\hat{i})] = 2 [0 - 2\hat{j} \times 2t^2\hat{i}] = 2 [4t^2\hat{k}] = 8t^2\hat{k}$.
Torque $\vec{\tau} = \frac{d\vec{L}}{dt} = \frac{d}{dt}(8t^2\hat{k}) = 16t\hat{k}$.
At $t = 2\, s$,$\vec{\tau} = 16(2)\hat{k} = 32\hat{k}\, N\cdot m$.

(C) Given: Radius $R = 0.2\, m$,Mass $M = 5\, kg$,Angular velocity $\omega = 2 + 6t$.
First,find the angular acceleration $\alpha$ by differentiating $\omega$ with respect to time $t$:
$\alpha = \frac{d\omega}{dt} = \frac{d}{dt}(2 + 6t) = 6\, rad/s^2$.
The moment of inertia $I$ of a disc about its central axis is $I = \frac{1}{2}MR^2$.
$I = \frac{1}{2} \times 5 \times (0.2)^2 = 0.5 \times 5 \times 0.04 = 0.1\, kg \cdot m^2$.
The torque $\tau$ is given by $\tau = I\alpha = 0.1 \times 6 = 0.6\, N \cdot m$.
Since $\tau = F \times R$,the tangential force $F$ is:
$F = \frac{\tau}{R} = \frac{0.6}{0.2} = 3\, N$.

(B) The torque $\tau$ is defined as the rate of change of angular momentum $L$ with respect to time $t$,given by the formula: $\tau = \frac{dL}{dt} = \frac{\Delta L}{\Delta t}$.
Given:
Initial angular momentum $L_i = 1\,J\cdot s$
Final angular momentum $L_f = 5\,J\cdot s$
Time interval $\Delta t = 5\,s$
Change in angular momentum $\Delta L = L_f - L_i = 5\,J\cdot s - 1\,J\cdot s = 4\,J\cdot s$.
Substituting the values into the formula:
$\tau = \frac{4\,J\cdot s}{5\,s} = 0.8\,N\cdot m$.
Thus,the torque is $0.8\,N\cdot m$,which is equivalent to $\frac{4}{5}\,N\cdot m$.

(D) The relationship between torque $(\tau)$,moment of inertia $(I)$,and angular acceleration $(\alpha)$ is given by the formula: $\tau = I \alpha$.
Given:
Torque $(\tau)$ = $31.4 \, Nm$
Angular acceleration $(\alpha)$ = $4\pi \, rad/s^2$
Using the value of $\pi \approx 3.14$,we have $\alpha = 4 \times 3.14 = 12.56 \, rad/s^2$.
Substituting the values into the formula:
$31.4 = I \times 12.56$
$I = \frac{31.4}{12.56} = 2.5 \, kg \cdot m^2$.

(B) Given,the initial frequency $f_0 = 1.3 \, rev/sec$. The initial angular velocity is $\omega_0 = 2\pi f_0 = 2\pi \times 1.3 = 2.6\pi \, rad/s$.
Since the disc comes to a stop,the final angular velocity $\omega = 0 \, rad/s$ in time $t = 30 \, s$.
Using the first equation of rotational motion: $\omega = \omega_0 + \alpha t$.
Substituting the values: $0 = 2.6\pi + \alpha \times 30$.
$\alpha = - \frac{2.6\pi}{30} \approx - \frac{2.6 \times 3.14}{30} \approx - 0.27 \, rad/s^2$.
The negative sign indicates retardation,so the magnitude of angular acceleration is approximately $0.27 \, rad/s^2$.

(A) Given: Mass $m = 2\; kg$,Radius $R = 4\; cm = 0.04\; m$,Initial angular velocity $\omega_0 = 3\; rpm = \frac{3 \times 2\pi}{60} = \frac{\pi}{10}\; rad/s$.
Total angular displacement $\theta = 2\pi \text{ revolutions} = 2\pi \times 2\pi = 4\pi^2\; rad$.
Final angular velocity $\omega = 0$.
Using the kinematic equation $\omega^2 = \omega_0^2 - 2\alpha\theta$:
$0 = (\frac{\pi}{10})^2 - 2\alpha(4\pi^2)$
$8\alpha\pi^2 = \frac{\pi^2}{100} \Rightarrow \alpha = \frac{1}{800}\; rad/s^2$.
The moment of inertia of a solid cylinder is $I = \frac{1}{2}mR^2 = \frac{1}{2} \times 2 \times (0.04)^2 = 0.0016\; kg\cdot m^2$.
The torque required is $\tau = I\alpha = 0.0016 \times \frac{1}{800} = 16 \times 10^{-4} \times \frac{1}{8} \times 10^{-2} = 2 \times 10^{-6}\; Nm$.

(B) Let $m$ and $r$ be the mass and radius of both the hollow cylinder and the solid sphere.
The moment of inertia of the hollow cylinder about its standard axis is $I_{c} = m r^{2}$.
The moment of inertia of the solid sphere about an axis passing through its centre is $I_{s} = \frac{2}{5} m r^{2}$.
We use the relation $\tau = I \alpha$,where $\tau$ is the torque,$I$ is the moment of inertia,and $\alpha$ is the angular acceleration.
Since the applied torque $\tau$ is the same for both,we have $\alpha = \frac{\tau}{I}$.
Thus,$\alpha \propto \frac{1}{I}$.
Since $I_{c} = m r^{2}$ and $I_{s} = 0.4 m r^{2}$,it is clear that $I_{s} < I_{c}$.
Therefore,$\alpha_{s} > \alpha_{c}$.
Using the relation $\omega = \omega_{0} + \alpha t$,for an initial angular velocity $\omega_{0} = 0$,we get $\omega = \alpha t$.
Since $\alpha_{s} > \alpha_{c}$,it follows that $\omega_{s} > \omega_{c}$ for any given time $t$.
Hence,the solid sphere will acquire a greater angular speed.

(A) Mass of the hollow cylinder,$m = 3\; kg$.
Radius of the hollow cylinder,$r = 40\; cm = 0.4\; m$.
Applied force,$F = 30\; N$.
The moment of inertia of the hollow cylinder about its geometric axis is given by $I = m r^2$.
$I = 3 \times (0.4)^2 = 3 \times 0.16 = 0.48\; kg\; m^2$.
Torque,$\tau = F \times r = 30 \times 0.4 = 12\; Nm$.
Using the relation $\tau = I \alpha$,the angular acceleration $\alpha$ is:
$\alpha = \frac{\tau}{I} = \frac{12}{0.48} = 25\; rad\; s^{-2}$.
The linear acceleration $a$ is given by $a = r \alpha$.
$a = 0.4 \times 25 = 10\; m\; s^{-2}$.

Angular acceleration is defined as the time rate of change of angular velocity of a rotating body.
It is denoted by the symbol $\alpha$ and is a vector quantity.
The $SI$ unit of angular acceleration is $\text{rad/s}^2$ and its dimensional formula is $[M^0 L^0 T^{-2}]$.
If the axis of rotation is fixed,the direction of the angular velocity vector $\vec{\omega}$ and the angular acceleration vector $\vec{\alpha}$ remains constant. In this specific situation,angular acceleration can be treated as a scalar quantity.
Mathematically,$\alpha = \frac{d\omega}{dt}$. Since angular velocity $\omega = \frac{d\theta}{dt}$,we can express angular acceleration as the second derivative of angular displacement:
$\alpha = \frac{d^2\theta}{dt^2}$.

By definition of angular momentum of a particle,$\vec{l} = \vec{r} \times \vec{p}$.
Differentiating this equation with respect to time $t$,we get $\frac{d\vec{l}}{dt} = \vec{r} \times \frac{d\vec{p}}{dt} + \frac{d\vec{r}}{dt} \times \vec{p}$.
Since $\frac{d\vec{p}}{dt}$ is the rate of change of linear momentum,which equals the force $\vec{F}$,and $\frac{d\vec{r}}{dt} = \vec{v}$,we have $\frac{d\vec{l}}{dt} = \vec{r} \times \vec{F} + \vec{v} \times \vec{p}$.
Because $\vec{v}$ and $\vec{p}$ are in the same direction,their cross product $\vec{v} \times \vec{p} = 0$.
Therefore,$\frac{d\vec{l}}{dt} = \vec{r} \times \vec{F} = \vec{\tau}$.
Thus,the time rate of change of angular momentum is equal to the torque acting on the particle.

(B) The angular momentum $L$ of a particle is given by $L = r \times p$.
The time rate of change of angular momentum is given by $\frac{dL}{dt}$.
According to Newton's second law for rotation,the torque $\tau$ acting on a system is equal to the time rate of change of angular momentum:
$\tau = \frac{dL}{dt}$.
Therefore,the physical quantity represented by the time rate of change of angular momentum is torque.

(N/A) For a system of particles performing rotational motion about a fixed axis,Newton's second law is expressed in terms of torque and angular acceleration.
The relationship is given by the equation: $\tau_{ext} = I \alpha$
Where:
- $\tau_{ext}$ is the net external torque acting on the system about the axis of rotation.
- $I$ is the moment of inertia of the system about the same axis.
- $\alpha$ is the angular acceleration of the system.
This equation is the rotational analogue of Newton's second law for linear motion,$F = ma$.

(N/A) The rotational kinetic energy of a rigid body is given by $K = \frac{1}{2} I \omega^{2}$.
The rate at which work is done on the body is equal to the rate at which its kinetic energy increases. The rate of increase of kinetic energy is:
$\frac{dK}{dt} = \frac{d}{dt} \left( \frac{1}{2} I \omega^{2} \right)$
$P = \frac{1}{2} I \frac{d}{dt} (\omega^{2})$
Since $I$ is constant for a rigid body:
$P = \frac{1}{2} I \times 2 \omega \frac{d\omega}{dt}$
$P = I \omega \alpha$,where $\alpha = \frac{d\omega}{dt}$ is the angular acceleration.
We also know that the power delivered by a torque is $P = \tau \omega$.
Equating the two expressions for power:
$\tau \omega = I \omega \alpha$
$\tau = I \alpha$
This equation is the rotational analogue of Newton's second law for linear motion,$F = ma$. Thus,the angular acceleration $\alpha$ is directly proportional to the applied torque $\tau$ and inversely proportional to the moment of inertia $I$ of the body.

(N/A) For a rigid body rotating about a fixed axis,Newton's second law of motion is expressed in terms of torque and angular acceleration.
It states that the net external torque $\tau_{ext}$ acting on the body about the axis of rotation is equal to the product of the moment of inertia $I$ of the body about that axis and its angular acceleration $\alpha$.
The mathematical expression is: $\tau_{ext} = I \alpha$.

(N/A) The total angular momentum of a rigid body rotating about a fixed axis is given by $\vec{L} = \vec{L}_{z} + \vec{L}_{\perp}$.
Differentiating with respect to time $t$:
$\frac{d\vec{L}}{dt} = \frac{d\vec{L}_{z}}{dt} + \frac{d\vec{L}_{\perp}}{dt}$.
Since the rotation is about a fixed axis,the component of angular momentum perpendicular to the axis $\vec{L}_{\perp}$ remains constant,so $\frac{d\vec{L}_{\perp}}{dt} = 0$.
The rate of change of angular momentum along the axis is equal to the external torque $\vec{\tau}$,so $\frac{d\vec{L}_{z}}{dt} = \vec{\tau}$.
Thus,$\frac{d\vec{L}}{dt} = \vec{\tau}$.
Since $\vec{L}_{z} = I\omega\hat{k}$,where $I$ is the moment of inertia and $\omega$ is the angular velocity:
$\vec{\tau} = \frac{d}{dt}(I\omega\hat{k}) = I\frac{d\omega}{dt}\hat{k}$.
Since $\alpha = \frac{d\omega}{dt}$,we get $\vec{\tau} = I\vec{\alpha}$.
In scalar form,this is $\tau = I\alpha$,which is the rotational analogue of Newton's second law $F = ma$.

(B) The time rate of change of angular momentum is equal to the net external torque acting on the system.
Mathematically,this is expressed as:
$\frac{d\vec{L}}{dt} = \vec{\tau}$
where $\vec{L}$ is the angular momentum and $\vec{\tau}$ is the external torque.

(A) The value of the torque is $0$.
Since the object moves with a constant angular speed, the angular acceleration $\alpha = 0$.
As the torque $\vec{\tau} = I\vec{\alpha}$, where $I$ is the moment of inertia, if $\alpha = 0$, then $\vec{\tau} = 0$.
Alternatively, in uniform circular motion, the net force is the centripetal force $\vec{F}_{c}$ directed towards the center.
The torque is defined as $\vec{\tau} = \vec{r} \times \vec{F}$. Since the centripetal force acts along the radius vector (or is anti-parallel to it), the angle $\theta$ between $\vec{r}$ and $\vec{F}$ is $0^{\circ}$ or $180^{\circ}$.
Thus, $|\vec{\tau}| = rF \sin(0^{\circ} \text{ or } 180^{\circ}) = 0$.

(D) The linear acceleration $(a)$ of the center of mass is given by Newton's second law:
$a = \frac{F}{m} = \frac{20 \, N}{2 \, kg} = 10 \, m/s^2$.
The torque $(\tau)$ about the center of mass is given by $\tau = F \times d$,where $d$ is the perpendicular distance from the center of mass to the line of action of the force. Since the force $F$ is applied at the center of mass,the perpendicular distance $d = 0$.
Therefore,$\tau = F \times 0 = 0$.
Using the relation $\tau = I \alpha$,where $I$ is the moment of inertia and $\alpha$ is the angular acceleration:
$0 = I \alpha \implies \alpha = 0 \, rad/s^2$.
However,if the question implies the force is applied at the edge (tangentially) to cause rotation,then $\tau = F \times r = 20 \times 0.1 = 2 \, N \cdot m$.
Assuming a solid disk,$I = \frac{1}{2} m r^2 = \frac{1}{2} \times 2 \times (0.1)^2 = 0.01 \, kg \cdot m^2$.
Then $\alpha = \frac{\tau}{I} = \frac{2}{0.01} = 200 \, rad/s^2$.
The ratio $\frac{a}{\alpha} = \frac{10}{200} = \frac{1}{20}$.

(C) Given: Mass $m = 20\, kg$,Radius $R = 0.2\, m$,Force $F = 20\, N$,Initial angular speed $\omega_0 = 0$,Final angular speed $\omega = 50\, rad/s$.
The torque $\tau$ applied by the force $F$ is $\tau = F \cdot R$.
The moment of inertia of the disk about its center is $I = \frac{1}{2} mR^2$.
Using $\tau = I\alpha$,the angular acceleration $\alpha$ is:
$\alpha = \frac{F \cdot R}{\frac{1}{2} mR^2} = \frac{2F}{mR} = \frac{2 \times 20}{20 \times 0.2} = \frac{2}{0.2} = 10\, rad/s^2$.
Using the kinematic equation $\omega^2 = \omega_0^2 + 2\alpha\Delta\theta$:
$(50)^2 = 0^2 + 2(10)\Delta\theta$
$2500 = 20\Delta\theta$
$\Delta\theta = 125\, rad$.
Since one revolution corresponds to $2\pi \approx 6.28\, rad$,the number of revolutions $n$ is:
$n = \frac{\Delta\theta}{2\pi} = \frac{125}{6.28} \approx 19.90$.
Rounding to the nearest integer,$n = 20$.

(D) Given: Radius $R = 20 \, \text{cm} = 0.2 \, \text{m}$,Mass $M = 10 \, \text{kg}$,Initial angular velocity $\omega_i = 600 \, \text{rpm} = \frac{600 \times 2\pi}{60} \, \text{rad/s} = 20\pi \, \text{rad/s}$.
Final angular velocity $\omega_f = 0 \, \text{rad/s}$,Time $\Delta t = 10 \, \text{s}$.
The moment of inertia of the disc is $I = \frac{1}{2}MR^2 = \frac{1}{2} \times 10 \times (0.2)^2 = 5 \times 0.04 = 0.2 \, \text{kg m}^2$.
The torque required is $\tau = \frac{\Delta L}{\Delta t} = \frac{I(\omega_f - \omega_i)}{\Delta t}$.
Magnitude of torque $|\tau| = \frac{0.2 \times (20\pi - 0)}{10} = \frac{4\pi}{10} = 0.4\pi = 4\pi \times 10^{-1} \, \text{Nm}$.
Comparing with $x\pi \times 10^{-1} \, \text{Nm}$,we get $x = 4$.

(A) The torque applied is $\tau = F \cdot r = (12t - 3t^2) \cdot 1.5 = 18t - 4.5t^2$.
Using $\tau = I\alpha$,we have $18t - 4.5t^2 = 4.5\alpha$,which gives $\alpha = 4t - t^2$.
Since $\alpha = \frac{d\omega}{dt}$,we integrate to find angular velocity: $\omega = \int (4t - t^2) dt = 2t^2 - \frac{t^3}{3}$.
The motion reverses when $\omega = 0$,so $t^2(2 - \frac{t}{3}) = 0$,giving $t = 6\,s$.
The angular displacement $\theta$ is $\int_{0}^{6} (2t^2 - \frac{t^3}{3}) dt = [\frac{2t^3}{3} - \frac{t^4}{12}]_{0}^{6} = \frac{2(216)}{3} - \frac{1296}{12} = 144 - 108 = 36\,rad$.
The number of rotations is $N = \frac{\theta}{2\pi} = \frac{36}{2\pi} = \frac{18}{\pi}$.
Comparing with $\frac{K}{\pi}$,we get $K = 18$.

(A) Given that the power $P$ is constant.
Work done $W = P \cdot t$.
By the work-energy theorem,$W = \Delta K = \frac{1}{2} m v^2$.
Since $v = r \omega$,we have $P t = \frac{1}{2} m (r \omega)^2 = \frac{1}{2} m r^2 \omega^2$.
Thus,$\omega^2 = \frac{2 P t}{m r^2}$,which implies $\omega = \sqrt{\frac{2 P}{m r^2}} \cdot t^{1/2}$.
Angular acceleration $\alpha = \frac{d\omega}{dt} = \frac{d}{dt} \left( \sqrt{\frac{2 P}{m r^2}} \cdot t^{1/2} \right)$.
$\alpha = \sqrt{\frac{2 P}{m r^2}} \cdot \frac{1}{2} t^{-1/2}$.
Therefore,$\alpha \propto t^{-1/2}$ or $\alpha \propto \frac{1}{\sqrt{t}}$.

(A) Given:
Angular acceleration $\alpha = 3 \, rad/s^2$
Initial angular velocity $\omega_1 = 10 \, rad/s$
Final angular velocity $\omega_2 = 20 \, rad/s$
Using the kinematic equation for rotational motion:
$\omega_2^2 = \omega_1^2 + 2 \alpha \theta$
Substituting the given values:
$(20)^2 = (10)^2 + 2 \times 3 \times \theta$
$400 = 100 + 6 \theta$
$300 = 6 \theta$
$\theta = \frac{300}{6} = 50 \, rad$
Thus,the angle rotated is $50 \, rad$.

(A) Given:
Moment of inertia,$I = 2 \, kg \cdot m^2$
Initial angular velocity,$\omega_0 = 30 \, rad/s$
Final angular velocity,$\omega_f = 0 \, rad/s$
Time taken to stop,$t = 15 \, s$
Using the equation of rotational motion:
$\omega_f = \omega_0 + \alpha t$
$0 = 30 + \alpha(15)$
$\alpha = -30 / 15 = -2 \, rad/s^2$
The magnitude of angular deceleration is $|\alpha| = 2 \, rad/s^2$.
Average torque is given by:
$\tau = I |\alpha|$
$\tau = 2 \, kg \cdot m^2 \times 2 \, rad/s^2 = 4 \, N \cdot m$.

(A) The moment of inertia of a disc about its central axis is $I = \frac{1}{2} M R^2$.
Using the rotational analogue of Newton's second law,the torque $\tau$ is given by $\tau = I \alpha$,where $\alpha$ is the angular acceleration.
The torque applied by the tangential force $F$ is $\tau = F \times R$.
Equating the two expressions for torque: $F R = \frac{1}{2} M R^2 \alpha$.
Solving for angular acceleration: $\alpha = \frac{2 F}{M R}$.
Since the disc comes to rest from an initial angular speed $\omega$ in time $t$,the magnitude of angular deceleration is $\alpha = \frac{\omega}{t}$.
Substituting $\alpha$ into the equation: $\frac{2 F}{M R} = \frac{\omega}{t}$.
Solving for $F$: $F = \frac{M R \omega}{2 t}$.

(A) The torque $\tau$ acting on the rod about the hinge point is given by $\tau = r_{\perp} F$,where $r_{\perp}$ is the perpendicular distance from the axis of rotation to the line of action of the force.
From the geometry,$r_{\perp} = L \sin 30^{\circ} = L \times \frac{1}{2} = \frac{L}{2}$.
Thus,the initial torque is $\tau = F \times \frac{L}{2} = \frac{FL}{2}$.
The moment of inertia $I$ of a thin rod of mass $M$ and length $L$ about an axis passing through one end is $I = \frac{ML^2}{3}$.
Using the relation $\tau = I \alpha$,the initial angular acceleration $\alpha$ is given by:
$\alpha = \frac{\tau}{I} = \frac{FL/2}{ML^2/3} = \frac{FL}{2} \times \frac{3}{ML^2} = \frac{3F}{2ML}$.
Therefore,the correct option is $A$.

(B) The moment of inertia of a hollow sphere about its diameter is given by $I = \frac{2}{3}MR^2$.
Given: Mass $M = 1 \,kg$,Radius $R = 10 \,cm = 0.1 \,m$,Force $F = 30 \,N$.
The torque applied is $\tau = F \times R$.
Using the relation $\tau = I \alpha$,where $\alpha$ is the angular acceleration:
$F \times R = \left(\frac{2}{3}MR^2\right) \alpha$
$30 \times 0.1 = \left(\frac{2}{3} \times 1 \times (0.1)^2\right) \alpha$
$3 = \left(\frac{2}{3} \times 0.01\right) \alpha$
$3 = \frac{0.02}{3} \alpha$
$\alpha = \frac{3 \times 3}{0.02} = \frac{9}{0.02} = 450 \,rad/s^2$.

(C) The net torque $\tau$ acting on the disc about its centre is the sum of the torques produced by the two forces.
Since both forces are applied tangentially at a distance $R$ from the centre,each force produces a torque of magnitude $\tau_i = F \times R$.
Both forces create a torque in the same rotational direction (counter-clockwise),so the total torque is $\tau_{net} = F R + F R = 2 F R$.
The moment of inertia of a uniform disc of mass $M$ and radius $R$ about an axis passing through its centre and perpendicular to its plane is $I = \frac{1}{2} M R^2$.
Using the rotational analogue of Newton's second law,$\tau_{net} = I \alpha$,where $\alpha$ is the angular acceleration:
$2 F R = (\frac{1}{2} M R^2) \alpha$
Solving for $\alpha$:
$\alpha = \frac{2 F R}{\frac{1}{2} M R^2} = \frac{4 F}{M R}$.

(B) Given: Moment of inertia $I = 4 \,kg \cdot m^2$,initial angular velocity $\omega_0 = 240 \,rpm$,time $t = 1 \,min = 60 \,s$,and final angular velocity $\omega = 0 \,rad/s$.
First,convert the angular velocity from $rpm$ to $rad/s$:
$\omega_0 = 240 \times \frac{2 \pi}{60} = 8 \pi \,rad/s$.
Using the equation of rotational motion $\omega = \omega_0 - \alpha t$,where $\alpha$ is the angular deceleration:
$0 = 8 \pi - \alpha(60) \implies \alpha = \frac{8 \pi}{60} = \frac{2 \pi}{15} \,rad/s^2$.
The torque $\tau$ required is given by $\tau = I \alpha$:
$\tau = 4 \times \frac{2 \pi}{15} = \frac{8 \pi}{15} \,N \cdot m$.

(A) The relationship between torque $\tau$ and the rate of change of angular momentum $L$ is given by the equation: $\tau = \frac{\Delta L}{\Delta t}$.
Given the initial angular momentum $L_i = A_0$ and the final angular momentum $L_f = 4 A_0$.
The time interval $\Delta t = 4 \text{ s}$.
Substituting these values into the formula:
$\tau = \frac{4 A_0 - A_0}{4} = \frac{3 A_0}{4}$.
Thus,the magnitude of the torque is $\frac{3 A_0}{4}$.

(A) The torque $\tau$ acting on the disc about the pivot point $P$ is due to the gravitational force $mg$ acting at the centre of mass $C$.
When the line $PC$ makes an angle $\theta$ with the horizontal,the perpendicular distance from $P$ to the line of action of the gravitational force is $R \cos \theta$.
Therefore,the torque is $\tau = mg(R \cos \theta)$.
The moment of inertia $I$ of the disc about an axis passing through the pivot point $P$ (which is on the rim) is given by the parallel axis theorem: $I = I_{cm} + mR^2 = \frac{1}{2} mR^2 + mR^2 = \frac{3}{2} mR^2$.
Using the relation $\tau = I \alpha$,we have:
$mg(R \cos \theta) = \left( \frac{3}{2} mR^2 \right) \alpha$
Solving for the angular acceleration $\alpha$:
$\alpha = \frac{mgR \cos \theta}{\frac{3}{2} mR^2} = \frac{2g \cos \theta}{3R}$.

(B) The rate of change of angular momentum is equal to the net torque acting on the system,i.e.,$\left|\frac{dL}{dt}\right| = \tau_{\text{net}}$.
For each mass $m$ at a distance $l$ from the centre,the radius of the circular path is $r = l\sin\alpha$.
The centripetal force acting on each mass is $F = mr\omega^2 = m(l\sin\alpha)\omega^2$.
The torque $\tau$ due to one mass about the centre is $\tau = F \times r_{\perp}$,where $r_{\perp} = l\cos\alpha$ is the perpendicular distance from the axis of rotation to the force vector.
Thus,$\tau = (ml\sin\alpha\omega^2) \times (l\cos\alpha) = ml^2\omega^2\sin\alpha\cos\alpha$.
Since there are two such masses on opposite sides of the centre,the torques produced by them are in the same direction.
Therefore,the net torque is $\tau_{\text{net}} = 2\tau = 2ml^2\omega^2\sin\alpha\cos\alpha$.
Using the trigonometric identity $\sin 2\alpha = 2\sin\alpha\cos\alpha$,we get:
$\tau_{\text{net}} = ml^2\omega^2\sin 2\alpha$.
Hence,$\left|\frac{dL}{dt}\right| = ml^2\omega^2\sin 2\alpha$.

(B) The torque $\tau$ acting on the cylinder is given by $\tau = F \times R$.
For a hollow cylinder,the moment of inertia $I$ about its central axis is $I = mR^2$.
Using the rotational form of Newton's second law,$\tau = I \alpha$,where $\alpha$ is the angular acceleration.
Substituting the expressions,we get $F \times R = (mR^2) \alpha$.
Simplifying for $\alpha$,we have $\alpha = \frac{F}{mR}$.
Given $F = 52.5\,N$,$m = 5\,kg$,and $R = 70\,cm = 0.7\,m$.
Calculating the value: $\alpha = \frac{52.5}{5 \times 0.7} = \frac{52.5}{3.5} = 15\,rad\,s^{-2}$.

(A) Given: Moment of inertia $I = 0.40 \ kg \cdot m^2$,radius $R = 10 \ cm = 0.1 \ m$,force $F = 40 \ N$,time $t = 10 \ s$,initial angular velocity $\omega_0 = 0 \ rad/s$.
The torque $\tau$ applied to the wheel is given by $\tau = F \times R$.
$\tau = 40 \times 0.1 = 4 \ N \cdot m$.
Using the relation $\tau = I \alpha$,where $\alpha$ is the angular acceleration:
$4 = 0.40 \times \alpha \Rightarrow \alpha = \frac{4}{0.40} = 10 \ rad/s^2$.
Using the equation of motion $\omega = \omega_0 + \alpha t$:
$\omega = 0 + (10 \times 10) = 100 \ rad/s$.
Thus,$x = 100$.

(A) The torque $\tau$ produced by the force $F$ acting at the rim of the wheel is given by $\tau = F \times R$.
Given: Force $F = 10 \ N$,Radius $R = 0.2 \ m$,and angular acceleration $\alpha = 2 \ rad/s^2$.
The relationship between torque,moment of inertia $I$,and angular acceleration $\alpha$ is $\tau = I \alpha$.
Substituting the values:
$F \times R = I \alpha$
$10 \times 0.2 = I \times 2$
$2 = 2I$
$I = 1 \ kg \ m^2$.
Thus,the moment of inertia of the wheel is $1 \ kg \ m^2$.

(C) The torque $\vec{\tau}$ is defined as the rate of change of angular momentum: $\vec{\tau} = \frac{d\vec{L}}{dt}$.
Since $\vec{L} = \vec{r} \times \vec{p}$,expression $B$ is correct: $\vec{\tau} = \frac{d}{dt}(\vec{r} \times \vec{p})$.
Using the product rule: $\frac{d}{dt}(\vec{r} \times \vec{p}) = (\frac{d\vec{r}}{dt} \times \vec{p}) + (\vec{r} \times \frac{d\vec{p}}{dt})$.
Since $\frac{d\vec{r}}{dt} = \vec{v}$ and $\vec{p} = m\vec{v}$,the term $(\vec{v} \times m\vec{v}) = 0$. Thus,$\vec{\tau} = \vec{r} \times \frac{d\vec{p}}{dt}$,so expression $C$ is correct.
Since $\frac{d\vec{p}}{dt} = \vec{F}$,expression $E$ is correct: $\vec{\tau} = \vec{r} \times \vec{F}$.
For a rigid body rotating about a fixed axis,$\vec{\tau} = I\vec{\alpha}$,so expression $D$ is correct.
Expression $A$ is incorrect as $\vec{\tau} = \frac{d\vec{L}}{dt}$,not $\vec{r} \times \vec{L}$.
Therefore,$B, C, D,$ and $E$ are correct.

(D) Given: Mass $m = 1 \ kg$.
Initial angular velocity $\omega_{i} = 1800 \ rpm = 1800 \times \frac{2 \pi}{60} = 60 \pi \ rad/s$.
Final angular velocity $\omega_{f} = 2100 \ rpm = 2100 \times \frac{2 \pi}{60} = 70 \pi \ rad/s$.
External torque $\tau_{ext} = 25 \pi \ Nm$.
Time $t = 40 \ s$.
Using the equation of rotational motion,$\omega_{f} = \omega_{i} + \alpha t$:
$70 \pi = 60 \pi + \alpha(40) \implies 10 \pi = 40 \alpha \implies \alpha = \frac{\pi}{4} \ rad/s^2$.
For a disk rotating about its diameter,the moment of inertia is $I = \frac{mR^2}{4}$.
Using $\tau = I \alpha$:
$25 \pi = \left( \frac{1 \times R^2}{4} \right) \times \frac{\pi}{4}$.
$25 = \frac{R^2}{16} \implies R^2 = 400 \implies R = 20 \ m$.
The diameter of the disk is $D = 2R = 2 \times 20 = 40 \ m$.

(D) The moment of inertia $I$ of a uniform solid disc about its central axis is given by:
$I = \frac{1}{2} mR^2$
Substituting the given values ($m = 2 \ kg$,$R = 0.5 \ m$):
$I = \frac{1}{2} \times 2 \times (0.5)^2 = 0.25 \ kg \cdot m^2$
The torque $\tau$ produced by the force $F = 2.5 \ N$ applied at the rim (radius $R = 0.5 \ m$) is:
$\tau = F \times R = 2.5 \times 0.5 = 1.25 \ N \cdot m$
Using the relation $\tau = I \alpha$,we find the angular acceleration $\alpha$:
$1.25 = 0.25 \times \alpha$
$\alpha = \frac{1.25}{0.25} = 5 \ rad/s^2$
Since the pulley starts from rest $(\omega_0 = 0)$,the angular velocity $\omega$ after $t = 10 \ s$ is:
$\omega = \omega_0 + \alpha t$
$\omega = 0 + 5 \times 10 = 50 \ rad/s$

(D) Using the third equation of rotational motion: $\omega^2 = \omega_0^2 - 2\alpha\theta$.
Initially,the angular velocity becomes $\omega = \frac{\omega_0}{2}$ after $\theta_1 = 36 \times 2\pi$ radians.
Substituting these values: $(\frac{\omega_0}{2})^2 = \omega_0^2 - 2\alpha(36 \times 2\pi)$.
$\frac{\omega_0^2}{4} = \omega_0^2 - 144\pi\alpha$,which gives $144\pi\alpha = \frac{3\omega_0^2}{4}$,so $\alpha = \frac{3\omega_0^2}{576\pi} = \frac{\omega_0^2}{192\pi}$.
Now,for the motion from $\omega = \frac{\omega_0}{2}$ to $\omega = 0$,let the additional rotations be $n$. The angle covered is $\theta_2 = n \times 2\pi$.
Using $0^2 = (\frac{\omega_0}{2})^2 - 2\alpha(n \times 2\pi)$.
$\frac{\omega_0^2}{4} = 2(\frac{\omega_0^2}{192\pi})(n \times 2\pi)$.
$\frac{1}{4} = \frac{4n\pi}{192\pi} = \frac{n}{48}$.
$n = \frac{48}{4} = 12$.
Thus,the fan will make $12$ more rotations before coming to rest.

(C) The impulse $J$ applied at one end provides both linear and angular momentum to the system.
$1$. Linear impulse equation: $J = M_{total} v_{cm} \Rightarrow MV = (2M) v_{cm} \Rightarrow v_{cm} = \frac{V}{2}$.
$2$. Angular impulse equation about the center of mass $(COM)$: $J \times r = I_{cm} \omega$.
Here,$r = \frac{L}{2}$ and $I_{cm} = M(\frac{L}{2})^2 + M(\frac{L}{2})^2 = \frac{ML^2}{2}$.
Substituting the values: $(MV) \times \frac{L}{2} = (\frac{ML^2}{2}) \omega$.
Solving for $\omega$: $\frac{MVL}{2} = \frac{ML^2 \omega}{2} \Rightarrow \omega = \frac{V}{L}$.