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

(C) According to the rotational analogue of Newton's second law,the torque $\tau$ acting on a rigid body is given by the relation $\tau = I\alpha$,where $I$ is the moment of inertia and $\alpha$ is the angular acceleration.
Since the torque $\tau$ is constant and the moment of inertia $I$ is a property of the body,the angular acceleration $\alpha = \frac{\tau}{I}$ must also be constant.
Therefore,the object undergoes angular acceleration.

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

(B) The relationship between torque and angular acceleration is given by $ \tau = I \alpha $, where $ I $ is the moment of inertia and $ \alpha $ is the angular acceleration.
Given that the external torque $ \tau = 0 $.
Since the moment of inertia $ I $ of a rigid body cannot be zero, it follows that $ \alpha = 0 $.
Therefore, the angular acceleration of the system is zero.

(C) Given: Power $P = 100 \, kW = 100 \times 10^3 \, W$.
Angular speed $n = 1800 \, rpm = \frac{1800}{60} \, rev/s = 30 \, rev/s$.
The angular velocity $\omega$ is given by $\omega = 2\pi n = 2\pi \times 30 = 60\pi \, rad/s$.
The relationship between power $P$,torque $\tau$,and angular velocity $\omega$ is $P = \tau \omega$.
Therefore,the torque $\tau = \frac{P}{\omega} = \frac{100 \times 10^3}{60\pi} \, N-m$.
Calculating the value: $\tau = \frac{100000}{188.495} \approx 530.5 \, N-m$,which rounds to $531 \, N-m$.

(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 change in angular momentum $\Delta L = L_2 - L_1 = 4A_0 - A_0 = 3A_0$.
The time interval is $\Delta t = 4 \ s$.
Therefore,the magnitude of the constant torque is $\tau = \frac{\Delta L}{\Delta t} = \frac{3A_0}{4}$.

(A) For a rigid body rotating about a fixed axis,the torque $\tau$ acting on the body is related to its moment of inertia $I$ and angular acceleration $\alpha$ by the equation $\tau = I\alpha$. This is the rotational analogue of Newton's second law of motion,$F = ma$.

(A) Given: Mass $m = 10 \ kg$,Radius $r = 30 \ cm = 0.3 \ m$,Initial angular velocity $\omega_1 = 10 \ rad/s$,Final angular velocity $\omega_2 = 0 \ rad/s$,Time $t = 10 \ s$.
First,calculate the angular acceleration $\alpha$:
$\alpha = \frac{\omega_2 - \omega_1}{t} = \frac{0 - 10}{10} = -1 \ rad/s^2$.
The magnitude of angular acceleration is $|\alpha| = 1 \ rad/s^2$.
Next,calculate the moment of inertia $I$ of the mass:
$I = m r^2 = 10 \times (0.3)^2 = 10 \times 0.09 = 0.9 \ kg \cdot m^2$.
Finally,calculate the torque $\tau$:
$\tau = I |\alpha| = 0.9 \times 1 = 0.9 \ N \cdot m$.

(C) Given: Moment of inertia $I = 2 \; kg \cdot m^2$,Initial angular velocity $\omega_i = 60 \; rpm = \frac{60 \times 2\pi}{60} \; rad/s = 2\pi \; rad/s$,Final angular velocity $\omega_f = 0 \; rad/s$,Time $t = 1 \; minute = 60 \; s$.
The angular acceleration $\alpha$ is given by $\alpha = \frac{\omega_f - \omega_i}{t} = \frac{0 - 2\pi}{60} = -\frac{\pi}{30} \; rad/s^2$.
The magnitude of the torque $\tau$ is given by $\tau = |I \alpha| = 2 \times \left| -\frac{\pi}{30} \right| = \frac{2\pi}{30} = \frac{\pi}{15} \; N \cdot m$.

(A) Given: Moment of inertia $I = 5 \times 10^{-3} \ kg \ m^2$, initial frequency $n_1 = 20 \ rev/s$, final frequency $n_2 = 0 \ rev/s$, time $t = 10 \ s$.
The angular acceleration $\alpha$ is given by $\alpha = \frac{\omega_2 - \omega_1}{t} = \frac{2\pi(n_2 - n_1)}{t}$.
Substituting the values: $\alpha = \frac{2\pi(0 - 20)}{10} = -4\pi \ rad/s^2$.
The magnitude of torque $\tau$ is given by $\tau = I|\alpha|$.
$\tau = (5 \times 10^{-3}) \times (4\pi) = 20\pi \times 10^{-3} = 2\pi \times 10^{-2} \ N \ m$.
Thus, the required value is $2\pi$.

(A) Given: Moment of inertia $I = 200 \, kg \cdot m^2$,Torque $\tau = 1000 \, N \cdot m$,Time $t = 3 \, s$,Initial angular velocity $\omega_0 = 0 \, rad/s$.
Using the relation $\tau = I \alpha$,the angular acceleration $\alpha$ is:
$\alpha = \frac{\tau}{I} = \frac{1000}{200} = 5 \, rad/s^2$.
Using the equation of rotational motion $\omega = \omega_0 + \alpha t$:
$\omega = 0 + (5 \times 3) = 15 \, rad/s$.

(A) The initial angular velocity of the Earth is $\omega_1 = \frac{2\pi}{T}$ rad/s,where $T = 86400 \, s$. The final angular velocity is $\omega_2 = 0$. The time taken is $t = 86400 \, s$.
The angular acceleration is $\alpha = \frac{\omega_2 - \omega_1}{t} = \frac{0 - \frac{2\pi}{86400}}{86400} = -\frac{2\pi}{(86400)^2} \, rad/s^2$.
The torque required is $\tau = I\alpha$,where $I = \frac{2}{5}MR^2$ is the moment of inertia of the Earth.
The tangential force $F$ is related to torque by $\tau = FR$,so $F = \frac{I\alpha}{R} = \frac{2}{5}MR\alpha$.
Substituting the values: $M = 6 \times 10^{24} \, kg$,$R = 6.4 \times 10^6 \, m$,and $\alpha = \frac{2\pi}{(86400)^2}$.
$F = \frac{2}{5} \times (6 \times 10^{24}) \times (6.4 \times 10^6) \times \frac{2\pi}{(86400)^2} \approx 1.3 \times 10^{22} \, N$.

(A) The torque $\tau$ acting on a body is equal to the rate of change of its angular momentum $L$,given by the formula: $\tau = \frac{dL}{dt}$.
Given:
Initial angular momentum $L_1 = 2L$.
Final angular momentum $L_2 = 5L$.
Time interval $\Delta t = 3 \ s$.
Substituting these values into the formula:
$\tau = \frac{L_2 - L_1}{\Delta t} = \frac{5L - 2L}{3} = \frac{3L}{3} = L$.
Thus,the magnitude of the torque acting on the wheel is $L$.

(A) Given: Torque $\tau = 30 \, N \cdot m$,Moment of inertia $I = 2 \, kg \cdot m^2$,Time $t = 10 \, s$,Initial angular velocity $\omega_0 = 0$.
First,calculate the angular acceleration $\alpha$ using the relation $\tau = I\alpha$:
$\alpha = \frac{\tau}{I} = \frac{30}{2} = 15 \, rad/s^2$.
Now,use the kinematic equation for rotational motion to find the angular displacement $\theta$:
$\theta = \omega_0 t + \frac{1}{2} \alpha t^2$.
Substituting the values:
$\theta = 0 \times 10 + \frac{1}{2} \times 15 \times (10)^2 = 0.5 \times 15 \times 100 = 750 \, rad$.

(C) When a force acts on a body at a point other than the center of mass,it provides both a net force and a net torque about the center of mass.
According to Newton's second law for linear motion,$F = ma$,the linear acceleration $a$ changes.
According to the rotational analogue of Newton's second law,$\tau = I\alpha$,the torque $\tau$ causes a change in angular acceleration $\alpha$.
Therefore,both linear and angular accelerations change.

(B) The rotational kinetic energy is given by $K = \frac{1}{2} I \omega^2$.
Given $K = 1500 \ J$ and $I = 1.2 \ kg \cdot m^2$,we have:
$1500 = \frac{1}{2} \times 1.2 \times \omega^2$
$1500 = 0.6 \times \omega^2$
$\omega^2 = \frac{1500}{0.6} = 2500$
$\omega = 50 \ rad/s$.
Since the body starts from rest,the initial angular velocity $\omega_0 = 0$.
Using the equation of motion $\omega = \omega_0 + \alpha t$,where $\alpha = 25 \ rad/s^2$:
$50 = 0 + 25 \times t$
$t = \frac{50}{25} = 2 \ s$.

(B) Given:
Initial frequency $n_1 = 20 \ rev/s$
Final frequency $n_2 = 0 \ rev/s$
Time $t = 20 \ s$
Initial angular velocity $\omega_1 = 2\pi n_1 = 2\pi \times 20 = 40\pi \ rad/s$
Final angular velocity $\omega_2 = 2\pi n_2 = 0 \ rad/s$
Using the equation of motion $\omega_2 = \omega_1 + \alpha t$:
$0 = 40\pi + \alpha(20)$
$\alpha = -\frac{40\pi}{20} = -2\pi \ rad/s^2$
The magnitude of angular deceleration is $2\pi \ rad/s^2$.

(A) Given: Moment of inertia $I = 5 \times 10^{-3} \ kg \cdot m^2$,initial frequency $n_1 = 20 \ rev/s$,final frequency $n_2 = 0 \ rev/s$,time $t = 10 \ s$.
Initial angular velocity $\omega_1 = 2\pi n_1 = 2\pi \times 20 = 40\pi \ rad/s$.
Final angular velocity $\omega_2 = 2\pi n_2 = 0 \ rad/s$.
Angular acceleration $\alpha = \frac{\omega_2 - \omega_1}{t} = \frac{0 - 40\pi}{10} = -4\pi \ rad/s^2$.
The magnitude of torque required is $\tau = I|\alpha|$.
$\tau = (5 \times 10^{-3}) \times 4\pi = 20\pi \times 10^{-3} = 2\pi \times 10^{-2} \ N \cdot m$.

(C) The center of mass $X_{cm}$ is given by:
$X_{cm} = \frac{\int x dm}{\int dm} = \frac{\int_{0}^{L} x (\lambda_0 x) dx}{\int_{0}^{L} (\lambda_0 x) dx} = \frac{\lambda_0 [x^3/3]_0^L}{\lambda_0 [x^2/2]_0^L} = \frac{L^3/3}{L^2/2} = \frac{2L}{3}$.
The total mass $M$ of the rod is:
$M = \int_{0}^{L} \lambda_0 x dx = \frac{\lambda_0 L^2}{2}$.
The moment of inertia $I$ about the hinge $O$ is:
$I = \int x^2 dm = \int_{0}^{L} x^2 (\lambda_0 x) dx = \lambda_0 \int_{0}^{L} x^3 dx = \frac{\lambda_0 L^4}{4}$.
Just after the string is cut,the torque $\tau$ about the hinge $O$ is due to gravity acting at the center of mass:
$\tau = Mg \times X_{cm} = \left( \frac{\lambda_0 L^2}{2} \right) g \left( \frac{2L}{3} \right) = \frac{\lambda_0 g L^3}{3}$.
Using the relation $\tau = I \alpha$:
$\frac{\lambda_0 g L^3}{3} = \left( \frac{\lambda_0 L^4}{4} \right) \alpha$.
Solving for $\alpha$:
$\alpha = \frac{\lambda_0 g L^3}{3} \times \frac{4}{\lambda_0 L^4} = \frac{4g}{3L}$.

(A) The torque $\tau$ acting on the rod about the hinge is given by $\tau = mg \cdot r_{\perp}$,where $r_{\perp}$ is the perpendicular distance from the hinge to the line of action of gravity.
When the rod makes an angle $\theta$ with the horizontal,the angle with the vertical is $(90^{\circ} - \theta)$. The center of mass is at $l/2$ from the hinge.
The perpendicular distance is $r_{\perp} = \frac{l}{2} \cos(90^{\circ} - \theta) = \frac{l}{2} \sin \theta$.
Using the rotational analog of Newton's second law,$\tau = I\alpha$,where $\alpha$ is the angular acceleration.
$mg \left( \frac{l}{2} \sin \theta \right) = I \alpha$.
Given $I = \frac{ml^2}{3}$,we have:
$mg \frac{l}{2} \sin \theta = \left( \frac{ml^2}{3} \right) \alpha$.
Solving for $\alpha$:
$\alpha = \frac{mg \frac{l}{2} \sin \theta}{\frac{ml^2}{3}} = \frac{3g \sin \theta}{2l}$.
For $\theta = 45^{\circ}$,$\sin 45^{\circ} = \frac{1}{\sqrt{2}}$.
$\alpha = \frac{3g}{2l} \cdot \frac{1}{\sqrt{2}} = \frac{3g}{2\sqrt{2}l}$.

(A) The torque $\tau$ applied to the pulley is $\tau = F \cdot R = (20t - 5t^2) \cdot 2 = 40t - 10t^2 \ N \cdot m$.
Using $\tau = I \alpha$,we get $\alpha = \frac{\tau}{I} = \frac{40t - 10t^2}{10} = 4t - t^2 \ rad/s^2$.
Since $\alpha = \frac{d\omega}{dt}$,we integrate to find angular velocity $\omega$:
$\omega = \int (4t - t^2) dt = 2t^2 - \frac{t^3}{3}$.
The direction of motion reverses when $\omega = 0$ (at $t > 0$):
$2t^2 - \frac{t^3}{3} = 0 \Rightarrow t^2(2 - \frac{t}{3}) = 0 \Rightarrow t = 6 \ s$.
Now,find the angular displacement $\theta$ by integrating $\omega$ from $t = 0$ to $t = 6$:
$\theta = \int_0^6 (2t^2 - \frac{t^3}{3}) dt = [\frac{2t^3}{3} - \frac{t^4}{12}]_0^6 = \frac{2(216)}{3} - \frac{1296}{12} = 144 - 108 = 36 \ rad$.
The number of rotations $n = \frac{\theta}{2\pi} = \frac{36}{2\pi} = \frac{18}{\pi} \approx \frac{18}{3.14} \approx 5.73$.
Since $5.73$ is between $3$ and $6$,the correct option is $A$.

(B) The moment of inertia of the space station about its central axis is $I = MR^2.$
Two rockets,each providing a thrust $T$ at a distance $R$ from the center,create a total torque $\tau = 2TR.$
According to the angular impulse-momentum theorem,the angular impulse equals the change in angular momentum: $\int \tau dt = L_f - L_i.$
Assuming the station starts from rest $(L_i = 0)$,we have $2TRt = I\omega = MR^2\omega.$
To simulate Earth's gravity,the radial acceleration at the rim must be $g,$ so $a_r = \omega^2R = g.$
Solving for angular velocity,we get $\omega = \sqrt{\frac{g}{R}}.$
Substituting $\omega$ into the impulse equation: $2TRt = MR^2 \sqrt{\frac{g}{R}} = M\sqrt{gR^3}.$
Solving for time $t,$ we get $t = \frac{M\sqrt{gR^3}}{2TR} = \frac{M\sqrt{gR}}{2T}.$
Thus,the correct option is $B.$

(A) Initial angular velocity $\omega_0 = 720 \text{ rpm} = \frac{720}{60} \text{ rev/s} = 12 \text{ rev/s}$.
Final angular velocity $\omega = 0 \text{ rev/s}$.
Time taken $t = 8 \text{ s}$.
Since the torque is constant,the angular acceleration is constant,and we can use the average angular velocity formula to find the total displacement (number of revolutions $\theta$):
$\theta = \left( \frac{\omega + \omega_0}{2} \right) \times t$
$\theta = \left( \frac{0 + 12}{2} \right) \times 8$
$\theta = 6 \times 8 = 48 \text{ revolutions}$.

(D) The angular acceleration $\alpha$ is defined as the rate of change of angular velocity $\omega$ with respect to time $t$,i.e.,$\alpha = \frac{d\omega}{dt}$.
From the given graph,we can determine the change in angular acceleration over time,but we have no information regarding the initial angular velocity $\omega_0$ at $t = 0$.
Since the sense of rotation depends on the sign of the angular velocity $\omega(t) = \omega_0 + \int \alpha dt$,and $\omega_0$ is unknown,it is impossible to determine the direction of rotation (clockwise or anti-clockwise) from the given graph alone.
Therefore,the data provided is insufficient to conclude the sense of rotation.

(A) The impulse $J$ applied by the cue provides both linear momentum and angular momentum to the ball.
Linear momentum: $J = mv_0$.
Angular impulse about the center of mass: $\tau \Delta t = J \times h = I \omega_0$.
Substituting $J = mv_0$,we get $mv_0h = I \omega_0$.
The moment of inertia of a solid sphere about its center is $I = \frac{2}{5}mr^2$.
Substituting $I$ into the equation: $mv_0h = (\frac{2}{5}mr^2) \omega_0$.
Solving for $\omega_0$: $\omega_0 = \frac{mv_0h}{\frac{2}{5}mr^2} = \frac{5v_0h}{2r^2}$.

(D) The center of mass $(COM)$ of a thin uniform hemispherical bowl of radius $R$ is at a distance $R/2$ from the center of its circular rim.
When a horizontal force $F$ is applied at the rim,the torque $\tau$ about the $COM$ is given by $\tau = F \cdot (R/2)$.
The moment of inertia $I_{COM}$ of the hemispherical bowl about an axis passing through its $COM$ and parallel to the rim is $I_{COM} = I_{axis} - m(R/2)^2$,where $I_{axis} = (2/3)MR^2$ is the moment of inertia about the axis passing through the center of the rim.
Thus,$I_{COM} = \frac{2}{3}MR^2 - m\left(\frac{R}{2}\right)^2 = \frac{2}{3}MR^2 - \frac{1}{4}MR^2 = \frac{8-3}{12}MR^2 = \frac{5}{12}MR^2$.
Using the relation $\tau = I_{COM} \cdot \alpha$,we have:
$F \cdot \frac{R}{2} = \left(\frac{5}{12}MR^2\right) \alpha$
$\alpha = \frac{F \cdot R / 2}{5/12 \cdot MR^2} = \frac{F \cdot R}{2} \cdot \frac{12}{5MR^2} = \frac{6F}{5MR}$.

(A) The torque $\tau$ is given by $\tau = F \cdot R = I \alpha$.
Given $F = (20t - 5t^2) \ N$,$R = 2 \ m$,and $I = 10 \ kg \ m^2$.
$(20t - 5t^2) \cdot 2 = 10 \alpha \implies \alpha = 4t - t^2$.
Integrating $\alpha$ with respect to time to find angular velocity $\omega$:
$\omega = \int (4t - t^2) dt = 2t^2 - \frac{t^3}{3}$.
Integrating $\omega$ with respect to time to find angular displacement $\theta$:
$\theta = \int (2t^2 - \frac{t^3}{3}) dt = \frac{2t^3}{3} - \frac{t^4}{12}$.
The direction of motion reverses when $\omega = 0$:
$2t^2 - \frac{t^3}{3} = 0 \implies t^2(2 - \frac{t}{3}) = 0 \implies t = 6 \ s$.
Calculating angular displacement at $t = 6 \ s$:
$\theta = \frac{2(6)^3}{3} - \frac{6^4}{12} = \frac{2 \cdot 216}{3} - \frac{1296}{12} = 144 - 108 = 36 \ rad$.
Number of rotations $N = \frac{\theta}{2\pi} = \frac{36}{2\pi} = \frac{18}{\pi} \approx 5.73$.

(A) The original rod has mass $m$ and length $2l$. After cutting a length $l/2$ (mass $m/4$),the remaining rod has length $3l/2$ and mass $3m/4$. The cut piece of length $l/2$ is placed on top of the remaining rod. The pivot is at the center of the original rod.
The moment of inertia $I$ about the pivot is:
$I = I_{\text{rod}} + I_{\text{piece}} = \left[ \frac{1}{12} \left( \frac{3m}{4} \right) \left( \frac{3l}{2} \right)^2 + \left( \frac{3m}{4} \right) \left( \frac{l}{4} \right)^2 \right] + \left[ \frac{1}{12} \left( \frac{m}{4} \right) \left( \frac{l}{2} \right)^2 + \left( \frac{m}{4} \right) \left( \frac{l}{4} \right)^2 \right]$
$= \frac{3m}{4} \left( \frac{3l^2}{16} + \frac{l^2}{16} \right) + \frac{m}{4} \left( \frac{l^2}{48} + \frac{l^2}{16} \right) = \frac{3m}{4} \left( \frac{4l^2}{16} \right) + \frac{m}{4} \left( \frac{4l^2}{48} \right) = \frac{3ml^2}{16} + \frac{ml^2}{48} = \frac{9ml^2 + ml^2}{48} = \frac{10ml^2}{48} = \frac{5ml^2}{24}$.
The torque $\tau$ about the pivot is:
$\tau = \left( \frac{3m}{4} \right) g \left( \frac{l}{4} \right) - \left( \frac{m}{4} \right) g \left( \frac{l}{4} \right) = \frac{2mg}{4} \cdot \frac{l}{4} = \frac{mgl}{8}$.
Using $\tau = I\alpha$:
$\frac{mgl}{8} = \left( \frac{5ml^2}{24} \right) \alpha \implies \alpha = \frac{mgl}{8} \cdot \frac{24}{5ml^2} = \frac{3g}{5l}$.

(C) Given:
Initial angular velocity,$\omega_{0} = 0\, rad/s$ (since the wheel is at rest).
Angular displacement,$\theta = 200\, rad$.
Time,$t = 5\, s$.
Using the kinematic equation for rotational motion:
$\theta = \omega_{0} t + \frac{1}{2} \alpha t^{2}$
Since $\omega_{0} = 0$,the equation simplifies to:
$\theta = \frac{1}{2} \alpha t^{2}$
Rearranging to solve for angular acceleration $\alpha$:
$\alpha = \frac{2 \theta}{t^{2}}$
Substituting the given values:
$\alpha = \frac{2 \times 200}{(5)^{2}} = \frac{400}{25} = 16\, rad/s^{2}$.

(A) The torque $\tau$ about the hinge point $B$ due to gravity acting at the centre of mass (at distance $l/2$ from $B$) is given by $\tau = mg \times \frac{l}{2}$.
Using the rotational form of Newton's second law,$\tau = I\alpha$,where $I$ is the moment of inertia of the stick about the hinge $B$ $(I = \frac{ml^2}{3})$.
Thus,$\alpha = \frac{\tau}{I} = \frac{mg(l/2)}{ml^2/3} = \frac{3g}{2l}$.
The linear acceleration $a$ of the centre of mass is given by $a = \alpha r$,where $r = l/2$ is the distance of the centre of mass from the hinge.
Therefore,$a = \left(\frac{3g}{2l}\right) \times \left(\frac{l}{2}\right) = \frac{3}{4} g$.

(B) When the string is cut,the only torque acting on the rod about the hinge is due to the gravitational force acting at the center of mass of the rod.
The torque $\tau$ about the hinge is given by $\tau = Mg \cdot \frac{L}{2}$.
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:
$Mg \cdot \frac{L}{2} = \left( \frac{ML^2}{3} \right) \alpha$
Solving for $\alpha$:
$\alpha = \frac{MgL/2}{ML^2/3} = \frac{3g}{2L}$.

(C) Given: Mass of the hollow cylinder $M = 3 \, kg$.
Radius of the hollow cylinder $R = 40 \, cm = 0.4 \, m$.
Force applied $F = 30 \, N$.
The torque $\tau$ produced by the force is given by $\tau = F \times R$.
$\tau = 30 \, N \times 0.4 \, m = 12 \, Nm$.
The moment of inertia $I$ of a hollow cylinder about its central axis is $I = MR^2$.
$I = 3 \, kg \times (0.4 \, m)^2 = 3 \times 0.16 = 0.48 \, kg \, m^2$.
Using the relation $\tau = I \alpha$,where $\alpha$ is the angular acceleration:
$\alpha = \frac{\tau}{I} = \frac{12 \, Nm}{0.48 \, kg \, m^2} = 25 \, rad \, s^{-2}$.

(C) Given: Mass of the disc $M = 20 \, kg$,Radius $R = 20 \, cm = 0.2 \, m$,Force $F = 25 \, N$.
The torque $\tau$ applied to the flywheel is given by $\tau = F \times R$.
$\tau = 25 \, N \times 0.2 \, m = 5 \, N \cdot m$.
The moment of inertia $I$ of a disc about its central axis is $I = \frac{1}{2} M R^2$.
$I = \frac{1}{2} \times 20 \times (0.2)^2 = 10 \times 0.04 = 0.4 \, kg \cdot m^2$.
Using the rotational analogue of Newton's second law,$\tau = I \alpha$,where $\alpha$ is the angular acceleration:
$5 = 0.4 \times \alpha$.
$\alpha = \frac{5}{0.4} = 12.5 \, rad/s^2$.

(C) The torque $\tau$ applied to the flywheel is given by $\tau = F \times r$.
Given $F = 10 \, N$ and $r = 0.2 \, m$,we have $\tau = 10 \times 0.2 = 2 \, N \cdot m$.
From the relation $\tau = I \alpha$,where $I = 0.4 \, kg \cdot m^2$ is the moment of inertia and $\alpha$ is the angular acceleration,we get $\alpha = \frac{\tau}{I} = \frac{2}{0.4} = 5 \, rad \cdot s^{-2}$.
Assuming the flywheel starts from rest $(\omega_1 = 0)$,the angular velocity $\omega_2$ after time $t = 4 \, s$ is given by $\omega_2 = \omega_1 + \alpha t$.
Substituting the values,$\omega_2 = 0 + (5 \times 4) = 20 \, rad \cdot s^{-1}$.

(A) Given: Moment of inertia $I = 50\,kg-m^2$,initial angular velocity $\omega_i = 20\,rad/s$,final angular velocity $\omega_f = 0\,rad/s$,and time $t = 10\,s$.
The angular acceleration $\alpha$ is given by $\alpha = \frac{\omega_f - \omega_i}{t} = \frac{0 - 20}{10} = -2\,rad/s^2$.
The torque $\tau$ is given by $\tau = I \alpha$.
Substituting the values: $\tau = 50 \times (-2) = -100\,N-m$.
The magnitude of the torque required to stop the object is $100\,N-m$.

(D) The angular momentum $\vec{L}$ is defined as $\vec{L} = \vec{r} \times \vec{p}$.
Differentiating both sides with respect to time $t$,we get:
$\frac{d\vec{L}}{dt} = \frac{d}{dt}(\vec{r} \times \vec{p})$
Using the product rule for cross products:
$\frac{d\vec{L}}{dt} = \left(\frac{d\vec{r}}{dt} \times \vec{p}\right) + \left(\vec{r} \times \frac{d\vec{p}}{dt}\right)$
Since velocity $\vec{v} = \frac{d\vec{r}}{dt}$ and linear momentum $\vec{p} = m\vec{v}$,the term $\frac{d\vec{r}}{dt} \times \vec{p}$ becomes $\vec{v} \times (m\vec{v}) = 0$ because the cross product of parallel vectors is zero.
Therefore,$\frac{d\vec{L}}{dt} = 0 + \vec{r} \times \frac{d\vec{p}}{dt}$.
Rearranging the terms,we get $\frac{d\vec{L}}{dt} - \vec{r} \times \frac{d\vec{p}}{dt} = 0$.

(A) The rotational analogue of Newton's second law of motion $(F = ma)$ relates torque,moment of inertia,and angular acceleration.
For a rigid body rotating about a fixed axis,the torque $\tau$ is defined as the product of the moment of inertia $I$ and the angular acceleration $\alpha$.
Mathematically,this is expressed as $\tau = I\alpha$.
Therefore,the correct relation is $\tau = I\alpha$.

(B) The angular acceleration $\alpha$ is given by the relation $\tau = I \alpha$,where $\tau$ is the torque and $I$ is the moment of inertia. Thus,$\alpha = \frac{\tau}{I}$.
In both cases,the torque $\tau = F \times L$ is the same,where $L$ is the total length of the scale.
Let $m_w$ be the mass of the wooden half and $m_s$ be the mass of the steel half. Since steel is denser than wood,$m_s > m_w$.
In case $(a)$,the pivot is at the center of the wooden part. The moment of inertia $I_a$ is the sum of the moment of inertia of the wooden part about $O$ and the steel part about $O$.
In case $(b)$,the pivot is at the center of the steel part. The moment of inertia $I_b$ is the sum of the moment of inertia of the steel part about $O'$ and the wooden part about $O'$.
Since the heavier mass (steel) is closer to the pivot in case $(b)$,the moment of inertia $I_b$ is less than $I_a$ $(I_b < I_a)$.
Since $\alpha \propto \frac{1}{I}$,a smaller moment of inertia results in a larger angular acceleration. Therefore,more angular acceleration will be produced in $(b)$.

(A) Given: Moment of inertia $I = 5 \times 10^{-3} \, kg \, m^2$, initial frequency $n = 20 \, rev/sec$, time $t = 10 \, sec$, final angular velocity $\omega_f = 0$.
Initial angular velocity $\omega_i = 2 \pi n = 2 \pi \times 20 = 40 \pi \, rad/sec$.
Using the equation of rotational motion $\omega_f = \omega_i + \alpha t$, we get $0 = 40 \pi + \alpha \times 10$.
Thus, angular acceleration $\alpha = -4 \pi \, rad/sec^2$.
The magnitude of torque required is $\tau = |I \alpha| = 5 \times 10^{-3} \times 4 \pi = 20 \pi \times 10^{-3} = 2 \pi \times 10^{-2} \, N-m$.
Therefore, the required value is $2 \pi$.

(B) Let the length of the rod be $\ell = 0.5\,m$. The center of mass of the rod is at a distance $\frac{\ell}{2}$ from the pivot.
When the rod is at an angle of $30^o$ with the horizontal,the vertical height of the center of mass above the horizontal position is $h = \frac{\ell}{2} \sin(30^o) = \frac{\ell}{2} \times \frac{1}{2} = \frac{\ell}{4}$.
By the law of conservation of mechanical energy,the loss in potential energy equals the gain in rotational kinetic energy:
$mg h = \frac{1}{2} I \omega^2$
Where $I = \frac{m\ell^2}{3}$ is the moment of inertia of the rod about the pivot.
$mg \left( \frac{\ell}{4} \right) = \frac{1}{2} \left( \frac{m\ell^2}{3} \right) \omega^2$
$g \frac{\ell}{4} = \frac{\ell^2}{6} \omega^2$
$\omega^2 = \frac{6g}{4\ell} = \frac{3g}{2\ell}$
Substituting $g = 10\,ms^{-2}$ and $\ell = 0.5\,m$:
$\omega^2 = \frac{3 \times 10}{2 \times 0.5} = \frac{30}{1} = 30$
$\omega = \sqrt{30}\,rad\,s^{-1}$.

(A) The net torque $\tau$ about the pivot point $P$ is given by the difference in torques due to the two masses.
$\tau = (5M_0g)(l) - (2M_0g)(2l) = 5M_0gl - 4M_0gl = M_0gl$.
The moment of inertia $I$ about the pivot point $P$ is the sum of the moments of inertia of the two masses.
$I = (5M_0)(l)^2 + (2M_0)(2l)^2 = 5M_0l^2 + 8M_0l^2 = 13M_0l^2$.
The angular acceleration $\alpha$ is given by $\alpha = \frac{\tau}{I}$.
Substituting the values,$\alpha = \frac{M_0gl}{13M_0l^2} = \frac{g}{13l}$.

(A) The angular impulse is equal to the change in angular momentum.
$\tau \Delta t = \Delta L$
Taking the edge of the table as the pivot point,the torque due to gravity is $\tau = mg \frac{\ell}{2}$.
$mg \frac{\ell}{2} \Delta t = I \omega$
Since the rod rotates about the edge,its moment of inertia is $I = \frac{m\ell^2}{3}$.
$mg \frac{\ell}{2} \Delta t = \frac{m\ell^2}{3} \omega$
$\omega = \frac{3g \Delta t}{2\ell} = \frac{3 \times 10 \times 0.01}{2 \times 0.3} = \frac{0.3}{0.6} = 0.5\, rad/s$
Time taken by the rod to hit the ground is $t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 5}{10}} = 1\, s$.
The angle rotated by the rod during this time is $\theta = \omega t = 0.5 \times 1 = 0.5\, radian$.

(D) The rotational kinetic energy of the disc is given by $KE = \frac{1}{2}I\omega^2 = K\theta^2.$
From this,we can express the angular velocity $\omega$ as:
$\omega^2 = \frac{2K\theta^2}{I} \implies \omega = \sqrt{\frac{2K}{I}}\theta \quad \dots(1)$
Angular acceleration $\alpha$ is the rate of change of angular velocity with respect to time:
$\alpha = \frac{d\omega}{dt} = \frac{d}{dt}\left( \sqrt{\frac{2K}{I}}\theta \right) = \sqrt{\frac{2K}{I}} \frac{d\theta}{dt}$
Since $\frac{d\theta}{dt} = \omega,$ we substitute the expression for $\omega$ from equation $(1)$:
$\alpha = \sqrt{\frac{2K}{I}} \left( \sqrt{\frac{2K}{I}}\theta \right)$
$\alpha = \frac{2K}{I}\theta.$

(A) Given: Moment of inertia $I = 1.5\, kg\, m^2$,Rotational kinetic energy $K = 1200\, J$,Angular acceleration $\alpha = 20\, rad/s^2$,Initial angular velocity $\omega_0 = 0$.
Formula for rotational kinetic energy is $K = \frac{1}{2}I\omega^2$.
Substituting the values: $1200 = \frac{1}{2} \times 1.5 \times \omega^2$.
$1200 = 0.75 \times \omega^2 \Rightarrow \omega^2 = \frac{1200}{0.75} = 1600$.
Thus,final angular velocity $\omega = \sqrt{1600} = 40\, rad/s$.
Using the equation of rotational motion: $\omega = \omega_0 + \alpha t$.
$40 = 0 + 20 \times t$.
$t = \frac{40}{20} = 2\, s$.

(A) Given: Mass $m = 5\,g = 5 \times 10^{-3}\,kg$,Radius $r = 1\,cm = 10^{-2}\,m$,Time $t = 5\,s$,Final frequency $f = 25\,rev/s$.
Angular velocity $\omega = 2\pi f = 2 \times \pi \times 25 = 50\pi\,rad/s$.
The moment of inertia of the disc about the axis $AB$ (tangent to the disc) 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 $\tau = I\alpha$,where $\alpha = \frac{\omega}{t}$:
$\tau = I \times \frac{\omega}{t} = \left(\frac{3}{2}mr^2\right) \times \frac{\omega}{t}$.
Substituting the values:
$\tau = \frac{3}{2} \times (5 \times 10^{-3}) \times (10^{-2})^2 \times \frac{50\pi}{5}$.
$\tau = \frac{3}{2} \times 5 \times 10^{-3} \times 10^{-4} \times 10\pi$.
$\tau = 7.5 \times 10^{-7} \times 10\pi = 7.5\pi \times 10^{-6} \approx 23.56 \times 10^{-6} = 2.356 \times 10^{-5}\,Nm$.
Re-evaluating based on the provided options,the closest value is $2.0 \times 10^{-5}\,Nm$.

(A) The angular velocity is given by $\omega(t) = \alpha - \beta t$.
The body comes to rest when $\omega(t) = 0$,which implies $\alpha - \beta t = 0$,so the time taken to come to rest is $t = \frac{\alpha}{\beta}$.
The angular displacement $\theta$ is the integral of angular velocity with respect to time: $\theta = \int_{0}^{t} \omega(t) dt$.
Substituting the expression for $\omega(t)$: $\theta = \int_{0}^{\alpha/\beta} (\alpha - \beta t) dt$.
Integrating,we get $\theta = [\alpha t - \frac{1}{2}\beta t^2]_{0}^{\alpha/\beta}$.
Evaluating at the limits: $\theta = \alpha(\frac{\alpha}{\beta}) - \frac{1}{2}\beta(\frac{\alpha}{\beta})^2 = \frac{\alpha^2}{\beta} - \frac{\alpha^2}{2\beta} = \frac{\alpha^2}{2\beta}$.

(B) The relationship between power $(P)$, torque $(\tau)$, and angular velocity $(\omega)$ is given by the formula: $P = \tau \times \omega$.
Given:
Power $(P) = 1.2 \, kW = 1200 \, W$.
Angular velocity $(\omega) = 600 \, rad/s$.
Substituting the values into the formula:
$1200 = \tau \times 600$.
Solving for torque $(\tau)$:
$\tau = \frac{1200}{600} = 2 \, N-m$.
Therefore, the effective torque on the stone is $2 \, N-m$.

(D) The retarding torque is constant,so the angular retardation $\alpha$ is also constant.
Using the rotational kinematic equation $\omega^2 = \omega_0^2 - 2\alpha\theta$:
For the first $n$ rotations,the angular displacement is $\theta_1 = 2\pi n$. The final angular velocity is $\frac{\omega_0}{2}$.
$\left(\frac{\omega_0}{2}\right)^2 = \omega_0^2 - 2\alpha(2\pi n) \implies \frac{\omega_0^2}{4} = \omega_0^2 - 4\pi n\alpha \implies 4\pi n\alpha = \frac{3\omega_0^2}{4} \implies \alpha = \frac{3\omega_0^2}{16\pi n} \quad ...(i)$
For the next $\theta_2$ rotations (where $\theta_2 = 2\pi n'$),the disc comes to rest $(\omega = 0)$:
$0^2 = \left(\frac{\omega_0}{2}\right)^2 - 2\alpha(2\pi n') \implies 0 = \frac{\omega_0^2}{4} - 4\pi n'\alpha \implies 4\pi n'\alpha = \frac{\omega_0^2}{4} \quad ...(ii)$
Dividing equation $(ii)$ by equation $(i)$:
$\frac{4\pi n'\alpha}{4\pi n\alpha} = \frac{\omega_0^2 / 4}{3\omega_0^2 / 4} \implies \frac{n'}{n} = \frac{1}{3} \implies n' = \frac{n}{3}$.
Thus,the disc will make $\frac{n}{3}$ more rotations before coming to rest.

(B) Given: $r_1 = 0.05 \, m$,$r_2 = 0.30 \, m$,$I = 5100 \, kg \cdot m^2$.
Forces acting on the wheel:
$1$. $A$ force of $10 \, N$ acts tangentially at $r_2$ (clockwise torque).
$2$. $A$ force of $9 \, N$ acts tangentially at $r_2$ (clockwise torque).
$3$. $A$ force of $12 \, N$ acts at $r_1$ at an angle of $30^\circ$ with the horizontal. The perpendicular component is $12 \sin(30^\circ) = 12 \times 0.5 = 6 \, N$ (counter-clockwise torque).
Net torque $\tau_{net} = (10 \times 0.30) + (9 \times 0.30) - (6 \times 0.05) = 3.0 + 2.7 - 0.3 = 5.4 \, N \cdot m$.
Using $\tau_{net} = I \alpha$:
$5.4 = 5100 \times \alpha$
$\alpha = \frac{5.4}{5100} = \frac{54}{51000} \approx 1.058 \times 10^{-3} \, rad/s^2$.
Given the options,the closest value is $10^{-3} \, rad/s^2$.

(D) Given:
Moment of inertia $I = 0.20\, kg-m^2$
Radius $r = 20\, cm = 0.2\, m$
Force $F = 20\, N$
Time $t = 5\, s$
Initial angular velocity $\omega_0 = 0$
The torque applied to the wheel is $\tau = r \times F = 0.2\, m \times 20\, N = 4\, N-m$.
Using the relation $\tau = I \alpha$,the angular acceleration is $\alpha = \frac{\tau}{I} = \frac{4}{0.20} = 20\, rad/s^2$.
Using the equation of motion $\omega = \omega_0 + \alpha t$:
$\omega = 0 + (20)(5) = 100\, rad/s$.