Energy conservation, Kinetic Energy,  Work and Power of Rotational Motion Questions in English

(B) The rotational kinetic energy $(K_r)$ is given by the formula $K_r = \frac{1}{2} I \omega^2$.
Given $I = 2.4 \ kg \cdot m^2$ and $K_r = 750 \ J$,we have:
$750 = \frac{1}{2} \times 2.4 \times \omega^2$
$750 = 1.2 \times \omega^2$
$\omega^2 = \frac{750}{1.2} = 625$
$\omega = \sqrt{625} = 25 \ rad/s$.
Assuming the body starts from rest,the initial angular velocity $\omega_0 = 0$.
Using the kinematic equation $\omega = \omega_0 + \alpha t$:
$25 = 0 + 5 \times t$
$t = \frac{25}{5} = 5 \ s$.

(D) The work done to stop the rotating disc is equal to the change in its rotational kinetic energy.
Given: Mass $M = 1\ kg$,Radius $R = 40\ cm = 0.4\ m$,Frequency $f = 10\ rev/s$.
The moment of inertia of the disc about its central axis is $I = \frac{1}{2}MR^2 = \frac{1}{2} \times 1 \times (0.4)^2 = 0.08\ kg\cdot m^2$.
The angular velocity is $\omega = 2\pi f = 2 \times \pi \times 10 = 20\pi\ rad/s$.
The initial rotational kinetic energy is $K_i = \frac{1}{2}I\omega^2 = \frac{1}{2} \times 0.08 \times (20\pi)^2$.
$K_i = 0.04 \times 400\pi^2 = 16\pi^2$.
Using $\pi^2 \approx 9.87$,$K_i = 16 \times 9.87 = 157.92\ J \approx 158\ J$.
Since the final kinetic energy is $0$,the work done to stop the disc is $W = K_f - K_i = 0 - 158 = -158\ J$.
The magnitude of the work done is $158\ J$.

(D) The work done by a force $F$ acting on a body is given by the product of the force and the displacement in the direction of the force.
For a tangential force $F$ applied to the edge of a disc of radius $R$,the displacement $s$ along the arc for an angular deflection $\theta$ (in radians) is given by $s = R\theta$.
Since the force $F$ is always tangential to the disc,it remains parallel to the displacement at every point along the arc.
Therefore,the work done $W$ is calculated as:
$W = F \times s$
$W = F \times (R\theta)$
$W = FR\theta$
Thus,the correct option is $D$.

(B) Given: Moment of inertia $I = 10 \ kg \cdot m^2$,initial angular speed $N_1 = 10 \ rpm$.
The initial angular velocity is $\omega_1 = \frac{2 \pi N_1}{60} = \frac{2 \pi \times 10}{60} = \frac{\pi}{3} \ rad/s$.
The final angular velocity is $\omega_2 = 5 \omega_1 = 5 \times \frac{\pi}{3} = \frac{5 \pi}{3} \ rad/s$.
According to the work-energy theorem,the work done is equal to the change in rotational kinetic energy:
$W = \Delta K = \frac{1}{2} I (\omega_2^2 - \omega_1^2)$.
Substituting the values: $W = \frac{1}{2} \times 10 \times \left[ (\frac{5 \pi}{3})^2 - (\frac{\pi}{3})^2 \right]$.
$W = 5 \times \left( \frac{25 \pi^2}{9} - \frac{\pi^2}{9} \right) = 5 \times \frac{24 \pi^2}{9} = 5 \times \frac{8 \pi^2}{3} = \frac{40 \pi^2}{3}$.
Using $\pi^2 \approx 9.8696$,$W = \frac{40 \times 9.8696}{3} \approx 131.59 \ J$. Given the options,the closest value is $131.4 \ J$.

(C) The rotational kinetic energy is given by $K.E. = \frac{1}{2} I \omega_i^2 = 200 \ J$.
Given $I = 4 \ kg \cdot m^2$,we have $\frac{1}{2} \times 4 \times \omega_i^2 = 200$,which simplifies to $2 \omega_i^2 = 200$,so $\omega_i^2 = 100$,and $\omega_i = 10 \ rad/s$.
The work done by the opposing torque $T$ to bring the flywheel to rest is equal to the initial kinetic energy: $W = T \theta = K.E.$
Here,$T = 5 \ N \cdot m$ and $K.E. = 200 \ J$.
So,$5 \times \theta = 200$,which gives $\theta = 40 \ radians$.
The number of revolutions $n$ is given by $n = \frac{\theta}{2 \pi} = \frac{40}{2 \pi} = \frac{20}{\pi}$.
Using $\pi \approx 3.14159$,$n \approx 6.366 \approx 6.4$ revolutions.

(C) Let the initial frequency be $n$.
The initial angular velocity is $\omega_i = 2\pi n$.
The initial rotational kinetic energy is $K_i = \frac{1}{2} I \omega_i^2 = \frac{1}{2} I (2\pi n)^2 = 2 I \pi^2 n^2$.
When the frequency is doubled,the new frequency is $n' = 2n$.
The new angular velocity is $\omega_f = 2\pi (2n) = 4\pi n$.
The final rotational kinetic energy is $K_f = \frac{1}{2} I \omega_f^2 = \frac{1}{2} I (4\pi n)^2 = 8 I \pi^2 n^2$.
According to the work-energy theorem,the work done $W$ is equal to the change in kinetic energy:
$W = K_f - K_i = 8 I \pi^2 n^2 - 2 I \pi^2 n^2 = 6 I \pi^2 n^2$.

(C) Initial angular speed $\omega_1 = 20\ rad/s$.
Final angular speed $\omega_f = 0\ rad/s$ at $t = 4\ s$.
Angular retardation $\alpha = \frac{\omega_1 - \omega_f}{t} = \frac{20 - 0}{4} = 5\ rad/s^2$.
Angular speed after $t' = 2\ s$ is $\omega_2 = \omega_1 - \alpha t' = 20 - (5 \times 2) = 10\ rad/s$.
Work done by the torque is equal to the change in rotational kinetic energy:
$W = \Delta K = \frac{1}{2} I (\omega_1^2 - \omega_2^2)$.
$W = \frac{1}{2} \times 0.20 \times (20^2 - 10^2)$.
$W = 0.1 \times (400 - 100) = 0.1 \times 300 = 30\ J$.

(A) The rotational power $(P)$ is given by the dot product of torque $(\vec{\tau})$ and angular velocity $(\vec{\omega})$.
$P = \vec{\tau} \cdot \vec{\omega}$
Given $\vec{\tau} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{\omega} = 2\hat{i} + 3\hat{j} + 4\hat{k}$.
$P = (1\hat{i} + 2\hat{j} + 3\hat{k}) \cdot (2\hat{i} + 3\hat{j} + 4\hat{k})$
$P = (1 \times 2) + (2 \times 3) + (3 \times 4)$
$P = 2 + 6 + 12$
$P = 20 \ W$

(B) The work done by a torque $\tau$ during an angular displacement $\theta$ is given by the formula $W = \tau \theta$.
Here,the force $F$ is applied tangentially to the rim of the ring of radius $R$.
The torque $\tau$ produced by this force is $\tau = F \times R$.
Substituting the value of torque into the work formula,we get $W = (F \times R) \times \theta$.
Therefore,the work done by the force is $W = FR\theta$.

(D) The moment of inertia of a rod of length $l$ and mass $m$ about an axis passing through one of its ends is $I = \frac{1}{3} ml^2$.
The rotational kinetic energy of the rod at its lowest position is $K = \frac{1}{2} I \omega^2 = \frac{1}{2} (\frac{1}{3} ml^2) \omega^2 = \frac{1}{6} ml^2 \omega^2$.
As the rod swings,this kinetic energy is converted into gravitational potential energy of the center of mass.
The center of mass of the rod is at a distance $l/2$ from the axis of rotation.
Let $h$ be the maximum height reached by the center of mass.
By the law of conservation of energy,the change in potential energy equals the initial rotational kinetic energy:
$mgh = \frac{1}{6} ml^2 \omega^2$.
Solving for $h$,we get $h = \frac{l^2 \omega^2}{6g}$.

(C) Let the length of the rod be $\ell$ and its mass be $m$. The rod rotates about the fixed end on the ground.
Initially,the center of mass $(CM)$ of the rod is at a height $h = \frac{\ell}{2}$ from the ground.
When the rod hits the ground,the potential energy lost by the rod is converted into rotational kinetic energy.
Loss in potential energy = $mg \Delta h = mg \frac{\ell}{2}$.
Rotational kinetic energy = $\frac{1}{2} I \omega^2$,where $I$ is the moment of inertia about the end,$I = \frac{m\ell^2}{3}$.
Equating the two: $mg \frac{\ell}{2} = \frac{1}{2} \left( \frac{m\ell^2}{3} \right) \omega^2$.
Simplifying,$g\ell = \frac{\ell^2}{3} \omega^2$,so $\omega^2 = \frac{3g}{\ell}$,which gives $\omega = \sqrt{\frac{3g}{\ell}}$.
The velocity $v$ of the upper end is $v = \omega \ell$.
Substituting $\omega$,we get $v = \sqrt{\frac{3g}{\ell}} \times \ell = \sqrt{3g\ell}$.

(B) By the principle of conservation of energy,the loss in potential energy equals the gain in rotational kinetic energy.
Let $m$ be the mass of each particle $A$,$B$,and $C$. The distances of $A$,$B$,and $C$ from the pivot are $\frac{\ell}{3}$,$\frac{2\ell}{3}$,and $\ell$ respectively.
Loss in potential energy = $mg(\frac{\ell}{3}) + mg(\frac{2\ell}{3}) + mg(\ell) = mg\ell(1/3 + 2/3 + 1) = 2mg\ell$.
Gain in rotational kinetic energy = $\frac{1}{2} I \omega^2$,where $I = m(\frac{\ell}{3})^2 + m(\frac{2\ell}{3})^2 + m(\ell)^2 = m\ell^2(\frac{1}{9} + \frac{4}{9} + 1) = m\ell^2(\frac{14}{9})$.
Equating the two: $2mg\ell = \frac{1}{2} (m\ell^2 \frac{14}{9}) \omega^2$.
$2g = \frac{7}{9} \ell \omega^2 \Rightarrow \omega^2 = \frac{18g}{7\ell} \Rightarrow \omega = \sqrt{\frac{18g}{7\ell}}$.
The velocity of mass $B$ is $v_B = \omega r_B = \omega (\frac{2\ell}{3})$.
$v_B = \frac{2\ell}{3} \sqrt{\frac{18g}{7\ell}} = \sqrt{\frac{4\ell^2}{9} \cdot \frac{18g}{7\ell}} = \sqrt{\frac{8g\ell}{7}}$.

(A) Let the mass of the rod be $m$ and its length be $l = 2L$.
The rod rotates about the end in contact with the ground. The moment of inertia of the rod about this end is $I = \frac{m l^2}{3} = \frac{m(2L)^2}{3} = \frac{4mL^2}{3}$.
Using the principle of conservation of mechanical energy,the loss in gravitational potential energy equals the gain in rotational kinetic energy.
The initial height of the center of mass is $h = \frac{l}{2} \sin \alpha = L \sin \alpha$.
Loss in potential energy = $mgh = mg(L \sin \alpha)$.
Gain in rotational kinetic energy = $\frac{1}{2} I \omega^2 = \frac{1}{2} \left( \frac{4mL^2}{3} \right) \omega^2 = \frac{2mL^2}{3} \omega^2$.
Equating the two: $mgL \sin \alpha = \frac{2mL^2}{3} \omega^2$.
Solving for $\omega$: $\omega^2 = \frac{3g \sin \alpha}{2L} \Rightarrow \omega = \sqrt{\frac{3g \sin \alpha}{2L}}$.

(A) The work done required to bring an object to rest is equal to its rotational kinetic energy: $W = \Delta KE = \frac{1}{2} I \omega^2$.
Since all objects have the same angular speed $\omega$,the work done is directly proportional to the moment of inertia: $W \propto I$.
The moments of inertia for the given objects about their symmetry axes are:
$1$. Solid sphere $(A)$: $I_A = \frac{2}{5} MR^2 = 0.4 MR^2$
$2$. Thin circular disk $(B)$: $I_B = \frac{1}{2} MR^2 = 0.5 MR^2$
$3$. Circular ring $(C)$: $I_C = MR^2 = 1.0 MR^2$
Comparing these values,we get $I_C > I_B > I_A$.
Therefore,the work required satisfies the relation $W_C > W_B > W_A$.

(D) When a body rolls without slipping,it possesses both translational kinetic energy and rotational kinetic energy.
Translational kinetic energy is given by $K_t = \frac{1}{2}mv^2$,where $m$ is the mass and $v$ is the velocity of the center of mass.
Rotational kinetic energy is given by $K_r = \frac{1}{2}I\omega^2$,where $I$ is the moment of inertia about the center of mass and $\omega$ is the angular velocity.
The total kinetic energy $K$ is the sum of these two energies:
$K = K_t + K_r = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.
Therefore,the correct option is $D$.

(B) In this process,the potential energy of the rod is converted into rotational kinetic energy.
Initial potential energy of the rod $U = mg \left( \frac{l}{2} \right)$,because the center of gravity of the rod is at its midpoint.
Rotational kinetic energy $K = \frac{1}{2} I \omega^2$.
Here,$I$ is the moment of inertia of the rod about the fixed end $A$,which is $I = \frac{ml^2}{3}$.
Let $\omega$ be the angular velocity of the rod when it hits the ground.
The velocity of the end $B$ when it hits the ground is $v_B = \omega l$,so $\omega = \frac{v_B}{l}$.
According to the law of conservation of energy:
$mg \left( \frac{l}{2} \right) = \frac{1}{2} I \omega^2$
$mg \left( \frac{l}{2} \right) = \frac{1}{2} \left( \frac{ml^2}{3} \right) \left( \frac{v_B}{l} \right)^2$
$mg \frac{l}{2} = \frac{1}{6} m v_B^2$
$v_B^2 = 3gl$
$v_B = \sqrt{3gl} = \sqrt{3 \times 9.8 \times 1} = \sqrt{29.4} \approx 5.42 \ m/s$.
Thus,the velocity is approximately $5.4 \ m/s$.

(A) The work done is equal to the change in rotational kinetic energy.
$W = \Delta K = K_f - K_i = \frac{1}{2} I (\omega_f^2 - \omega_i^2)$.
Given $I = \frac{9.8}{\pi^2} \ kg \ m^2$,$n_i = 600 \ rpm = 10 \ rev/s$,and $n_f = 300 \ rpm = 5 \ rev/s$.
Since $\omega = 2\pi n$,we have $\omega_i = 20\pi \ rad/s$ and $\omega_f = 10\pi \ rad/s$.
$W = \frac{1}{2} \times \frac{9.8}{\pi^2} \times ((10\pi)^2 - (20\pi)^2)$.
$W = \frac{1}{2} \times \frac{9.8}{\pi^2} \times (100\pi^2 - 400\pi^2) = \frac{1}{2} \times 9.8 \times (-300) = 4.9 \times (-300) = -1470 \ J$.
The magnitude of work done is $1470 \ J$ (as the system loses energy).

(B) The rotational kinetic energy is given by $KE = \frac{L^2}{2I}$, where $L$ is the angular momentum and $I$ is the moment of inertia.
When the man folds his arms, the distribution of mass changes such that the moment of inertia $I$ decreases.
Since there are no external torques acting on the system, the angular momentum $L$ remains constant.
As $I$ decreases, the rotational kinetic energy $KE$ increases.
According to the work-energy theorem, the work done by the man is equal to the change in kinetic energy $(\Delta KE = KE_f - KE_i)$.
Since the final kinetic energy is greater than the initial kinetic energy, the work done is positive.

(C) Let the center of the bar be the origin $c$. The moment of inertia of the bar about $c$ is $I_{bar} = \frac{(8m)(6l)^2}{12} = 24ml^2$.
After the collision,the masses $m$ and $2m$ stick to the bar at distances $2l$ and $l$ from the center respectively.
The total moment of inertia of the system about $c$ is $I = I_{bar} + m(2l)^2 + (2m)(l)^2 = 24ml^2 + 4ml^2 + 2ml^2 = 30ml^2$.
The angular momentum of the system about $c$ before the collision is $L = (2m)(v)(l) + (m)(2v)(2l) = 2mvl + 4mvl = 6mvl$.
Since there are no external torques,angular momentum is conserved: $L = I\omega$.
$6mvl = (30ml^2)\omega \Rightarrow \omega = \frac{6mvl}{30ml^2} = \frac{v}{5l}$.
The total energy about the center of mass is the rotational kinetic energy: $E = \frac{1}{2}I\omega^2$.
$E = \frac{1}{2}(30ml^2)\left(\frac{v}{5l}\right)^2 = 15ml^2 \times \frac{v^2}{25l^2} = \frac{15}{25}mv^2 = \frac{3mv^2}{5}$.

(A) The initial rotational kinetic energy of the cylinder is $K_i = \frac{1}{2} I \omega_0^2 = \frac{1}{2} (\frac{1}{2} M R^2) \omega_0^2 = \frac{1}{4} M R^2 \omega_0^2$.
When the cylinder rotates by an angle $\theta = 2\pi$,the length of the string wound around the cylinder is $x = R \theta = R(2\pi) = 2\pi R$.
The elastic potential energy stored in the string when it is stretched by length $x$ is $U = \frac{1}{2} K x^2 = \frac{1}{2} K (2\pi R)^2 = \frac{1}{2} K (4\pi^2 R^2) = 2\pi^2 K R^2$.
By the law of conservation of energy,the initial kinetic energy must equal the potential energy stored in the string at the point where the cylinder momentarily stops (at $\theta = 2\pi$):
$\frac{1}{4} M R^2 \omega_0^2 = 2\pi^2 K R^2$.
Solving for $\omega_0$:
$\omega_0^2 = \frac{2 \pi^2 K R^2 \times 4}{M R^2} = \frac{8 \pi^2 K}{M}$.
Therefore,$\omega_0 = \sqrt{\frac{8 \pi^2 K}{M}}$.

(C) The rod oscillates about one end. The moment of inertia of the rod about an axis passing through one end is $I = \frac{ml^2}{3}$.
By the principle of conservation of mechanical energy,the rotational kinetic energy at the lowest point is equal to the gravitational potential energy gained by the centre of mass at the maximum height $h$:
$\frac{1}{2} I \omega^2 = mgh$
Substituting the value of $I$:
$\frac{1}{2} \left( \frac{ml^2}{3} \right) \omega^2 = mgh$
$\frac{1}{6} ml^2 \omega^2 = mgh$
Solving for $h$:
$h = \frac{l^2 \omega^2}{6g}$

(B) Given: Moment of Inertia $I = 1.2 \, kg \cdot m^2$,Initial angular velocity $\omega_0 = 0$,Rotational Kinetic Energy $K_r = 1500 \, J$,Angular acceleration $\alpha = 25 \, rad/s^2$.
Step $1$: Calculate the final angular velocity $\omega$ using the formula for rotational kinetic energy:
$K_r = \frac{1}{2} I \omega^2$
$1500 = \frac{1}{2} \times 1.2 \times \omega^2$
$1500 = 0.6 \times \omega^2$
$\omega^2 = \frac{1500}{0.6} = 2500$
$\omega = \sqrt{2500} = 50 \, rad/s$.
Step $2$: Calculate the time $t$ required using the kinematic equation $\omega = \omega_0 + \alpha t$:
$50 = 0 + 25 \times t$
$t = \frac{50}{25} = 2 \, s$.

(D) Let $l$ be the length of the rod and $m$ be its mass.
When the rod is horizontal,its potential energy is taken as zero.
When the rod makes an angle $\alpha$ with the horizontal,the center of mass of the rod (at distance $l/2$ from $N$) descends by a vertical distance $h = (l/2) \sin \alpha$.
By the law of conservation of energy,the loss in potential energy equals the gain in rotational kinetic energy:
$mg(l/2) \sin \alpha = \frac{1}{2} I \omega^2$
Here,$I$ is the moment of inertia of the rod about the fixed end $N$,which is $I = \frac{ml^2}{3}$.
Substituting $I$ into the energy equation:
$mg(l/2) \sin \alpha = \frac{1}{2} (\frac{ml^2}{3}) \omega^2$
$mg(l/2) \sin \alpha = \frac{ml^2}{6} \omega^2$
$\omega^2 = \frac{3g \sin \alpha}{l}$
$\omega = \sqrt{\frac{3g \sin \alpha}{l}}$
The linear speed $v$ of the end $M$ is given by $v = \omega l$:
$v = l \sqrt{\frac{3g \sin \alpha}{l}} = \sqrt{3gl \sin \alpha}$
Thus,the speed $v$ is proportional to $\sqrt{\sin \alpha}$.

(C) The mass of the rod is concentrated at its center of mass. When the rod is in the horizontal position,its center of mass is at a height $h = \frac{l}{2}$ from the lowest point. When it reaches the vertical position,the center of mass is at the lowest point.
Applying the principle of conservation of mechanical energy:
Loss in potential energy $(PE)$ = Gain in rotational kinetic energy $(KE_{rot})$
$mg \left(\frac{l}{2}\right) = \frac{1}{2} I \omega^2$
Here,$I$ is the moment of inertia of the rod about the hinge at one end,given by $I = \frac{ml^2}{3}$.
Substituting the value of $I$:
$mg \left(\frac{l}{2}\right) = \frac{1}{2} \left(\frac{ml^2}{3}\right) \omega^2$
$mg \frac{l}{2} = \frac{ml^2}{6} \omega^2$
$g = \frac{l}{3} \omega^2$
$\omega^2 = \frac{3g}{l}$
$\omega = \sqrt{\frac{3g}{l}}$

(D) The total kinetic energy $(KE)$ of a solid sphere rolling without slipping is the sum of its translational and rotational kinetic energies.
$KE_{total} = KE_{trans} + KE_{rot} = \frac{1}{2} Mv^2 + \frac{1}{2} I\omega^2$
For a solid sphere,the moment of inertia $I = \frac{2}{5} MR^2$ and for rolling without slipping,$\omega = \frac{v}{R}$.
Substituting these values:
$KE_{total} = \frac{1}{2} Mv^2 + \frac{1}{2} (\frac{2}{5} MR^2) (\frac{v}{R})^2 = \frac{1}{2} Mv^2 + \frac{1}{5} Mv^2 = \frac{7}{10} Mv^2$
When the sphere compresses the spring by a maximum distance $x_m$,the entire kinetic energy is converted into the potential energy of the spring $(U = \frac{1}{2} kx_m^2)$.
Equating the two:
$\frac{1}{2} kx_m^2 = \frac{7}{10} Mv^2$
$x_m^2 = \frac{14}{10} \frac{Mv^2}{k} = \frac{7Mv^2}{5k}$
$x_m = v \sqrt{\frac{7M}{5k}}$

(B) When the rod rotates through an angle $\theta$,the vertical fall $h$ of the centre of gravity is given by geometry.
From the figure,the distance of the centre of gravity from the pivot $B$ is $L/2$. The vertical component of this distance after rotation is $(L/2) \cos \theta$.
The vertical fall $h$ of the centre of gravity is $h = L/2 - (L/2) \cos \theta = \frac{L}{2}(1 - \cos \theta)$.
According to the law of conservation of energy,the decrease in potential energy equals the increase in rotational kinetic energy.
Decrease in potential energy $= Mgh = Mg \frac{L}{2}(1 - \cos \theta)$.
Rotational kinetic energy $= \frac{1}{2} I \omega^2$,where $I = \frac{ML^2}{3}$ is the moment of inertia about the end $B$.
Equating the two: $Mg \frac{L}{2}(1 - \cos \theta) = \frac{1}{2} \left( \frac{ML^2}{3} \right) \omega^2$.
$Mg \frac{L}{2}(1 - \cos \theta) = \frac{ML^2}{6} \omega^2$.
Using the trigonometric identity $1 - \cos \theta = 2 \sin^2(\theta/2)$:
$Mg \frac{L}{2} \cdot 2 \sin^2(\theta/2) = \frac{ML^2}{6} \omega^2$.
$MgL \sin^2(\theta/2) = \frac{ML^2}{6} \omega^2$.
$\omega^2 = \frac{6g}{L} \sin^2(\theta/2)$.
$\omega = \sqrt{\frac{6g}{L}} \sin \left( \frac{\theta}{2} \right)$.

(A) Power $P$ is given by $P = \tau \cdot \omega$.
Since $\tau = I \alpha$ and $\alpha = \omega \frac{d\omega}{d\theta}$,we have $P = I \omega \frac{d\omega}{d\theta} \cdot \omega = I \omega^2 \frac{d\omega}{d\theta}$.
Rearranging the terms,we get $\omega^2 d\omega = \frac{P}{I} d\theta$.
Integrating both sides,$\int \omega^2 d\omega = \int \frac{P}{I} d\theta$,which gives $\frac{\omega^3}{3} = \frac{P}{I} \theta$.
Thus,$\omega^3 \propto \theta$,or $\omega \propto \theta^{1/3}$.
Since the number of rotations $n = \frac{\theta}{2\pi}$,we have $\theta \propto n$.
Therefore,$\omega \propto n^{1/3}$.

(B) By the principle of conservation of energy,the potential energy of the rod at the initial vertical position is converted into rotational kinetic energy at the moment it hits the floor.
Initial potential energy $U_i = Mg \frac{L}{2}$.
Final rotational kinetic energy $K_f = \frac{1}{2} I \omega^2$,where $I$ is the moment of inertia of the rod about the end on the floor,$I = \frac{ML^2}{3}$.
Equating $U_i = K_f$:
$Mg \frac{L}{2} = \frac{1}{2} (\frac{ML^2}{3}) \omega^2$.
Simplifying,$gL = \frac{L^2}{3} \omega^2$,which gives $\omega^2 = \frac{3g}{L}$,so $\omega = \sqrt{\frac{3g}{L}}$.
The linear velocity $v$ of the other end is given by $v = L \omega$.
Substituting $\omega$,$v = L \sqrt{\frac{3g}{L}} = \sqrt{3gL}$.

(A) The angular speed of the rotor is $\omega = 200 \; rad \; s^{-1}$.
The torque required to maintain this speed is $\tau = 180 \; N \; m$.
The power $P$ delivered by an engine providing a torque $\tau$ at an angular velocity $\omega$ is given by the formula:
$P = \tau \times \omega$
Substituting the given values:
$P = 180 \; N \; m \times 200 \; rad \; s^{-1}$
$P = 36000 \; W$
Converting to kilowatts:
$P = 36 \; kW$
Therefore,the power required by the engine is $36 \; kW$.

(N/A) In rotational motion, the power $(P)$ delivered by a torque $(\tau)$ to a rotating body with angular velocity $(\omega)$ is given by the formula:
$P = \tau \cdot \omega$
Where:
$P$ is the power in Watts $(W)$,
$\tau$ is the torque in Newton-meters $(N \cdot m)$,
$\omega$ is the angular velocity in radians per second $(rad/s)$.
The angular momentum $(L)$ of a rigid body rotating about a fixed axis with moment of inertia $(I)$ and angular velocity $(\omega)$ is given by the formula:
$L = I \cdot \omega$
Where:
$L$ is the angular momentum in kilogram-meters squared per second $(kg \cdot m^2/s)$,
$I$ is the moment of inertia in kilogram-meters squared $(kg \cdot m^2)$,
$\omega$ is the angular velocity in radians per second $(rad/s)$.

(N/A) Consider a rigid body rotating about a fixed axis,which is taken as the $Z$-axis. This axis is perpendicular to the plane $X^{\prime} Y^{\prime}$.
Let a force $\overrightarrow{F}_{1}$ act on a particle of the body at point $P_{1}$. The particle rotates on a circle of radius $r_{1}$ with center $C$ on the axis,such that $CP_{1} = r_{1}$.
In a small time interval $\Delta t$,the particle moves from position $P_{1}$ to $P_{1}^{\prime}$. The displacement of the particle is $\Delta S_{1} = r_{1} \Delta \theta$,which is in the tangential direction at $P_{1}$.
Here,$\Delta \theta = \angle P_{1} C P_{1}^{\prime}$ is the angular displacement of the particle.
The work done by the force $\overrightarrow{F}_{1}$ on the particle is given by:
$dW_{1} = \overrightarrow{F}_{1} \cdot d\overrightarrow{S}_{1} = F_{1} dS_{1} \cos \phi_{1}$
Since $dS_{1} = r_{1} d\theta$ and $\phi_{1} = 90^{\circ} - \alpha_{1}$ (where $\alpha_{1}$ is the angle between $\overrightarrow{F}_{1}$ and the radius vector $\overrightarrow{r}_{1}$),we have:
$dW_{1} = F_{1} (r_{1} d\theta) \cos(90^{\circ} - \alpha_{1}) = F_{1} r_{1} \sin \alpha_{1} d\theta$
Since the torque $\tau_{1} = r_{1} F_{1} \sin \alpha_{1}$,the work done is:
$dW_{1} = \tau_{1} d\theta$
The total work done on the rigid body is the sum of the work done on all particles:
$dW = \sum dW_{1} = \sum \tau_{1} d\theta = \tau d\theta$
where $\tau$ is the net torque acting on the body about the axis of rotation.

(N/A) Consider a rigid body rotating about a fixed axis,which is taken as the $Z$-axis. This axis is perpendicular to the plane $X^{\prime} Y^{\prime}$.
Let a force $\overrightarrow{F}_{1}$ act on a particle of the body at point $P_{1}$. The particle rotates on a circle of radius $r_{1}$ with center $C$ on the axis,such that $CP_{1} = r_{1}$.
In a small time interval $\Delta t$,the point moves from $P_{1}$ to $P_{1}^{\prime}$. The displacement of the particle is $\Delta S_{1} = r_{1} \Delta \theta$,and it is in the tangential direction at $P_{1}$.
Here,$\Delta \theta = \angle P_{1} C P_{1}^{\prime}$ is the angular displacement of the particle. The work done by the force $\overrightarrow{F}_{1}$ on the particle is given by:
$dW_{1} = \overrightarrow{F}_{1} \cdot d\overrightarrow{S}_{1}$
$dW_{1} = F_{1} \Delta S_{1} \cos \phi_{1}$
Since $\Delta S_{1} = r_{1} \Delta \theta$ and $\phi_{1} = 90^{\circ} - \alpha_{1}$,where $\alpha_{1}$ is the angle between the force $\overrightarrow{F}_{1}$ and the radius vector $\overrightarrow{r}_{1}$,we have:
$dW_{1} = F_{1} (r_{1} \Delta \theta) \cos(90^{\circ} - \alpha_{1})$
$dW_{1} = F_{1} r_{1} \sin \alpha_{1} \Delta \theta$
Since the torque $\tau_{1} = r_{1} F_{1} \sin \alpha_{1}$,the work done is:
$dW_{1} = \tau_{1} \Delta \theta$
For the entire body,the total work done is the sum of the work done on all particles:
$dW = \sum dW_{i} = \sum \tau_{i} \Delta \theta = \tau \Delta \theta$
where $\tau$ is the total torque acting on the body.

(N/A) For a rigid body rotating about a fixed axis,the work done $W$ by a torque $\tau$ through an angular displacement $d\theta$ is given by the integral of the torque over the angular displacement.
The infinitesimal work done is $dW = \tau \cdot d\theta$.
For a finite angular displacement from $\theta_1$ to $\theta_2$,the total work done is:
$W = \int_{\theta_1}^{\theta_2} \tau \, d\theta$.
If the torque $\tau$ is constant,the formula simplifies to:
$W = \tau (\theta_2 - \theta_1) = \tau \Delta \theta$.

(N/A) For a rigid body rotating about a fixed axis,the power $P$ delivered by a torque $\tau$ is given by the product of the torque and the angular velocity $\omega$ of the body.
Mathematically,the formula is expressed as:
$P = \tau \omega$
Where:
$P$ is the instantaneous power,
$\tau$ is the torque applied to the body,
$\omega$ is the angular velocity of the body.

(C) In linear motion,work done is given by $W = F \Delta x$. The rotational analogue of force $F$ is torque $\tau$,and the analogue of displacement $\Delta x$ is angular displacement $\Delta \theta$. Thus,the rotational work is $W = \tau \Delta \theta$. This corresponds to $(1-b)$.
In linear motion,power is given by $P = Fv$. The rotational analogue of force $F$ is torque $\tau$,and the analogue of velocity $v$ is angular velocity $\omega$. Thus,the rotational power is $P = \tau \omega$. This corresponds to $(2-a)$.
Therefore,the correct matching is $(1-b), (2-a)$.

(A) By the law of conservation of mechanical energy,the loss in potential energy of the rod is equal to the gain in its rotational kinetic energy.
The center of mass of the rod is at a distance $L/2$ from the pivot.
When the rod moves from the horizontal position to the vertical position,the center of mass falls by a vertical distance $h = L/2$.
Loss in potential energy = $Mgh = Mg(L/2) = MgL/2$.
The moment of inertia of a rod about an axis passing through one end is $I = ML^2/3$.
Gain in rotational kinetic energy = $\frac{1}{2} I \omega^2 = \frac{1}{2} (ML^2/3) \omega^2 = \frac{ML^2 \omega^2}{6}$.
Equating the two: $MgL/2 = \frac{ML^2 \omega^2}{6}$.
Solving for $\omega$: $\omega^2 = \frac{6gL}{2L^2} = \frac{3g}{L}$.
Therefore,$\omega = \sqrt{\frac{3g}{L}}$.

(C) Let $m$ be the mass of the disc and the particle. The moment of inertia of the disc about the axis is $I_{\text{disc}} = \frac{1}{2} mR^2$. The moment of inertia of the particle at distance $R$ is $I_{\text{particle}} = mR^2$. The total moment of inertia is $I = I_{\text{disc}} + I_{\text{particle}} = \frac{1}{2} mR^2 + mR^2 = \frac{3}{2} mR^2$.
Using the law of conservation of mechanical energy,the loss in potential energy of the particle equals the gain in rotational kinetic energy of the system.
The particle moves from the highest point to the lowest point,a vertical distance of $2R$.
Loss in potential energy $= mg(2R) = 2mgR$.
Gain in rotational kinetic energy $= \frac{1}{2} I \omega^2 = \frac{1}{2} \left( \frac{3}{2} mR^2 \right) \omega^2 = \frac{3}{4} mR^2 \omega^2$.
Equating the two: $2mgR = \frac{3}{4} mR^2 \omega^2$.
Solving for $\omega^2$: $\omega^2 = \frac{8g}{3R}$.
Given $\omega = 4 \sqrt{\frac{x}{3R}}$,so $\omega^2 = 16 \left( \frac{x}{3R} \right) = \frac{16x}{3R}$.
Equating the expressions for $\omega^2$: $\frac{16x}{3R} = \frac{8g}{3R} \implies 16x = 8g$.
Substituting $g = 10 \text{ m s}^{-2}$: $16x = 8(10) = 80$.
$x = \frac{80}{16} = 5$.

(A) The power $P$ delivered by a force $\vec{F}$ acting on a particle moving with velocity $\vec{v}$ is given by $P = \vec{F} \cdot \vec{v}$.
For a rotating body,the velocity of a point with position vector $\vec{r}$ relative to the axis of rotation is $\vec{v} = \vec{\omega} \times \vec{r}$.
Substituting this into the power formula: $P = \vec{F} \cdot (\vec{\omega} \times \vec{r})$.
Using the property of the scalar triple product,$\vec{A} \cdot (\vec{B} \times \vec{C}) = (\vec{A} \times \vec{B}) \cdot \vec{C}$,we can rewrite the expression:
$P = (\vec{F} \times \vec{\omega}) \cdot \vec{r} = \vec{r} \cdot (\vec{F} \times \vec{\omega})$.
Alternatively,using the cyclic property of the scalar triple product $[\vec{a} \vec{b} \vec{c}] = [\vec{b} \vec{c} \vec{a}] = [\vec{c} \vec{a} \vec{b}]$,we have:
$P = \vec{F} \cdot (\vec{\omega} \times \vec{r}) = [\vec{F} \vec{\omega} \vec{r}] = [\vec{\omega} \vec{r} \vec{F}] = \vec{\omega} \cdot (\vec{r} \times \vec{F})$.
Since the torque $\vec{\tau} = \vec{r} \times \vec{F}$,the power is $P = \vec{\tau} \cdot \vec{\omega} = (\vec{r} \times \vec{F}) \cdot \vec{\omega}$.

(A) The angular position is given by $\theta(t) = 5t^2 - 8t$.
Angular velocity $\omega$ is the derivative of angular position with respect to time: $\omega = \frac{d\theta}{dt} = 10t - 8$.
Angular acceleration $\alpha$ is the derivative of angular velocity with respect to time: $\alpha = \frac{d\omega}{dt} = 10 \text{ rad/s}^2$.
The moment of inertia of a circular disc about an axis perpendicular to its plane and passing through its center is $I = \frac{1}{2}MR^2$.
The torque applied is $\tau = I\alpha = (\frac{1}{2}MR^2)(10) = 5MR^2$.
Power $P$ delivered by the torque is $P = \tau \omega$.
At $t = 2$ s,$\omega = 10(2) - 8 = 12 \text{ rad/s}$.
Therefore,$P = (5MR^2)(12) = 60 MR^2$ $W$.

(A) Given: Mass of the wheel $M = 10 \ kg$,Radius $R = 10 \ cm = 0.1 \ m$,Force $F = 20 \ N$,Displacement $d = 1 \ m$.
Since the wheel is supported by spokes of negligible mass,its moment of inertia is $I = MR^2 = 10 \times (0.1)^2 = 0.1 \ kg \ m^2$.
The work done by the force $F$ is $W = F \times d = 20 \times 1 = 20 \ J$.
According to the work-energy theorem,the work done is equal to the change in rotational kinetic energy: $W = \Delta KE = \frac{1}{2} I \omega^2$.
Substituting the values: $20 = \frac{1}{2} \times 0.1 \times \omega^2$.
$20 = 0.05 \times \omega^2$.
$\omega^2 = \frac{20}{0.05} = 400$.
$\omega = \sqrt{400} = 20 \ rad/s$.

(B) Given: Moment of Inertia $I = 1.2 \ kg \cdot m^2$,Rotational Kinetic Energy $K_r = 1500 \ J$,Angular acceleration $\alpha = 25 \ rad/s^2$,Initial angular velocity $\omega_0 = 0$.
The formula for rotational kinetic energy is $K_r = \frac{1}{2} I \omega^2$.
Substituting the values: $1500 = \frac{1}{2} \times 1.2 \times \omega^2$.
$1500 = 0.6 \times \omega^2 \Rightarrow \omega^2 = \frac{1500}{0.6} = 2500$.
Therefore,the final angular velocity $\omega = \sqrt{2500} = 50 \ rad/s$.
Using the kinematic equation for rotation: $\omega = \omega_0 + \alpha t$.
$50 = 0 + 25 \times t$.
$t = \frac{50}{25} = 2 \ s$.

(C) According to the law of conservation of energy,the loss in potential energy $(P.E.)$ is equal to the gain in rotational kinetic energy $(K.E.)$.
When the rod falls from a vertical position to the ground,its center of mass falls through a height of $h = \frac{l}{2}$.
Loss in $P.E. = mgh = mg \left( \frac{l}{2} \right) = \frac{mgl}{2}$.
The moment of inertia of the rod about the hinged end $A$ is $I = \frac{ml^2}{3}$.
Gain in rotational $K.E. = \frac{1}{2} I \omega^2 = \frac{1}{2} \left( \frac{ml^2}{3} \right) \omega^2 = \frac{ml^2 \omega^2}{6}$.
Equating the loss in $P.E.$ to the gain in rotational $K.E.$:
$\frac{mgl}{2} = \frac{ml^2 \omega^2}{6}$.
Solving for $\omega^2$:
$\omega^2 = \frac{mgl}{2} \times \frac{6}{ml^2} = \frac{3g}{l}$.
Therefore,the angular velocity is $\omega = \sqrt{\frac{3g}{l}}$.

(A) The rod swings about a horizontal axis passing through its end. By the principle of conservation of energy,the maximum rotational kinetic energy at the lowest point is equal to the maximum gravitational potential energy gained by the centre of mass at the highest point.
$1$. The rotational kinetic energy is given by $K = \frac{1}{2} I \omega^{2}$.
$2$. The moment of inertia of a rod of mass $M$ and length $L$ about an axis passing through its end is $I = \frac{ML^{2}}{3}$.
$3$. The potential energy gained by the centre of mass is $U = Mgh$,where $h$ is the maximum height reached by the centre of mass.
$4$. Equating kinetic energy and potential energy: $Mgh = \frac{1}{2} I \omega^{2}$.
$5$. Substituting the value of $I$: $Mgh = \frac{1}{2} \left( \frac{ML^{2}}{3} \right) \omega^{2}$.
$6$. Simplifying the equation: $Mgh = \frac{ML^{2} \omega^{2}}{6}$.
$7$. Solving for $h$: $h = \frac{L^{2} \omega^{2}}{6g}$.

(B) Given: Work done $W = 12000 \ J$,Initial frequency $f_1 = 10 \ Hz$,Final frequency $f_2 = 20 \ Hz$.
Angular velocity is given by $\omega = 2\pi f$.
Initial angular velocity $\omega_1 = 2\pi(10) = 20\pi \ rad/s$.
Final angular velocity $\omega_2 = 2\pi(20) = 40\pi \ rad/s$.
According to the work-energy theorem,the work done equals the change in rotational kinetic energy:
$W = \Delta K = \frac{1}{2} I \omega_2^2 - \frac{1}{2} I \omega_1^2 = \frac{1}{2} I (\omega_2^2 - \omega_1^2)$.
Substituting the values:
$12000 = \frac{1}{2} I [(40\pi)^2 - (20\pi)^2]$.
$12000 = \frac{1}{2} I [1600\pi^2 - 400\pi^2]$.
$12000 = \frac{1}{2} I [1200\pi^2]$.
Using $\pi^2 = 10$:
$12000 = \frac{1}{2} I [1200 \times 10]$.
$12000 = I [6000]$.
$I = \frac{12000}{6000} = 2 \ kg \cdot m^2$.

(D) The rod rotates about an axis passing through its end. The moment of inertia of the rod about this axis is $I = \frac{1}{3}ML^2$.
The rotational kinetic energy of the rod at its lowest point (maximum angular speed) is $K = \frac{1}{2}I\omega^2 = \frac{1}{2}(\frac{1}{3}ML^2)\omega^2 = \frac{1}{6}ML^2\omega^2$.
At the maximum height $h$,the centre of mass of the rod rises by $h$. The potential energy gained by the rod is $U = Mgh$.
By the law of conservation of energy,the rotational kinetic energy at the bottom equals the potential energy at the maximum height:
$\frac{1}{6}ML^2\omega^2 = Mgh$.
Solving for $h$:
$h = \frac{L^2\omega^2}{6g}$.

(C) By the principle of conservation of energy,the rotational kinetic energy at the lowest point is equal to the gravitational potential energy at the maximum height.
$\frac{1}{2} I \omega^2 = Mgh$
$\therefore h = \frac{I \omega^2}{2Mg} \dots (i)$
The moment of inertia $(I)$ of a uniform rod about an axis passing through its end is given by the parallel axis theorem:
$I = I_{cm} + Md^2 = \frac{ML^2}{12} + M\left(\frac{L}{2}\right)^2 = \frac{ML^2}{12} + \frac{ML^2}{4} = \frac{ML^2}{3} \dots (ii)$
Substituting equation $(ii)$ into equation $(i)$:
$h = \frac{(\frac{ML^2}{3}) \omega^2}{2Mg} = \frac{ML^2 \omega^2}{6Mg} = \frac{L^2 \omega^2}{6g}$

(A) The total kinetic energy of the Moon is the sum of its translational kinetic energy (due to orbital motion) and its rotational kinetic energy (due to rotation about its own axis).
$KE_{total} = KE_{translational} + KE_{rotational}$
$KE_{total} = \frac{1}{2} M v^2 + \frac{1}{2} I \omega^2$
Given that the orbital speed $v = R \omega$ and the angular velocity $\omega = \frac{2 \pi}{T}$.
The moment of inertia of the Moon (assuming it is a solid sphere) is $I = \frac{2}{5} M r^2$.
Substituting these values:
$KE_{total} = \frac{1}{2} M (R \omega)^2 + \frac{1}{2} (\frac{2}{5} M r^2) \omega^2$
$KE_{total} = \frac{1}{2} M R^2 \omega^2 + \frac{1}{5} M r^2 \omega^2$
Substituting $\omega^2 = (\frac{2 \pi}{T})^2 = \frac{4 \pi^2}{T^2}$:
$KE_{total} = \frac{1}{2} M R^2 (\frac{4 \pi^2}{T^2}) + \frac{1}{5} M r^2 (\frac{4 \pi^2}{T^2})$
$KE_{total} = \frac{2 M \pi^2 R^2}{T^2} + \frac{4 M r^2 \pi^2}{5 T^2}$

(B) The moment of inertia $I$ of a circular disc about an axis passing through its centre and perpendicular to its plane is given by $I = \frac{1}{2}MR^2$.
Given $M = \frac{10}{\pi^2} \,kg$ and $R = 2 \,m$,we have $I = \frac{1}{2} \times \frac{10}{\pi^2} \times (2)^2 = \frac{20}{\pi^2} \,kg \cdot m^2$.
The initial angular speed $\omega_1 = 90 \,rev/min = 90 \times \frac{2\pi}{60} \,rad/s = 3\pi \,rad/s$.
The final angular speed $\omega_2 = 120 \,rev/min = 120 \times \frac{2\pi}{60} \,rad/s = 4\pi \,rad/s$.
According to the work-energy theorem,the work done $W$ is equal to the change in rotational kinetic energy: $W = \Delta K = \frac{1}{2}I(\omega_2^2 - \omega_1^2)$.
Substituting the values: $W = \frac{1}{2} \times \frac{20}{\pi^2} \times ((4\pi)^2 - (3\pi)^2) = \frac{10}{\pi^2} \times (16\pi^2 - 9\pi^2) = \frac{10}{\pi^2} \times 7\pi^2 = 70 \,J$.

(A) By the principle of conservation of energy,the potential energy lost by the rod is equal to the rotational kinetic energy gained by it.
Initial potential energy of the rod (taking the center of mass at height $L/2$) is $U_i = mg(L/2)$.
Final rotational kinetic energy when the rod hits the ground is $K_f = \frac{1}{2} I \omega^2$.
Here,$I$ is the moment of inertia of the rod about the hinged end,given by $I = \frac{mL^2}{3}$.
Equating the two: $mg \frac{L}{2} = \frac{1}{2} (\frac{mL^2}{3}) \omega^2$.
Simplifying the equation: $g = \frac{L}{3} \omega^2$,which gives $\omega = \sqrt{\frac{3g}{L}}$.
Substituting the given values $g = 10 \ m \ s^{-2}$ and $L = 1 \ m$:
$\omega = \sqrt{\frac{3 \times 10}{1}} = \sqrt{30} \ rad \ s^{-1}$.