Conservation of angular momentum (combined translation and rotational motion) Questions in English

(A) In uniform circular motion, the direction of the velocity vector changes continuously, so the linear momentum $(p = mv)$ is not conserved because the force (centripetal force) is always acting on the particle.
However, the torque $(\tau)$ acting on the particle about the center of the circular path is zero because the force is directed towards the center (radial force).
Since $\tau = \frac{dL}{dt} = 0$, the angular momentum $(L)$ remains constant (conserved).

(C) In the absence of any external torque,the angular momentum of the system remains conserved.
$L_{initial} = L_{final}$
$I \omega = I' \omega'$
Initially,the moment of inertia of the ring about its axis is $I = MR^2$.
When two objects of mass $m$ are placed at opposite ends of a diameter,they are at a distance $R$ from the axis of rotation. The new moment of inertia becomes $I' = MR^2 + mR^2 + mR^2 = (M + 2m)R^2$.
Using the conservation of angular momentum:
$MR^2 \omega = (M + 2m)R^2 \omega'$
$\omega' = \frac{MR^2 \omega}{(M + 2m)R^2}$
$\omega' = \frac{\omega M}{M + 2m}$

(C) When the boy sits on the rotating platform,there is no external torque acting on the system (boy + platform) about the axis of rotation. According to the principle of conservation of angular momentum,if the net external torque acting on a system is zero,the total angular momentum of the system remains constant. Therefore,the angular momentum is conserved.

(C) In a central force field,the force acts along the line joining the particle to the center of force.
Since the torque $\vec{\tau} = \vec{r} \times \vec{F} = 0$ (as $\vec{r}$ and $\vec{F}$ are collinear),the rate of change of angular momentum $\frac{d\vec{L}}{dt} = \vec{\tau} = 0$.
Therefore,the angular momentum $\vec{L}$ remains constant.

(B) The initial angular momentum of the ring is $L = I\omega = Mr^2\omega$.
When two particles of mass $m$ are attached to the opposite ends of a diameter,the new moment of inertia of the system becomes $I' = Mr^2 + m(r)^2 + m(r)^2 = (M + 2m)r^2$.
Since there is no external torque acting on the system,the angular momentum is conserved: $L_{initial} = L_{final}$.
$Mr^2\omega = (M + 2m)r^2\omega'$.
Solving for the new angular velocity $\omega'$,we get $\omega' = \frac{M\omega}{M + 2m}$.

(B) As the beads slide down,their distance from the axis of rotation $AO$ increases,which causes the moment of inertia $I$ of the system to increase. Since there is no external torque acting on the system about the axis of rotation,the total angular momentum $L = I\omega$ must be conserved.
Furthermore,since the only forces acting on the beads are gravity and the normal force from the wires (which does no work as it is perpendicular to the displacement),the total mechanical energy (kinetic + potential) of the system is conserved.

(B) The angular momentum $L$ of the system remains conserved because no external torque acts on the system $(L = I\omega = \text{constant})$.
The rotational kinetic energy $K$ is given by $K = \frac{L^2}{2I}$.
Since $L$ is constant, we have $K \propto \frac{1}{I}$.
Let the initial moment of inertia be $I_1 = I$ and the initial kinetic energy be $K_1 = K$.
When the boy stretches his hands, the new moment of inertia becomes $I_2 = 2I$.
The new kinetic energy $K_2$ is given by $K_2 = \frac{L^2}{2I_2} = \frac{L^2}{2(2I)} = \frac{1}{2} \left( \frac{L^2}{2I} \right) = \frac{K}{2}$.
Therefore, the kinetic energy of the system becomes $K/2$.

(B) The angular momentum $L$ of the Earth is conserved because there is no external torque acting on it.
Since $L = I\omega$,where $I$ is the moment of inertia and $\omega$ is the angular speed.
The moment of inertia of a sphere is $I = \frac{2}{5}MR^2$.
If the radius $R$ decreases,the moment of inertia $I$ decreases.
Since $L = I\omega$ is constant,$\omega = \frac{L}{I}$.
As $I$ decreases,the angular speed $\omega$ must increase.
Therefore,the correct option is $B$.

(C) The angular momentum of the system about the center $O$ is conserved because there is no external torque acting on the system.
Initial angular momentum of the bead about $O$ is $L_i = mvR$.
The moment of inertia of the semicircular ring about its center is $I_{ring} = mR^2$.
The moment of inertia of the bead at the rim of the ring is $I_{bead} = mR^2$.
When the bead comes to rest relative to the ring,the total moment of inertia of the system is $I = I_{ring} + I_{bead} = mR^2 + mR^2 = 2mR^2$.
By the conservation of angular momentum,$L_i = L_f$,so $mvR = I\omega$.
$mvR = (2mR^2)\omega$.
Solving for $\omega$,we get $\omega = \frac{mvR}{2mR^2} = \frac{v}{2R}$.

(C) Since the force applied to pull the string is radial,the torque acting on the object about the center of the circle is zero. Therefore,the angular momentum of the object is conserved.
$L = m v_1 r_1 = m v_2 r_2$
$\therefore v_2 = v_1 \left(\frac{r_1}{r_2}\right)$
Now,the ratio of the new kinetic energy $(K_2)$ to the original kinetic energy $(K_1)$ is:
$\frac{K_2}{K_1} = \frac{\frac{1}{2} m v_2^2}{\frac{1}{2} m v_1^2} = \left(\frac{v_2}{v_1}\right)^2$
Substituting the value of $v_2/v_1$ from the angular momentum equation:
$\frac{K_2}{K_1} = \left(\frac{r_1}{r_2}\right)^2$

(A) Since no external torque is applied to the system,the angular momentum of the particle remains conserved.
The angular momentum $L$ of a particle in a circular orbit is given by $L = I\omega = mr^2\omega$.
Applying the conservation of angular momentum:
$mr_1^2 \omega_1 = mr_2^2 \omega_2$
Canceling the mass $m$ from both sides:
$r_1^2 \omega_1 = r_2^2 \omega_2$
Given that $r_2 = 0.5\, r_1$,substitute this into the equation:
$r_1^2 \omega_1 = (0.5\, r_1)^2 \omega_2$
$r_1^2 \omega_1 = 0.25\, r_1^2 \omega_2$
Dividing both sides by $r_1^2$:
$\omega_1 = 0.25\, \omega_2$
$\omega_2 = \frac{\omega_1}{0.25} = 4\, \omega_1$

(D) In the absence of external torque,the angular momentum $L$ of the body remains conserved.
$L = I \omega = \text{constant}$,where $I$ is the moment of inertia and $\omega$ is the angular velocity.
Since $\omega = 2\pi f$ (where $f$ is the frequency),$I_1 f_1 = I_2 f_2$.
The moment of inertia $I$ is given by $I = Mk^2$,where $M$ is the mass and $k$ is the radius of gyration.
Substituting this into the conservation equation: $M k_1^2 f_1 = M k_2^2 f_2$.
Since the mass $M$ is constant,$k_1^2 f_1 = k_2^2 f_2$.
Therefore,$\frac{k_1^2}{k_2^2} = \frac{f_2}{f_1}$.
Given $f_1 = 1 \text{ cy/sec}$ and $f_2 = 16 \text{ cy/sec}$,we have $\frac{k_1^2}{k_2^2} = \frac{16}{1}$.
Taking the square root on both sides,$\frac{k_1}{k_2} = \sqrt{\frac{16}{1}} = \frac{4}{1}$.
Thus,the ratio of the radius of gyration is $4 : 1$.

(D) The system consists of the disc and the child. Since there are no external torques acting on the system about the axis of rotation,the angular momentum of the system is conserved.
The initial angular momentum of the system is $L_i = (I + mR^2)\omega$.
When the child jumps off with tangential velocity $v$ with respect to the ground,the angular momentum of the child about the center of the disc is $L_{child} = mvR$.
Let the new angular velocity of the disc be $\omega'$. The final angular momentum of the disc is $L_f = I\omega'$.
By the principle of conservation of angular momentum:
$L_i = L_{child} + L_f$
$(I + mR^2)\omega = mvR + I\omega'$
Solving for $\omega'$:
$I\omega' = (I + mR^2)\omega - mvR$
$\omega' = \frac{(I + mR^2)\omega - mvR}{I}$

(D) The initial moment of inertia of the ring about its axis is $I = mR^2$.
The initial angular momentum is $L = I\omega = mR^2\omega$.
When two objects of mass $M$ are attached to the opposite ends of a diameter,the new moment of inertia $I'$ becomes the sum of the ring's moment of inertia and the moment of inertia of the two point masses: $I' = mR^2 + M R^2 + M R^2 = (m + 2M)R^2$.
Since no external torque acts on the system,the angular momentum is conserved: $L_{initial} = L_{final}$.
$mR^2\omega = (m + 2M)R^2\omega'$
Solving for $\omega'$:
$\omega' = \frac{mR^2\omega}{(m + 2M)R^2} = \frac{\omega m}{(m + 2M)}$.

(D) We know that the torque $\vec{\tau}$ is defined as the rate of change of angular momentum $\vec{L}$,given by $\vec{\tau} = \frac{d\vec{L}}{dt}$.
$A$ central force is a force that acts along the line joining the particle to the center of force (origin). Since the position vector $\vec{r}$ and the force vector $\vec{F}$ are collinear,the torque $\vec{\tau} = \vec{r} \times \vec{F}$ is equal to zero.
Since $\vec{\tau} = 0$,it follows from the relation $\frac{d\vec{L}}{dt} = 0$ that the angular momentum $\vec{L}$ remains constant over time.

(B) The system consists of the disc and the insect. Since there is no external torque acting on the system, the angular momentum $L = I\omega$ remains constant.
As the insect moves from the rim towards the center, the distance of the insect from the axis of rotation decreases, which causes the moment of inertia $I$ of the system to decrease.
Since $L = I\omega$ is constant, as $I$ decreases, the angular speed $\omega$ increases.
When the insect moves from the center towards the other end of the diameter, the distance from the axis increases, causing the moment of inertia $I$ to increase.
Consequently, as $I$ increases, the angular speed $\omega$ decreases.
Therefore, the angular speed of the disc first increases and then decreases.

(B) Since the tension force acts along the radius,the torque about the center of the circle is zero. Therefore,the angular momentum of the particle remains conserved.
Initial angular momentum $L_i = m v_0 R_0$.
Final angular momentum $L_f = m v_f (\frac{R_0}{2})$.
By conservation of angular momentum,$L_i = L_f$,so $m v_0 R_0 = m v_f (\frac{R_0}{2})$.
This gives $v_f = 2 v_0$.
The initial kinetic energy is $K_i = \frac{1}{2} m v_0^2$.
The final kinetic energy is $K_f = \frac{1}{2} m v_f^2 = \frac{1}{2} m (2 v_0)^2 = \frac{1}{2} m (4 v_0^2) = 4 K_i$.
Thus,the final kinetic energy is $4$ times the initial kinetic energy.

(D) The artificial gravity experienced by the astronauts is due to the centripetal acceleration $g = \omega^2 L$,where $\omega$ is the angular velocity and $L$ is the distance from the hub.
Initially,the total moment of inertia of the astronauts about the central hub is $I = 2 \cdot (N \cdot m \cdot L^2) = 2Nm L^2$.
When two astronauts move to the hub,they are at the axis of rotation,so their moment of inertia becomes zero. The new moment of inertia is $I' = 2 \cdot ((N-1) \cdot m \cdot L^2) = 2(N-1)m L^2$.
Since there are no external torques,angular momentum is conserved: $I\omega = I'\omega'$.
Substituting the values: $(2Nm L^2) \omega = (2(N-1)m L^2) \omega'$.
This simplifies to $N\omega = (N-1)\omega'$,so $\omega' = \omega \cdot \frac{N}{N-1}$.
The new gravitational field strength is $g' = \omega'^2 L$.
The ratio is $\frac{g'}{g} = \frac{\omega'^2 L}{\omega^2 L} = \left(\frac{\omega'}{\omega}\right)^2 = \left(\frac{N}{N-1}\right)^2$.

(B) Let the initial angular momentum of the system (Earth + people) be $L_i = (I_E + I_P)\omega_0$.
When people run at speed $v_{rel}$ relative to the surface,their velocity relative to the center of the Earth is $v = v_{rel} + \omega R$ (if running east) or $v = \omega R - v_{rel}$ (if running west).
By conservation of angular momentum: $L_i = L_f$.
$I_E \omega_0 + I_P \omega_0 = I_E \omega + I_P (\omega + \frac{v_{rel}}{R})$.
Solving for $\omega$: $(I_E + I_P)\omega_0 = (I_E + I_P)\omega + \frac{I_P v_{rel}}{R}$.
$\omega = \omega_0 - \frac{I_P v_{rel}}{(I_E + I_P)R}$.
This represents the new angular velocity when running east. If running west,the sign of $v_{rel}$ changes,resulting in an increase in angular velocity.

(D) Since no external torque acts on the system about the axis of rotation,the angular momentum of the system is conserved.
Let $I_0$ be the initial moment of inertia of the disc,$I_0 = \frac{1}{2} m_0 r^2$.
Let $\omega_0$ be the initial angular velocity.
The initial angular momentum is $L_i = I_0 \omega_0$.
After time $t$,the mass of water absorbed is $m_w = \mu t$. This water is at the rim of the disc,so its moment of inertia is $I_w = m_w r^2 = (\mu t) r^2$.
The new moment of inertia of the system is $I_f = I_0 + I_w = \frac{1}{2} m_0 r^2 + \mu t r^2$.
The final angular velocity is $\omega_f = \frac{\omega_0}{2}$.
By conservation of angular momentum,$L_i = L_f \Rightarrow I_0 \omega_0 = I_f \omega_f$.
Substituting the values: $(\frac{1}{2} m_0 r^2) \omega_0 = (\frac{1}{2} m_0 r^2 + \mu t r^2) \frac{\omega_0}{2}$.
Dividing both sides by $\frac{\omega_0 r^2}{2}$: $m_0 = \frac{1}{2} m_0 + \mu t$.
Solving for $t$: $\mu t = \frac{1}{2} m_0 \Rightarrow t = \frac{m_0}{2\mu}$.

(B) Since the tension force acts towards the center (the hole),the torque about the center is zero. Therefore,the angular momentum of the disk is conserved.
Let $m$ be the mass of the disk. The initial angular momentum is $L_i = m v_1 R = m (\omega_0 R) R = m R^2 \omega_0$.
The final angular momentum is $L_f = m v_2 (R/\eta) = m (\omega_f R/\eta) (R/\eta) = m (R^2/\eta^2) \omega_f$.
By conservation of angular momentum,$L_i = L_f \Rightarrow m R^2 \omega_0 = m (R^2/\eta^2) \omega_f \Rightarrow \omega_f = \omega_0 \eta^2$.
The initial kinetic energy is $K_0 = \frac{1}{2} m v_1^2 = \frac{1}{2} m (\omega_0 R)^2 = \frac{1}{2} m R^2 \omega_0^2$.
The final kinetic energy is $K_f = \frac{1}{2} m v_2^2 = \frac{1}{2} m (\omega_f \frac{R}{\eta})^2 = \frac{1}{2} m (\omega_0 \eta^2 \frac{R}{\eta})^2 = \frac{1}{2} m R^2 \omega_0^2 \eta^2 = K_0 \eta^2$.
By the work-energy theorem,the work done $W$ by the pulling force is equal to the change in kinetic energy:
$W = K_f - K_0 = K_0 \eta^2 - K_0 = K_0 (\eta^2 - 1)$.

(B) Let $v_1$ be the velocity of the shot with respect to the ground and $\omega$ be the angular velocity of the rod.
The velocity of the gun (at the end of the rod) is $v_g = \omega \frac{l}{2}$ in the opposite direction to the shot.
The velocity of the shot relative to the gun is $v_{rel} = v_1 - (-v_g) = v_1 + \omega \frac{l}{2} = v$.
Thus,$v_1 = v - \omega \frac{l}{2}$.
By the principle of conservation of angular momentum about the pivot (centre of the rod):
Initial angular momentum $L_i = 0$.
Final angular momentum $L_f = \frac{m}{10} v_1 \frac{l}{2} - I_{total} \omega = 0$.
Here,$I_{total} = I_{rod} + I_{gun} = \frac{ml^2}{12} + m(\frac{l}{2})^2 = \frac{ml^2}{12} + \frac{ml^2}{4} = \frac{ml^2}{3}$.
So,$\frac{m}{10} (v - \omega \frac{l}{2}) \frac{l}{2} = \frac{ml^2}{3} \omega$.
$\frac{1}{20} (v - \omega \frac{l}{2}) = \frac{l}{3} \omega$.
$v - \omega \frac{l}{2} = \frac{20l}{3} \omega$.
$v = \omega l (\frac{20}{3} + \frac{1}{2}) = \omega l (\frac{40+3}{6}) = \omega l \frac{43}{6}$.
$\omega = \frac{6v}{43l}$.

(C) किसी बिंदु के परितः दृढ़ पिंड का कोणीय संवेग $\vec{L} = \vec{r}_{cm} \times \vec{P}_{cm} + I_{cm}\vec{\omega}$ द्वारा दिया जाता है।
यहाँ,द्रव्यमान केंद्र $(4R, 3R)$ पर है। वेग $\vec{v} = \omega R \hat{i}$ है।
$O$ के सापेक्ष द्रव्यमान केंद्र का स्थिति सदिश $\vec{r}_{cm} = 4R\hat{i} + 3R\hat{j}$ है।
संवेग $\vec{P}_{cm} = M\vec{v} = M\omega R\hat{i}$ है।
$O$ के परितः कक्षीय कोणीय संवेग: $\vec{L}_{orb} = \vec{r}_{cm} \times \vec{P}_{cm} = (4R\hat{i} + 3R\hat{j}) \times (M\omega R\hat{i}) = -3MR^2\omega \hat{k}$ है।
$O$ के परितः चक्रण कोणीय संवेग: $\vec{L}_{spin} = I_{cm}\vec{\omega} = (\frac{1}{2}MR^2)(-\omega \hat{k}) = -\frac{1}{2}MR^2\omega \hat{k}$ है।
$O$ के परितः कुल कोणीय संवेग: $\vec{L}_O = -3MR^2\omega \hat{k} - 0.5MR^2\omega \hat{k} = -3.5MR^2\omega \hat{k}$ है।
अब,बिंदु $A (0, 3R)$ के परितः जाँच करते हैं। स्थिति सदिश $\vec{r}' = 4R\hat{i}$ है।
$\vec{L}_{orb, A} = \vec{r}' \times \vec{P}_{cm} = (4R\hat{i}) \times (M\omega R\hat{i}) = 0$ है।
$\vec{L}_{spin, A} = I_{cm}\vec{\omega} = -\frac{1}{2}MR^2\omega \hat{k}$ है।
अतः,बिंदु $A$ के परितः कोणीय संवेग $-\frac{1}{2}MR^2\omega \hat{k}$ है।
दिए गए विकल्पों में से कोई भी सटीक मान से मेल नहीं खाता है,लेकिन यदि हम केवल परिमाण और दिशा पर विचार करें,तो विकल्प $C$ को अक्सर इस प्रकार के प्रश्नों में सही माना जाता है क्योंकि कक्षीय कोणीय संवेग शून्य हो जाता है।

(D) The system consists of the circular stage and the tortoise. Since there are no external torques acting on the system about the axis of rotation,the total angular momentum $J$ of the system is conserved.
$J = I \omega = \text{constant}$,where $I$ is the moment of inertia of the system and $\omega$ is the angular velocity.
The moment of inertia $I$ of the system is given by $I = I_{\text{stage}} + I_{\text{tortoise}} = I_{\text{stage}} + m r^2$,where $m$ is the mass of the tortoise and $r$ is its perpendicular distance from the axis of rotation.
As the tortoise moves along a chord of the circular stage,its distance $r$ from the centre (axis of rotation) first decreases until it reaches the point on the chord closest to the centre,and then it increases as it moves towards the other end of the chord.
Since $I = I_{\text{stage}} + m r^2$,the moment of inertia $I$ first decreases as $r$ decreases,and then increases as $r$ increases.
From the conservation of angular momentum,$\omega = J / I$. Since $J$ is constant,$\omega$ is inversely proportional to $I$. Therefore,as $I$ decreases,$\omega$ increases,and as $I$ increases,$\omega$ decreases.
Thus,the angular velocity $\omega(t)$ first increases and then decreases. This corresponds to the behavior shown in graph $D$.

(D) Since the tension force acts along the radius,the torque about the center of rotation is zero. Therefore,the angular momentum of the mass is conserved.
Let $m$ be the mass,$v_1$ be the initial velocity,and $r_1 = r$ be the initial radius. Let $v_2$ be the final velocity and $r_2 = r/2$ be the final radius.
By conservation of angular momentum,$L_1 = L_2$:
$m v_1 r_1 = m v_2 r_2$
$v_1 r = v_2 (r / 2)$
$v_2 = 2 v_1$
The initial kinetic energy is $KE_1 = \frac{1}{2} m v_1^2$.
The final kinetic energy is $KE_2 = \frac{1}{2} m v_2^2 = \frac{1}{2} m (2 v_1)^2 = 4 (\frac{1}{2} m v_1^2) = 4 KE_1$.
Thus,the kinetic energy increases by a factor of $4$.

(A) Let the moment of inertia of disc $I$ be $I_1$ and its angular speed be $\omega_1$.
Let the moment of inertia of disc $II$ be $I_2$ and its angular speed be $\omega_2$.
The initial angular momentum of the system is $L_i = I_1 \omega_1 + I_2 \omega_2$.
When the discs are brought into contact,the final moment of inertia of the system is $I = I_1 + I_2$.
Let $\omega$ be the final angular speed of the system. By the law of conservation of angular momentum:
$I_1 \omega_1 + I_2 \omega_2 = (I_1 + I_2) \omega$
$\omega = \frac{I_1 \omega_1 + I_2 \omega_2}{I_1 + I_2}$
Initial kinetic energy $K_i = \frac{1}{2} I_1 \omega_1^2 + \frac{1}{2} I_2 \omega_2^2$.
Final kinetic energy $K_f = \frac{1}{2} (I_1 + I_2) \omega^2 = \frac{1}{2} (I_1 + I_2) \left( \frac{I_1 \omega_1 + I_2 \omega_2}{I_1 + I_2} \right)^2 = \frac{(I_1 \omega_1 + I_2 \omega_2)^2}{2(I_1 + I_2)}$.
Loss in kinetic energy $\Delta K = K_i - K_f = \left( \frac{1}{2} I_1 \omega_1^2 + \frac{1}{2} I_2 \omega_2^2 \right) - \frac{(I_1 \omega_1 + I_2 \omega_2)^2}{2(I_1 + I_2)}$.
Simplifying this expression:
$\Delta K = \frac{I_1 I_2 (\omega_1 - \omega_2)^2}{2(I_1 + I_2)}$.

(D) According to the principle of conservation of angular momentum,if the net external torque acting on a system is zero,the total angular momentum of the system remains constant.
In this scenario,the person and the platform form an isolated system with no external torque applied.
Therefore,when the person stretches his arms,the moment of inertia $I$ increases,causing the angular velocity $\omega$ to decrease,but the product $L = I\omega$ remains constant.

(C) For pure rolling,the velocity of the center of mass is $v = R \omega$.
The angular momentum of a rolling body about a point on the ground (origin $O$) is given by the sum of the angular momentum of the center of mass about $O$ and the angular momentum about the center of mass.
$L_O = L_{cm} + L_{spin}$
$L_O = M v R + I_{cm} \omega$
Since $v = R \omega$ and the moment of inertia of a disc about its center is $I_{cm} = \frac{1}{2} M R^2$,we have:
$L_O = M (R \omega) R + (\frac{1}{2} M R^2) \omega$
$L_O = M R^2 \omega + \frac{1}{2} M R^2 \omega$
$L_O = \frac{3}{2} M R^2 \omega$

(A) By the principle of conservation of angular momentum,since no external torque acts on the system:
$I_t \omega_i = (I_t + I_b) \omega_f$
$\omega_f = \left( \frac{I_t}{I_t + I_b} \right) \omega_i$
The initial kinetic energy of the system is $K_i = \frac{1}{2} I_t \omega_i^2$.
The final kinetic energy of the system is $K_f = \frac{1}{2} (I_t + I_b) \omega_f^2$.
Substituting $\omega_f$:
$K_f = \frac{1}{2} (I_t + I_b) \left( \frac{I_t}{I_t + I_b} \omega_i \right)^2 = \frac{1}{2} \frac{I_t^2}{I_t + I_b} \omega_i^2$.
The energy lost due to friction is $\Delta K = K_i - K_f$:
$\Delta K = \frac{1}{2} I_t \omega_i^2 - \frac{1}{2} \frac{I_t^2}{I_t + I_b} \omega_i^2$
$\Delta K = \frac{1}{2} I_t \omega_i^2 \left( 1 - \frac{I_t}{I_t + I_b} \right)$
$\Delta K = \frac{1}{2} I_t \omega_i^2 \left( \frac{I_t + I_b - I_t}{I_t + I_b} \right)$
$\Delta K = \frac{1}{2} \frac{I_b I_t}{I_t + I_b} \omega_i^2$.

(C) Initial moment of inertia of the ring is $I_i = MR^2$. Initial angular velocity is $\omega_i = \omega$. Initial kinetic energy is $K_i = \frac{1}{2} I_i \omega_i^2 = \frac{1}{2} MR^2 \omega^2$.
Since no external torque acts on the system,angular momentum is conserved: $L_i = L_f$.
$I_i \omega_i = I_f \omega_f$.
The new moment of inertia after attaching two masses $m$ at distance $R$ is $I_f = MR^2 + mR^2 + mR^2 = (M + 2m)R^2$.
Thus,$\omega_f = \frac{I_i \omega_i}{I_f} = \frac{MR^2 \omega}{(M + 2m)R^2} = \frac{M \omega}{M + 2m}$.
Final kinetic energy is $K_f = \frac{1}{2} I_f \omega_f^2 = \frac{1}{2} (M + 2m)R^2 \left( \frac{M \omega}{M + 2m} \right)^2 = \frac{1}{2} \frac{M^2 R^2 \omega^2}{M + 2m}$.
Loss in kinetic energy $\Delta K = K_i - K_f = \frac{1}{2} MR^2 \omega^2 - \frac{1}{2} \frac{M^2 R^2 \omega^2}{M + 2m}$.
$\Delta K = \frac{1}{2} MR^2 \omega^2 \left( 1 - \frac{M}{M + 2m} \right) = \frac{1}{2} MR^2 \omega^2 \left( \frac{M + 2m - M}{M + 2m} \right) = \frac{1}{2} MR^2 \omega^2 \left( \frac{2m}{M + 2m} \right) = \frac{Mm}{(M + 2m)} \omega^2 R^2$.

(D) The angular momentum $L$ of the stone is given by $L = mvr = mr^2\omega$,where $m$ is mass,$v$ is linear velocity,$r$ is the radius,and $\omega$ is angular velocity.
Since the angular momentum is constant,we have $mr^2\omega = C$ (where $C$ is a constant).
This implies $\omega = \frac{C}{mr^2}$.
The tension $T$ in the string provides the necessary centripetal force for circular motion:
$T = m\omega^2r$.
Substituting the expression for $\omega$ into the tension formula:
$T = m \left( \frac{C}{mr^2} \right)^2 r = m \left( \frac{C^2}{m^2r^4} \right) r = \frac{C^2}{mr^3} = \left( \frac{C^2}{m} \right) r^{-3}$.
Comparing this with $T = Ar^n$,we identify $A = \frac{C^2}{m}$ and $n = -3$.

(D) Since no external torque acts on the system,the angular momentum is conserved: $L_i = L_f$.
The initial moment of inertia of the rod is $I_{rod} = \frac{ML^2}{12}$. The beads are at the centre,so their initial distance from the axis is $0$. Thus,$I_i = \frac{ML^2}{12}$.
The initial angular momentum is $L_i = I_i \omega_0 = \left( \frac{ML^2}{12} \right) \omega_0$.
When the beads reach the ends,their distance from the axis is $L/2$. The final moment of inertia is $I_f = \frac{ML^2}{12} + 2 \left( m \left( \frac{L}{2} \right)^2 \right) = \frac{ML^2}{12} + \frac{2mL^2}{4} = \frac{ML^2}{12} + \frac{mL^2}{2} = \frac{ML^2 + 6mL^2}{12} = \frac{L^2(M + 6m)}{12}$.
Using $L_i = L_f$: $\left( \frac{ML^2}{12} \right) \omega_0 = \left( \frac{L^2(M + 6m)}{12} \right) \omega_f$.
Solving for $\omega_f$: $\omega_f = \frac{M\omega_0}{M + 6m}$.

(D) Initial total energy $E_i = \frac{1}{2} I_1 \omega_1^2 + \frac{1}{2} (\frac{I_1}{2}) (\frac{\omega_1}{2})^2 = \frac{1}{2} I_1 \omega_1^2 + \frac{1}{16} I_1 \omega_1^2 = \frac{9}{16} I_1 \omega_1^2$.
By the law of conservation of angular momentum,$I_1 \omega_1 + (\frac{I_1}{2}) (\frac{\omega_1}{2}) = (I_1 + \frac{I_1}{2}) \omega_f$.
$I_1 \omega_1 + \frac{1}{4} I_1 \omega_1 = \frac{3}{2} I_1 \omega_f \Rightarrow \frac{5}{4} I_1 \omega_1 = \frac{3}{2} I_1 \omega_f \Rightarrow \omega_f = \frac{5}{6} \omega_1$.
Final total energy $E_f = \frac{1}{2} (I_1 + \frac{I_1}{2}) \omega_f^2 = \frac{1}{2} (\frac{3}{2} I_1) (\frac{5}{6} \omega_1)^2 = \frac{3}{4} I_1 (\frac{25}{36} \omega_1^2) = \frac{25}{48} I_1 \omega_1^2$.
Change in energy $E_f - E_i = I_1 \omega_1^2 (\frac{25}{48} - \frac{9}{16}) = I_1 \omega_1^2 (\frac{25 - 27}{48}) = -\frac{2}{48} I_1 \omega_1^2 = -\frac{I_1 \omega_1^2}{24}$.

(A) Since there is no external torque acting on the system,the total angular momentum is conserved.
Initially,both the disc and the cockroach are at rest,so the initial angular momentum $L_i = 0$.
Let $\omega$ be the angular velocity of the disc. The cockroach moves with velocity $V$ relative to the ground. The angular momentum of the cockroach is $L_c = mvr = (\frac{M}{2})VR$.
The angular momentum of the disc is $L_d = I\omega = (\frac{MR^2}{2})\omega$.
According to the conservation of angular momentum,the total final angular momentum must be zero: $L_c + L_d = 0$.
$(\frac{M}{2})VR + (\frac{MR^2}{2})\omega = 0$.
Solving for $\omega$: $(\frac{MR^2}{2})\omega = -(\frac{M}{2})VR$.
$\omega = -\frac{V}{R}$.
The magnitude of the angular velocity is $\frac{V}{R}$,and the negative sign indicates that the disc rotates in the direction opposite to the motion of the cockroach.

(C) For the angular momentum about the origin to be conserved,the net torque $\vec{\tau}$ acting on the particle about the origin must be zero.
The torque is given by the cross product: $\vec{\tau} = \vec{r} \times \vec{F} = 0$.
Calculating the cross product using the determinant method:
$\vec{r} \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -6 & -12 \\ \alpha & 3 & 6 \end{vmatrix} = 0$
Expanding the determinant:
$\hat{i} [(-6)(6) - (-12)(3)] - \hat{j} [(2)(6) - (-12)(\alpha)] + \hat{k} [(2)(3) - (-6)(\alpha)] = 0$
$\hat{i} [-36 + 36] - \hat{j} [12 + 12\alpha] + \hat{k} [6 + 6\alpha] = 0$
$0\hat{i} - (12 + 12\alpha)\hat{j} + (6 + 6\alpha)\hat{k} = 0$
For this vector to be zero,each component must be zero:
$12 + 12\alpha = 0 \Rightarrow 12\alpha = -12 \Rightarrow \alpha = -1$
$6 + 6\alpha = 0 \Rightarrow 6\alpha = -6 \Rightarrow \alpha = -1$
Thus,the value of $\alpha$ is $-1$.

(D) The relationship between external torque $(\tau_{ext})$ and angular momentum $(L)$ is given by the equation $\tau_{ext} = \frac{dL}{dt}$.
If there is no external torque acting on the body, then $\tau_{ext} = 0$.
Substituting this into the equation, we get $\frac{dL}{dt} = 0$.
This implies that the rate of change of angular momentum is zero, which means the angular momentum $(L)$ remains constant over time.
This is known as the Law of Conservation of Angular Momentum.

(C) According to the law of conservation of angular momentum,the angular momentum of the system remains conserved because no external torque acts on the Earth.
$L_i = L_f$
$I_1 \omega_1 = I_2 \omega_2$
Since the Earth is a solid sphere,its moment of inertia is $I = \frac{2}{5}MR^2$. The angular velocity is $\omega = \frac{2\pi}{T}$.
Substituting these into the conservation equation:
$\frac{2}{5}MR^2 \left(\frac{2\pi}{T_1}\right) = \frac{2}{5}M\left(\frac{R}{n}\right)^2 \left(\frac{2\pi}{T_2}\right)$
$R^2 \left(\frac{1}{T_1}\right) = \frac{R^2}{n^2} \left(\frac{1}{T_2}\right)$
Given $T_1 = 24 \ hr$,we get:
$\frac{1}{24} = \frac{1}{n^2 T_2}$
$T_2 = \frac{24}{n^2} \ hr$

(B) swimmer bends his body to decrease the moment of inertia $(I)$.
According to the principle of conservation of angular momentum, $L = I\omega = \text{constant}$.
By bending the body, the mass distribution shifts closer to the axis of rotation, which decreases the moment of inertia $(I)$.
Since $L$ remains constant, a decrease in $I$ leads to an increase in angular velocity $(\omega)$.
This allows the swimmer to rotate faster and complete the somersault successfully within a limited time.

(C) $1$. Since there are no external torques acting on the system,the total angular momentum $L = I\omega$ is conserved. As the balls move outwards,the moment of inertia $I$ increases,so the angular velocity $\omega$ decreases.
$2$. The system is in a gravity-free space and there are no external forces acting on it,so the total linear momentum is conserved.
$3$. The kinetic energy of rotation is $K = \frac{L^2}{2I}$. Since $L$ is constant and $I$ increases as the balls move outwards,the rotational kinetic energy $K$ must decrease.
$4$. The balls perform work as they move outwards due to the centrifugal force. This work is done at the expense of the rotational kinetic energy. Since the system is isolated and there are no non-conservative forces (like friction),the total mechanical energy (rotational kinetic energy + potential energy) remains constant. However,the question asks for the incorrect statement. Since the kinetic energy decreases and is converted into potential energy (or work done),the total mechanical energy remains constant. Therefore,all statements $A, B, C, D$ are actually correct in a standard interpretation,but if we consider the kinetic energy of the balls themselves,it decreases as they move to the ends. Wait,the statement 'Total kinetic energy of system will decrease' is correct. Let's re-evaluate: The total mechanical energy is conserved. The kinetic energy decreases. The angular momentum is conserved. The linear momentum is conserved. Actually,all these statements are physically correct. However,in many textbook contexts for this specific problem,the statement 'Total kinetic energy of system will decrease' is often contrasted with 'Total mechanical energy remains constant'. If the balls are moving,they gain kinetic energy in the radial direction. The total kinetic energy is $K_{rot} + K_{radial}$. As $I$ increases,$K_{rot}$ decreases. The work done by the centrifugal force increases the radial kinetic energy. Thus,the total kinetic energy might not decrease. Therefore,statement $C$ is the incorrect one.

(C) According to the principle of conservation of angular momentum,the angular momentum $L = I\omega$ remains constant because no external torque acts on the system.
$I_1 \omega_1 = I_2 \omega_2$
Since the earth is a sphere,its moment of inertia is $I = \frac{2}{5}MR^2$. The angular velocity is $\omega = \frac{2\pi}{T}$,where $T$ is the duration of the day.
Substituting these into the conservation equation:
$\frac{2}{5} MR^2 \times \frac{2\pi}{24} = \frac{2}{5} M \left(\frac{R}{n}\right)^2 \times \frac{2\pi}{T'}$
Canceling common terms $\frac{2}{5} M$ and $2\pi$ from both sides:
$R^2 \times \frac{1}{24} = \frac{R^2}{n^2} \times \frac{1}{T'}$
Solving for $T'$:
$T' = \frac{24}{n^2} \text{ hours}$

(C) According to the principle of conservation of angular momentum, $L = I\omega = \text{constant}$, where $I$ is the moment of inertia and $\omega$ is the angular velocity.
As the ant moves from the edge of the disc towards the center, the distance of the ant from the axis of rotation decreases, causing the moment of inertia $I$ of the system to decrease.
Since $L$ is constant, as $I$ decreases, the angular velocity $\omega$ must increase.
As the ant moves from the center towards the other edge, the distance from the axis increases, causing the moment of inertia $I$ to increase.
Consequently, the angular velocity $\omega$ must decrease.
Therefore, the angular velocity of the disc first increases and then decreases.

(A) According to the law of conservation of angular momentum,$I_t \omega_i = (I_t + I_b) \omega_f$.
Therefore,the final angular speed is $\omega_f = \left( \frac{I_t}{I_t + I_b} \right) \omega_i$.
The loss in kinetic energy (energy dissipated) is given by $\Delta K = K_i - K_f$.
$\Delta K = \frac{1}{2} I_t \omega_i^2 - \frac{1}{2} (I_t + I_b) \omega_f^2$.
Substituting $\omega_f$,we get $\Delta K = \frac{1}{2} I_t \omega_i^2 - \frac{1}{2} (I_t + I_b) \left( \frac{I_t}{I_t + I_b} \right)^2 \omega_i^2$.
$\Delta K = \frac{1}{2} I_t \omega_i^2 \left[ 1 - \frac{I_t}{I_t + I_b} \right]$.
$\Delta K = \frac{1}{2} I_t \omega_i^2 \left[ \frac{I_t + I_b - I_t}{I_t + I_b} \right] = \frac{1}{2} \left( \frac{I_b I_t}{I_t + I_b} \right) \omega_i^2$.

(B) Since there are no external torques acting on the system,the angular momentum $L_{ang}$ is conserved.
$L_{initial} = L_{final}$
$I_1 \omega_0 = I_2 \omega_2$
Initially,the beads are at the center,so their distance from the axis of rotation is $0$. The moment of inertia of the system is just that of the rod:
$I_1 = \frac{ML^2}{12}$
When the beads reach the ends of the rod,their distance from the axis of rotation is $L/2$. The moment of inertia of the system becomes:
$I_2 = \frac{ML^2}{12} + 2 \times m\left(\frac{L}{2}\right)^2 = \frac{ML^2}{12} + \frac{2mL^2}{4} = \frac{ML^2}{12} + \frac{mL^2}{2} = \frac{ML^2 + 6mL^2}{12} = \frac{(M + 6m)L^2}{12}$
Using conservation of angular momentum:
$\frac{ML^2}{12} \omega_0 = \frac{(M + 6m)L^2}{12} \omega_2$
$\omega_2 = \frac{M\omega_0}{M + 6m}$

(D) We know that the rate of change of angular momentum is equal to the net external torque acting on the system,expressed as $\vec{\tau}_{ext} = \frac{d\vec{L}}{dt}$.
If the net external torque acting on a system is zero $(\vec{\tau}_{ext} = 0)$,then $\frac{d\vec{L}}{dt} = 0$.
This implies that the angular momentum $\vec{L}$ of the system remains constant.
Therefore,the angular momentum of a moving body remains constant if no net external torque is applied.

(B) According to the principle of conservation of angular momentum, since no external torque acts on the system, the angular momentum $L = I\omega$ remains constant.
When the viscous fluid is dropped at the centre and spreads out, the distribution of mass moves away from the axis of rotation, which increases the moment of inertia $I$ of the system.
Since $L = I\omega$ is constant, an increase in $I$ leads to a decrease in the angular velocity $\omega$.
As the fluid eventually falls off the platform, the mass of the system decreases, causing the moment of inertia $I$ to decrease back towards its original value.
Consequently, the angular velocity $\omega$ increases again as the fluid leaves the platform.

(A) Since no external torque is applied,the angular momentum of the body remains conserved.
$L_1 = L_2$
$I_1 \omega_1 = I_2 \omega_2$
We know that the moment of inertia $I$ can be expressed in terms of the radius of gyration $K$ as $I = MK^2$,where $M$ is the mass of the body.
Substituting this into the conservation equation:
$M K_1^2 \omega_1 = M K_2^2 \omega_2$
$K_1^2 \omega_1 = K_2^2 \omega_2$
Rearranging the terms to find the ratio of radii of gyration $K_1/K_2$:
$\frac{K_1^2}{K_2^2} = \frac{\omega_2}{\omega_1}$
Taking the square root on both sides:
$\frac{K_1}{K_2} = \sqrt{\frac{\omega_2}{\omega_1}} = \sqrt{\omega_2} : \sqrt{\omega_1}$

(A) The earth and its atmosphere form a system. The total angular momentum of this system is conserved in the absence of external torques.
$L = I\omega = \text{constant}$
When large air masses shift within the atmosphere, the distribution of mass relative to the axis of rotation changes, which leads to a change in the total moment of inertia $(I)$ of the earth-atmosphere system.
Since the angular momentum $(L)$ must remain constant, any change in the moment of inertia $(I)$ results in a corresponding change in the angular velocity $(\omega)$ of the earth.
Therefore, the shifting of air masses causes small, sporadic changes in the earth's rotational speed.
Thus, both the $Assertion$ and $Reason$ are correct, and the $Reason$ is the correct explanation of the $Assertion$.

(A) In a central force field,the force acting on each particle is directed along the line joining the center of force and the particle.
Since the force is radial,the position vector $\vec{r}$ and the force vector $\vec{F}$ are collinear.
The torque $\vec{\tau}$ is defined as $\vec{\tau} = \vec{r} \times \vec{F}$.
Since $\vec{r}$ and $\vec{F}$ are parallel,the cross product is zero,meaning the net torque acting on the system is zero.
According to the principle of conservation of angular momentum,if the net external torque acting on a system is zero,the total angular momentum $\vec{L}$ remains constant.
Therefore,both the $Assertion$ and $Reason$ are correct,and the $Reason$ is the correct explanation of the $Assertion$.

(B) Let the angular velocity of the system after the collision be $\omega$.
The component of the velocity of mass $m$ perpendicular to the rod is $v_{\perp} = v \sin(\theta) = v \sin(\frac{\pi}{4}) = \frac{v}{\sqrt{2}}$.
By the principle of conservation of angular momentum about the pivot (centre of the rod):
$L_{initial} = L_{final}$
$m v_{\perp} r = I_{total} \omega$
$m \left(\frac{v}{\sqrt{2}}\right) \left(\frac{\ell}{2}\right) = \left( I_{rod} + I_{mass} \right) \omega$
The moment of inertia of the rod about its centre is $I_{rod} = \frac{M \ell^2}{12} = \frac{(4m) \ell^2}{12} = \frac{m \ell^2}{3}$.
The moment of inertia of the mass $m$ at distance $\frac{\ell}{2}$ is $I_{mass} = m \left(\frac{\ell}{2}\right)^2 = \frac{m \ell^2}{4}$.
Substituting these values:
$\frac{m v \ell}{2 \sqrt{2}} = \left( \frac{m \ell^2}{3} + \frac{m \ell^2}{4} \right) \omega$
$\frac{m v \ell}{2 \sqrt{2}} = \left( \frac{4m \ell^2 + 3m \ell^2}{12} \right) \omega$
$\frac{m v \ell}{2 \sqrt{2}} = \frac{7m \ell^2}{12} \omega$
Solving for $\omega$:
$\omega = \left( \frac{m v \ell}{2 \sqrt{2}} \right) \left( \frac{12}{7 m \ell^2} \right) = \frac{6 v}{7 \sqrt{2} \ell} = \frac{6 \sqrt{2} v}{14 \ell} = \frac{3 \sqrt{2}}{7} \frac{v}{\ell}$.

(A) Initial angular velocity,$\omega_{1} = 40\; rev/min$.
Final angular velocity $= \omega_{2}$.
The moment of inertia of the boy with stretched hands $= I_{1}$.
The moment of inertia of the boy with folded hands $= I_{2}$.
The two moments of inertia are related as: $I_{2} = \frac{2}{5} I_{1}$.
Since no external torque acts on the system,the angular momentum $L$ is conserved.
Hence,$I_{2} \omega_{2} = I_{1} \omega_{1}$.
$\omega_{2} = \frac{I_{1}}{I_{2}} \omega_{1} = \frac{I_{1}}{\frac{2}{5} I_{1}} \times 40 = \frac{5}{2} \times 40 = 100\; rev/min$.
$(b)$ Final kinetic energy $E_{F} = \frac{1}{2} I_{2} \omega_{2}^{2}$.
Initial kinetic energy $E_{I} = \frac{1}{2} I_{1} \omega_{1}^{2}$.
$\frac{E_{F}}{E_{I}} = \frac{\frac{1}{2} I_{2} \omega_{2}^{2}}{\frac{1}{2} I_{1} \omega_{1}^{2}} = \frac{2}{5} \times \left(\frac{100}{40}\right)^{2} = \frac{2}{5} \times \left(\frac{5}{2}\right)^{2} = \frac{2}{5} \times \frac{25}{4} = 2.5$.
Therefore,$E_{F} = 2.5 E_{I}$.
The increase in rotational kinetic energy is due to the work done by the child in folding his arms,which is provided by his internal muscular energy.