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

(B) According to the law of conservation of angular momentum,$L = I\omega = \text{constant}$.
Therefore,$\omega \propto \frac{1}{I}$.
The rotational kinetic energy is given by $E_R = \frac{1}{2} I \omega^2$.
Substituting $\omega \propto \frac{1}{I}$,we get $E_R \propto I \times \left(\frac{1}{I}\right)^2 = \frac{1}{I}$.
Given that the initial moment of inertia is $I_1 = I$ and the final moment of inertia is $I_2 = 2I$.
Thus,$\frac{E_2}{E_1} = \frac{I_1}{I_2} = \frac{I}{2I} = \frac{1}{2}$.
Therefore,the new kinetic energy $E_2 = \frac{E_1}{2} = \frac{K}{2}$.

(A) The initial moment of inertia of the ring is $I = MR^2$. The initial angular momentum is $L = I\omega = MR^2\omega$.
When $4$ point masses of mass $m$ are placed at the ends of two mutually perpendicular diameters,they are all at a distance $R$ from the axis of rotation.
The new moment of inertia of the system becomes $I' = MR^2 + 4(mR^2) = (M + 4m)R^2$.
Since there is no external torque acting on the system,the angular momentum remains conserved $(L_{initial} = L_{final})$.
$MR^2\omega = (M + 4m)R^2\omega'$
Dividing both sides by $R^2$,we get $M\omega = (M + 4m)\omega'$.
Therefore,the new angular velocity is $\omega' = \left( \frac{M}{M + 4m} \right)\omega$.

(A) When ice at the poles melts and flows towards the equator,the mass distribution shifts further away from the axis of rotation.
Since the moment of inertia $I = \sum mr^2$,as mass moves away from the axis,the moment of inertia $I$ increases.
According to the law of conservation of angular momentum,$L = I\omega = \text{constant}$.
If $I$ increases,the angular velocity $\omega$ must decrease.
Since the time period $T = \frac{2\pi}{\omega}$,a decrease in $\omega$ leads to an increase in the duration of the day $T$ (the day becomes longer).

(B) According to the law of conservation of angular momentum,since no external torque acts on the system,the initial angular momentum is equal to the final angular momentum.
Initial angular momentum $L_i = I_1 \omega$.
When the man moves to a distance $r$ from the center,the new moment of inertia of the system becomes $I_2 = I_1 + mr^2$.
Let the final angular velocity be $\omega_2$.
Final angular momentum $L_f = (I_1 + mr^2) \omega_2$.
Equating $L_i = L_f$:
$I_1 \omega = (I_1 + mr^2) \omega_2$.
Therefore,the final angular velocity is $\omega_2 = \frac{I_1 \omega}{I_1 + mr^2}$.

(A) Angular momentum is conserved when the net torque acting on the particle is zero,i.e.,$\vec{\tau} = \vec{r} \times \vec{F} = 0$.
Calculating the cross product:
$\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -6 & -12 \\ p & 3 & 6 \end{vmatrix}$
$\vec{\tau} = \hat{i}(-36 - (-36)) - \hat{j}(12 - (-12p)) + \hat{k}(6 - (-6p))$
$\vec{\tau} = \hat{i}(0) - \hat{j}(12 + 12p) + \hat{k}(6 + 6p)$
For angular momentum to be conserved,$\vec{\tau} = 0$,which implies:
$12 + 12p = 0$ and $6 + 6p = 0$
Solving $12 + 12p = 0$,we get $12p = -12$,so $p = -1$.
Similarly,$6 + 6p = 0$ also gives $p = -1$.

(A) Initial angular velocity $\omega_1 = \frac{2\pi}{T} = \frac{2\pi}{10} = \frac{\pi}{5} \ rad/s$.
Initial moment of inertia $I_1 = I_{table} + I_{man} = 100 + 50(0)^2 = 100 \ kg \cdot m^2$.
Final moment of inertia $I_2 = I_{table} + I_{man} = 100 + 50(2)^2 = 100 + 200 = 300 \ kg \cdot m^2$.
By the law of conservation of angular momentum,$L_1 = L_2 \Rightarrow I_1 \omega_1 = I_2 \omega_2$.
$100 \times \frac{\pi}{5} = 300 \times \omega_2$.
$20\pi = 300 \omega_2$.
$\omega_2 = \frac{20\pi}{300} = \frac{\pi}{15} \ rad/s$.

(D) Initially,the velocity of the center of mass of the ring is $v_0 = 0$ and the initial angular velocity is $\omega_0$.
When the ring stops slipping,let the velocity of the center of mass be $v$ and the angular velocity be $\omega$.
Since the ring is rolling without slipping,the condition $v = r\omega$ must be satisfied.
By the principle of conservation of angular momentum about a point $P$ on the surface:
$L_i = L_f$
$I_{cm}\omega_0 + mvr_0 = I_{cm}\omega + mvr$
Since $I_{cm} = mr^2$ for a ring:
$mr^2\omega_0 + 0 = mr^2\omega + mvr$
$mr^2\omega_0 = mr^2(\frac{v}{r}) + mvr$
$mr^2\omega_0 = mrv + mvr$
$mr^2\omega_0 = 2mvr$
$v = \frac{r\omega_0}{2}$

(C) Since there are no external torques about the pivot point $O$ during the collision,the angular momentum of the system is conserved.
Initial angular momentum of the particle about $O$ is $L_i = Mv \times r$,where $r = OC = L/4$.
So,$L_i = Mv(L/4)$.
After the collision,the particle sticks to the rod (assuming an inelastic collision),and the system rotates with angular velocity $\omega$.
The total moment of inertia $I$ of the system about $O$ is the sum of the moment of inertia of the rod and the particle.
$I = I_{rod} + I_{particle} = \frac{ML^2}{12} + M(L/4)^2 = \frac{ML^2}{12} + \frac{ML^2}{16} = \frac{4ML^2 + 3ML^2}{48} = \frac{7ML^2}{48}$.
Applying conservation of angular momentum: $L_i = I\omega$.
$Mv(L/4) = (7ML^2/48) \omega$.
Solving for $\omega$: $\omega = \frac{MvL}{4} \times \frac{48}{7ML^2} = \frac{12v}{7L}$.

(A) Let the initial volume be $V_1$ and the final volume be $V_2 = \frac{1}{64} V_1$.
Since $V = \frac{4}{3} \pi R^3$,we have $\frac{R_2^3}{R_1^3} = \frac{V_2}{V_1} = \frac{1}{64}$.
Taking the cube root,we get $\frac{R_2}{R_1} = \frac{1}{4}$,which implies $R_2 = \frac{R_1}{4}$.
Since there is no external torque acting on the Earth,the angular momentum $L = I\omega$ is conserved.
$I_1 \omega_1 = I_2 \omega_2 \implies I_1 \left( \frac{2\pi}{T_1} \right) = I_2 \left( \frac{2\pi}{T_2} \right)$.
Since the Earth is a solid sphere,$I = \frac{2}{5}MR^2$.
Substituting this,we get $\left( \frac{2}{5} M R_1^2 \right) \frac{1}{T_1} = \left( \frac{2}{5} M R_2^2 \right) \frac{1}{T_2}$.
$T_2 = T_1 \left( \frac{R_2}{R_1} \right)^2$.
Given $T_1 = 24 \text{ hours}$ and $\frac{R_2}{R_1} = \frac{1}{4}$,we have $T_2 = 24 \times \left( \frac{1}{4} \right)^2 = 24 \times \frac{1}{16} = 1.5 \text{ hours}$.

(C) According to the principle of conservation of angular momentum:
$L_{initial} = L_{final}$
$I\omega + mvR = (I + mR^2)\omega'$
Where $I$ is the moment of inertia of the solid cylinder,$I = \frac{1}{2}MR^2$.
$I = \frac{1}{2} \times 2 \times (0.2)^2 = 0.04 \ kg \cdot m^2$.
Initial angular momentum $L_i = I\omega + mvR = (0.04 \times 3) + (0.5 \times 5 \times 0.2) = 0.12 + 0.5 = 0.62 \ kg \cdot m^2/s$.
Final moment of inertia $I_f = I + mR^2 = 0.04 + (0.5 \times (0.2)^2) = 0.04 + 0.02 = 0.06 \ kg \cdot m^2$.
Using $L_i = I_f \omega'$:
$0.62 = 0.06 \times \omega'$
$\omega' = \frac{0.62}{0.06} = \frac{62}{6} \approx 10.33 \ rad/s$.
Thus,the angular velocity is $10.3 \ rad/s$.

(A) Let $V$ be the velocity of the center of mass of the rod after the collision. By conservation of linear momentum:
$mv = MV \implies V = \frac{mv}{M}$
By conservation of angular momentum about the center of mass of the rod:
$mvd = I\omega \implies \omega = \frac{mvd}{I}$,where $I = \frac{ML^2}{12}$
Assuming the collision is elastic,by conservation of kinetic energy:
$\frac{1}{2}mv^2 = \frac{1}{2}MV^2 + \frac{1}{2}I\omega^2$
Substituting $V$ and $\omega$:
$mv^2 = M(\frac{mv}{M})^2 + I(\frac{mvd}{I})^2$
$mv^2 = \frac{m^2v^2}{M} + \frac{m^2v^2d^2}{I}$
Dividing by $mv^2$:
$1 = \frac{m}{M} + \frac{md^2}{I} = \frac{m}{M} + \frac{md^2}{ML^2/12} = \frac{m}{M}(1 + \frac{12d^2}{L^2})$
$1 = \frac{m}{M}(\frac{L^2 + 12d^2}{L^2})$
$m = \frac{ML^2}{L^2 + 12d^2}$

(A) The angular momentum of a particle moving in a circular path is given by $L = mr^2\omega$.
Since no external torque acts on the system,the angular momentum remains conserved $(L_1 = L_2)$.
For the initial state: $L_1 = mr_1^2\omega_1$.
For the final state: $L_2 = mr_2^2\omega_2$.
Equating the two: $mr_1^2\omega_1 = mr_2^2\omega_2$.
This simplifies to $\omega_2 = \omega_1 \left( \frac{r_1}{r_2} \right)^2$.
Given $r_1 = 0.8 \ m$,$r_2 = 1 \ m$,and $\omega_1 = 44 \ rad \ s^{-1}$.
Substituting the values: $\omega_2 = 44 \times \left( \frac{0.8}{1} \right)^2 = 44 \times 0.64$.
$\omega_2 = 28.16 \ rad \ s^{-1}$.

(B) The initial moment of inertia of the ring about the vertical axis passing through its center is $I = M R^2$.
The initial angular momentum of the ring is $L = I \omega = M R^2 \omega$.
When two objects each of mass $m$ are attached to the opposite ends of a diameter,the new moment of inertia $I'$ becomes:
$I' = I + 2(m R^2) = M R^2 + 2m R^2 = (M + 2m) R^2$.
Since no external torque acts on the system,the angular momentum is conserved,so $L' = L$.
$(M + 2m) R^2 \omega' = M R^2 \omega$.
Solving for the new angular velocity $\omega'$:
$\omega' = \frac{M \omega}{M + 2m}$.

(D) According to the law of conservation of angular momentum,the initial angular momentum equals the final angular momentum: $I_t \omega_i = (I_t + I_b) \omega_f$.
Thus,the final angular speed is $\omega_f = \frac{I_t \omega_i}{I_t + I_b}$.
The initial kinetic energy is $K_i = \frac{1}{2} I_t \omega_i^2$.
The final kinetic energy is $K_f = \frac{1}{2} (I_t + I_b) \omega_f^2 = \frac{1}{2} (I_t + I_b) \left( \frac{I_t \omega_i}{I_t + I_b} \right)^2 = \frac{1}{2} \frac{I_t^2 \omega_i^2}{I_t + I_b}$.
The energy lost is $\Delta E = K_i - K_f = \frac{1}{2} I_t \omega_i^2 - \frac{1}{2} \frac{I_t^2 \omega_i^2}{I_t + I_b}$.
Simplifying this expression: $\Delta E = \frac{1}{2} I_t \omega_i^2 \left( 1 - \frac{I_t}{I_t + I_b} \right) = \frac{1}{2} I_t \omega_i^2 \left( \frac{I_t + I_b - I_t}{I_t + I_b} \right) = \frac{1}{2} \frac{I_t I_b}{I_t + I_b} \omega_i^2$.

(C) Since the system is initially at rest,the initial angular momentum $L_i = 0$.
According to the principle of conservation of angular momentum,the final angular momentum $L_f$ must also be $0$.
Let $\omega$ be the angular velocity of the platform. The angular momentum of the man is $L_m = m v R$ and the angular momentum of the platform is $L_p = I \omega$.
Since the total angular momentum is zero,the man and the platform move in opposite directions: $m v R - I \omega = 0$.
$\omega = \frac{m v R}{I} = \frac{50 \times 1 \times 2}{200} = 0.5 \, rad/s$.
The angular velocity of the man relative to the ground is $\omega_m = \frac{v}{R} = \frac{1}{2} = 0.5 \, rad/s$.
The angular velocity of the man relative to the platform is $\omega_r = \omega_m + \omega = 0.5 + 0.5 = 1 \, rad/s$.
The time taken to complete one revolution is $T = \frac{2\pi}{\omega_r} = \frac{2\pi}{1} = 2\pi \, s$.

(C) Since the tension force acts towards the center of the circle,the torque about the center is zero. Therefore,the angular momentum of the mass $m$ is conserved.
Initial angular momentum $L_i = m v_0 R_0$.
Final angular momentum $L_f = m v R$,where $R = \frac{R_0}{2}$.
By conservation of angular momentum,$L_i = L_f$:
$m v_0 R_0 = m v \left( \frac{R_0}{2} \right)$
$v = 2 v_0$.
The initial kinetic energy is $K_i = \frac{1}{2} m v_0^2$.
The final kinetic energy is $K_f = \frac{1}{2} m v^2 = \frac{1}{2} m (2 v_0)^2 = \frac{1}{2} m (4 v_0^2) = 2 m v_0^2$.

(B) For the conservation of angular momentum about the origin,the net torque $\overrightarrow{\tau}$ acting on the particle must be zero.
By definition,$\overrightarrow{\tau} = \overrightarrow{R} \times \overrightarrow{F}$.
Given $\overrightarrow{R} = 2\hat{i} - 6\hat{j} - 12\hat{k}$ and $\overrightarrow{F} = \alpha \hat{i} + 3\hat{j} + 6\hat{k}$.
Calculating the cross product:
$\overrightarrow{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -6 & -12 \\ \alpha & 3 & 6 \end{vmatrix}$
$\overrightarrow{\tau} = \hat{i}(-36 - (-36)) - \hat{j}(12 - (-12\alpha)) + \hat{k}(6 - (-6\alpha))$
$\overrightarrow{\tau} = \hat{i}(0) - \hat{j}(12 + 12\alpha) + \hat{k}(6 + 6\alpha)$
For $\overrightarrow{\tau} = 0$,both the $\hat{j}$ and $\hat{k}$ components must be zero:
$12 + 12\alpha = 0 \implies \alpha = -1$
$6 + 6\alpha = 0 \implies \alpha = -1$
Thus,the value of $\alpha$ is $-1$.

(D) Initial moment of inertia of the rod about the axis passing through its centre is $I_1 = \frac{ML^2}{12}$.
Initial angular momentum $L_1 = I_1 \omega = \frac{ML^2}{12} \omega$.
When two objects of mass $\frac{M}{3}$ are attached to the ends,the new moment of inertia $I_2$ is:
$I_2 = I_{\text{rod}} + I_{\text{objects}} = \frac{ML^2}{12} + \left(\frac{M}{3}\right)\left(\frac{L}{2}\right)^2 + \left(\frac{M}{3}\right)\left(\frac{L}{2}\right)^2$
$I_2 = \frac{ML^2}{12} + \frac{ML^2}{12} + \frac{ML^2}{12} = \frac{3ML^2}{12} = \frac{ML^2}{4}$.
By the law of conservation of angular momentum,$L_1 = L_2$:
$\frac{ML^2}{12} \omega = \frac{ML^2}{4} \omega'$
$\omega' = \frac{4}{12} \omega = \frac{1}{3} \omega$.

(D) Initial angular momentum $L_i = I\omega_1 + I\omega_2$.
Let $\omega$ be the final angular speed of the combined system.
Final angular momentum $L_f = (I + I)\omega = 2I\omega$.
According to the law of conservation of angular momentum,$L_i = L_f$:
$I\omega_1 + I\omega_2 = 2I\omega \implies \omega = \frac{\omega_1 + \omega_2}{2}$.
Initial rotational kinetic energy $E_i = \frac{1}{2}I\omega_1^2 + \frac{1}{2}I\omega_2^2 = \frac{1}{2}I(\omega_1^2 + \omega_2^2)$.
Final rotational kinetic energy $E_f = \frac{1}{2}(2I)\omega^2 = I \left( \frac{\omega_1 + \omega_2}{2} \right)^2 = \frac{I}{4}(\omega_1^2 + \omega_2^2 + 2\omega_1\omega_2)$.
Loss of energy $\Delta E = E_i - E_f = \frac{I}{2}(\omega_1^2 + \omega_2^2) - \frac{I}{4}(\omega_1^2 + \omega_2^2 + 2\omega_1\omega_2)$.
$\Delta E = \frac{I}{4} [2\omega_1^2 + 2\omega_2^2 - \omega_1^2 - \omega_2^2 - 2\omega_1\omega_2] = \frac{I}{4}(\omega_1^2 + \omega_2^2 - 2\omega_1\omega_2) = \frac{I}{4}(\omega_1 - \omega_2)^2$.

(C) 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 case,the sphere is rotating freely in free space,meaning there is no external torque acting on it $({\tau_{ext}} = 0)$.
Since ${\tau_{ext}} = \frac{dL}{dt} = 0$,the angular momentum $(L)$ must remain constant.
When the radius of the sphere increases,its moment of inertia $(I = \frac{2}{5}MR^2)$ increases. Since $L = I\omega$ is constant,the angular velocity $(\omega)$ will decrease,and the rotational kinetic energy $(K = \frac{L^2}{2I})$ will also change. Therefore,only the angular momentum remains constant.

(C) According to the relation between torque and angular momentum,the rate of change of angular momentum is equal to the external torque applied to the system: $\tau = \frac{dL}{dt}$.
If the external torque $\tau = 0$,then $\frac{dL}{dt} = 0$.
This implies that the angular momentum $L$ remains constant in both magnitude and direction.

(C) The relationship between torque $\vec{\tau}$ and angular momentum $\vec{L}$ is given by $\vec{\tau} = \frac{d\vec{L}}{dt}$.
If the net torque $\vec{\tau} = 0$,then $\frac{d\vec{L}}{dt} = 0$.
This implies that $\vec{L}$ is a constant vector.
$A$ constant vector has both a constant magnitude and a constant direction.
Therefore,the angular momentum remains constant in both magnitude and direction.

(A) According to the law of conservation of angular momentum,$L = I\omega = mr^2\omega = \text{constant}$.
Since the mass $m$ remains constant,we have $r_1^2\omega_1 = r_2^2\omega_2$.
Given: $r_1 = 0.8 \ m$,$\omega_1 = 44 \ rad/s$,$r_2 = 1 \ m$.
Substituting the values: $(0.8)^2 \times 44 = (1)^2 \times \omega_2$.
$0.64 \times 44 = \omega_2$.
$\omega_2 = 28.16 \ rad/s$.

(B) According to the principle of conservation of angular momentum,the angular momentum $\vec{L}$ of a system is conserved if the net external torque $\vec{\tau}_{ext}$ acting on the system is zero.
This is derived from the relation $\vec{\tau}_{ext} = \frac{d\vec{L}}{dt}$.
If $\vec{\tau}_{ext} = 0$,then $\frac{d\vec{L}}{dt} = 0$,which implies $\vec{L} = \text{constant}$.
Therefore,the correct condition is that no external torque acts on the system.

(C) According to the law of conservation of angular momentum,the initial angular momentum equals the final angular momentum.
$L_i = L_f$
$I_1 \omega_1 = I_2 \omega_2$
Here,the moment of inertia of the first ring about the central axis is $I_1 = MR^2$.
When the second ring is placed coaxially,the total moment of inertia becomes $I_2 = I_1 + I_{ring2} = MR^2 + (\frac{M}{4})R^2 = \frac{5}{4}MR^2$.
Substituting these into the conservation equation:
$MR^2 \omega = (\frac{5}{4}MR^2) \omega_2$
$\omega_2 = \frac{4}{5} \omega$.

(A) Since there is no external torque acting on the dancer,the angular momentum of the system remains conserved.
According to the law of conservation of angular momentum: $L = I_1 \omega_1 = I_2 \omega_2$.
Given: Initial moment of inertia $I_1 = I$,initial angular velocity $\omega_1 = 20 \ rad/s$,and final angular velocity $\omega_2 = 10 \ rad/s$.
Substituting the values into the equation: $I \times 20 = I_2 \times 10$.
Solving for the new moment of inertia $I_2$: $I_2 = \frac{20}{10} I = 2I$.

(D) When the child sits on the rotating disc,there is no external torque acting on the system (disc + child) about the axis of rotation. According to the law of conservation of angular momentum,if the net external torque $\tau_{ext} = 0$,then the angular momentum $L$ of the system remains constant. Therefore,the angular momentum is conserved.

(A) Since no external torque is applied,the angular momentum $L$ of the body remains conserved.
$L = I_1 \omega_1 = I_2 \omega_2$
We know that the moment of inertia $I = MK^2$,where $M$ is the mass and $K$ is the radius of gyration.
Substituting this into the conservation equation:
$M K_1^2 \omega_1 = M K_2^2 \omega_2$
$K_1^2 \omega_1 = K_2^2 \omega_2$
$\frac{K_1^2}{K_2^2} = \frac{\omega_2}{\omega_1}$
Taking the square root on both sides,we get the ratio of the radii of gyration:
$\frac{K_1}{K_2} = \sqrt{\frac{\omega_2}{\omega_1}}$
Thus,the ratio $K_1 : K_2 = \sqrt{\omega_2} : \sqrt{\omega_1}$.

(A) According to the law of conservation of angular momentum, $L = I \omega = \text{constant}$.
Since the Earth is treated as a solid sphere, its moment of inertia is $I = \frac{2}{5}MR^2$.
Initially, $I_1 = \frac{2}{5}MR^2$ and the time period is $T_1 = 24 \text{ hours}$.
When the radius is halved, $R_2 = \frac{R}{2}$, the new moment of inertia is $I_2 = \frac{2}{5}M(\frac{R}{2})^2 = \frac{1}{4}I_1$.
Using $I_1 \omega_1 = I_2 \omega_2$, we have $I_1 (\frac{2\pi}{T_1}) = I_2 (\frac{2\pi}{T_2})$.
Substituting the values: $I_1 (\frac{2\pi}{24}) = (\frac{1}{4}I_1) (\frac{2\pi}{T_2})$.
This simplifies to $\frac{1}{24} = \frac{1}{4T_2}$, which gives $T_2 = \frac{24}{4} = 6 \text{ hours}$.

(C) According to the principle of conservation of angular momentum,if no external torque acts on the system,the angular momentum $L$ remains constant.
Since the plastic drop sticks to the disc,the moment of inertia $I$ of the system increases.
As $L = I\omega$ and $L$ is constant,an increase in $I$ results in a decrease in the angular velocity $\omega$.
Therefore,only the angular momentum $L$ remains constant.

(B) According to the law of conservation of angular momentum, $L = I\omega = \text{constant}$.
When the person folds their arms inward, the distribution of mass moves closer to the axis of rotation, which decreases the moment of inertia $(I)$.
Since $L$ is constant, if $I$ decreases, the angular velocity $(\omega)$ must increase to compensate.
Therefore, the moment of inertia decreases, and the angular velocity increases.

(C) According to the principle of conservation of angular momentum,the total angular momentum before joining equals the total angular momentum after joining.
$I_1\omega_1 + I_2\omega_2 = (I_1 + I_2)\omega$
where $\omega$ is the common angular velocity of the system.
Thus,the common angular velocity is $\omega = \frac{I_1\omega_1 + I_2\omega_2}{I_1 + I_2}$.
The rotational kinetic energy of the system is given by $K = \frac{1}{2} I_{total} \omega^2$.
Substituting the values,$K = \frac{1}{2} (I_1 + I_2) \left( \frac{I_1\omega_1 + I_2\omega_2}{I_1 + I_2} \right)^2$.
Simplifying this,we get $K = \frac{(I_1\omega_1 + I_2\omega_2)^2}{2(I_1 + I_2)}$.

(D) The principle of conservation of angular momentum states that the angular momentum of a system remains constant if the net external torque acting on the system is zero.
Mathematically, if $\tau_{ext} = 0$, then $\frac{dL}{dt} = 0$, which implies $L = \text{constant}$.
Option $A$ is incorrect because it is only conserved in the absence of external torque.
Option $B$ is incorrect as angular momentum is the product of moment of inertia and angular velocity $(L = I\omega)$.
Option $C$ is incorrect because it requires the torque to be zero, not just constant.
Therefore, the correct answer is $D$.

(D) According to the law of conservation of angular momentum,if no external torque acts on the system,the angular momentum $(L)$ remains constant.
$L = I \omega$,where $I$ is the moment of inertia and $\omega$ is the angular velocity.
When the person spreads their arms,the moment of inertia $(I)$ increases because the mass is distributed further from the axis of rotation.
To keep the angular momentum $(L)$ constant,the angular velocity $(\omega)$ decreases.
Therefore,the angular momentum remains unchanged.

(A) According to the law of conservation of angular momentum, $L = I\omega = \text{constant}$.
Since the Earth is a sphere, its moment of inertia is $I = \frac{2}{5}MR^2$.
Thus, $I \propto R^2$.
Since $L = I\omega = I \cdot \frac{2\pi}{T}$, we have $I_1 \cdot \frac{1}{T_1} = I_2 \cdot \frac{1}{T_2}$, which implies $T \propto I \propto R^2$.
If the radius $R$ becomes $R' = \frac{R}{2}$, then the new time period $T'$ is $T' = T \cdot (\frac{R'}{R})^2 = T \cdot (\frac{1}{2})^2 = \frac{T}{4}$.
Given the current duration of a day $T = 24 \text{ hours}$, the new duration is $T' = \frac{24}{4} = 6 \text{ hours}$.
The decrease in the duration of the day is $\Delta T = T - T' = 24 - 6 = 18 \text{ hours}$.

(A) The angular momentum $L$ of a rotating disc is given by $L = I\omega$,where $I = \frac{1}{2}MR^2$ is the moment of inertia.
Since no external torque is applied,the angular momentum remains conserved,i.e.,$L = \text{constant}$.
Therefore,$I_1\omega_1 = I_2\omega_2$.
Substituting the expression for moment of inertia: $(\frac{1}{2}MR_1^2)\omega_1 = (\frac{1}{2}MR_2^2)\omega_2$.
Given $R_2 = \frac{R_1}{2}$,we have $R_1^2\omega_1 = (\frac{R_1}{2})^2\omega_2$.
$R_1^2\omega_1 = \frac{R_1^2}{4}\omega_2$.
Solving for $\omega_2$,we get $\omega_2 = 4\omega_1$.
Thus,the angular velocity becomes four times the original value.

(D) According to the law of conservation of angular momentum,$L = I\omega = \text{constant}$.
When the ice melts,the water spreads out over the surface of the table,increasing the distance of the mass from the axis of rotation.
This increase in the distribution of mass leads to an increase in the moment of inertia $(I)$ of the system.
Since $L$ remains constant and $I$ increases,the angular velocity $\omega$ must decrease to satisfy the equation $\omega = L/I$.
Therefore,the rotational speed of the system will become less than $\omega$.

(B) The rotational kinetic energy $E$ is given by the formula $E = \frac{L^2}{2I}$,where $L$ is the angular momentum and $I$ is the moment of inertia.
Since there are no external torques acting on the system,the angular momentum $L$ remains constant.
From the relation $E \propto \frac{1}{I}$,it is clear that if the moment of inertia $I$ increases,the kinetic energy $E$ must decrease.
In the first state,$I_1 = 1 \ kg \cdot m^2$. In the second state,$I_2 = 2 \ kg \cdot m^2$.
Since $I_2 > I_1$,the kinetic energy in the second state will be less than in the first state.

(D) According to the principle of conservation of angular momentum, $L = I\omega = \text{constant}$ when no external torque acts on the system.
When the person pulls their hands in, the moment of inertia $(I)$ of the system decreases because the mass distribution moves closer to the axis of rotation.
Since $L$ remains constant, if $I$ decreases, the angular velocity $(\omega)$ must increase to compensate.
Therefore, the correct statement is that the angular velocity increases.

(C) According to the principle of conservation of angular momentum,if no external torque acts on a system,the angular momentum $(L = I\omega)$ remains constant.
When the person pulls their arms inward,the moment of inertia $(I)$ decreases.
Since $L = I\omega$ is constant,if $I$ decreases,the angular velocity $(\omega)$ must increase,causing the person to spin faster.
The rotational kinetic energy is given by $K = \frac{L^2}{2I}$. Since $L$ is constant and $I$ decreases,the kinetic energy $(K)$ increases.

(B) According to the principle of conservation of angular momentum,$L = I\omega = \text{constant}$.
When the person drops the masses,the moment of inertia $(I)$ of the system decreases because the mass is removed from the system at a certain distance from the axis of rotation.
Since $L = I\omega$ remains constant,if $I$ decreases,the angular velocity $\omega$ must increase to compensate.
Therefore,the new angular velocity will be greater than the initial angular velocity $\omega$.

(A) When the string is pulled downwards,the force applied by the hand acts along the radius of the circular path (towards the center).
Since the force is radial,the torque $\vec{\tau} = \vec{r} \times \vec{F}$ about the center of the circle is zero because the angle between the position vector $\vec{r}$ and the force $\vec{F}$ is $180^{\circ}$ (or $0^{\circ}$ depending on the reference).
According to the principle of conservation of angular momentum,if the net external torque acting on a system is zero,the angular momentum $L = mvr$ remains conserved.
Therefore,the angular momentum of the object remains constant.

(A) By pulling his hands and legs inward,the swimmer decreases his moment of inertia $(I)$.
According to the principle of conservation of angular momentum,$L = I\omega$ remains constant.
Since $I$ decreases,the angular velocity $(\omega)$ increases,which allows the swimmer to perform somersaults more easily.

(C) According to the principle of conservation of angular momentum,since no external torque acts on the system of two discs,the total angular momentum remains constant.
Initial angular momentum of the system: $L_{i} = I_{1}\omega + I_{2}(0) = I_{1}\omega$.
Final angular momentum of the system after the discs are combined: $L_{f} = (I_{1} + I_{2})\omega_{f}$,where $\omega_{f}$ is the final angular velocity.
Equating initial and final angular momentum: $I_{1}\omega = (I_{1} + I_{2})\omega_{f}$.
Solving for the final angular velocity: $\omega_{f} = \frac{I_{1}\omega}{I_{1} + I_{2}}$.

(B) According to the law of conservation of angular momentum,if no external torque acts on a system,the total angular momentum remains constant. Since the sphere is rotating in free space,there is no external torque acting on it. Therefore,the angular momentum will remain unaffected even if the radius is changed.

(A) The relationship between torque $\vec{\tau}$ and angular momentum $\vec{L}$ is given by the equation: $\vec{\tau} = \frac{d\vec{L}}{dt}$.
If the external torque $\vec{\tau} = 0$,then $\frac{d\vec{L}}{dt} = 0$.
This implies that $\vec{L} = \text{constant}$.
Therefore,the angular momentum of the system remains conserved.

(A) According to the principle of conservation of angular momentum, $L = I\omega = \text{constant}$.
When a swimmer curls their body, they bring their mass closer to the axis of rotation.
This action significantly decreases their moment of inertia $(I)$.
Since $L$ remains constant, a decrease in $I$ must result in an increase in angular velocity $(\omega)$, which allows the swimmer to perform somersaults more effectively.
Therefore, the correct option is $A$.

(B) The rate of change of angular momentum of a system is equal to the net external torque acting on it,given by the equation $\vec{\tau}_{ext} = \frac{d\vec{L}}{dt}$.
If the net external torque $\vec{\tau}_{ext} = 0$,then $\frac{d\vec{L}}{dt} = 0$,which implies that the angular momentum $\vec{L}$ is constant (conserved).
Therefore,if a net external torque acts on the system,the angular momentum is not conserved.
Thus,the correct option is $B$.

(C) According to the law of conservation of angular momentum, $L = I\omega = \text{constant}$.
As the person moves from the edge towards the center of the rotating plate, the distance of the person from the axis of rotation decreases.
This results in a decrease in the moment of inertia $(I)$ of the system $(I = \sum mr^2)$.
Since the angular momentum $(L)$ remains conserved and $I$ decreases, the angular velocity $(\omega = L/I)$ must increase to keep $L$ constant.