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

(D) In a central force field,the force acts along the line joining the center of force and the particle.
Therefore,the torque $\tau$ exerted by the central force about the center of force is zero.
Since torque is the rate of change of angular momentum,we have $\tau = \frac{dL}{dt} = 0$.
This implies that the angular momentum $L$ remains constant over time.

(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.
When a child sits on a rotating disc,the weight of the child acts vertically downwards,passing through the axis of rotation or creating no torque about the axis of rotation.
Since there is no external torque applied to the system (disc + child),the angular momentum of the system remains conserved.

(C) The system consists of the disc and the plasticine lump.
Since the plasticine is dropped vertically,there is no external torque acting on the system 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 angular momentum $L$ of the system remains constant.
However,when the plasticine sticks to the disc,the moment of inertia $I$ of the system increases.
Since $L = I\omega$ and $L$ is constant,the angular velocity $\omega$ must decrease.
Therefore,only the angular momentum $L$ remains constant.

(B) $1$. The system consists of the triangle and the two beads. Since there is no external torque acting on the system about the vertical axis $AO$,the total angular momentum $L$ of the system remains conserved.
$2$. As the beads slide down,the gravitational potential energy of the beads is converted into kinetic energy (both translational and rotational). Since there is no friction,the total mechanical energy (kinetic + potential) of the system is conserved.
$3$. The moment of inertia $I$ of the system about the axis $AO$ increases as the beads move away from the axis. Since $L = I\omega$ is constant,the angular velocity $\omega$ must decrease.
$4$. Therefore,the quantities that are conserved are the total angular momentum and the total energy.

(B) 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: $I' = Mr^2 + m(r)^2 + m(r)^2 = (M + 2m)r^2$.
According to the principle of conservation of angular momentum,the external torque is zero,so $L_{initial} = L_{final}$.
$Mr^2\omega = (M + 2m)r^2\omega'$
Solving for the new angular velocity $\omega'$:
$\omega' = \frac{Mr^2\omega}{(M + 2m)r^2} = \frac{M\omega}{M + 2m}$.

(C) The volume of a spherical body is given by $V = \frac{4}{3} \pi R^{3}$.
Taking the logarithmic derivative,we get $\frac{\Delta V}{V} = 3 \frac{\Delta R}{R}$.
Given $\frac{\Delta V}{V} = 1\% = 0.01$,we have $3 \frac{\Delta R}{R} = 0.01$,which implies $\frac{\Delta R}{R} = \frac{0.01}{3} \approx 0.0033$ or $0.33\%$.
Since there is no external torque,the angular momentum $L$ is conserved,where $L = I \omega$.
For a solid sphere,$I = \frac{2}{5} M R^{2}$,so $L = \frac{2}{5} M R^{2} \omega$.
Since $L$ and $M$ are constant,we have $R^{2} \omega = \text{constant}$.
Taking the derivative,$2 \frac{\Delta R}{R} + \frac{\Delta \omega}{\omega} = 0$.
Therefore,$\frac{\Delta \omega}{\omega} = -2 \frac{\Delta R}{R} = -2(0.33\%) = -0.66\% \approx -0.67\%$.
Thus,the angular speed decreases by approximately $0.67\%$.

(D) The angular momentum of the system about the axis of rotation is conserved because no external torque acts on the system.
Initial angular momentum $L_i = I_i \omega = (\frac{1}{2}MR^2)\omega$.
When a piece of mass $m$ breaks off,the mass of the remaining disc is $(M - m)$.
The moment of inertia of the remaining disc is $I_f = \frac{1}{2}(M - m)R^2$.
Since the piece $m$ breaks off and flies vertically upwards,it does not exert any torque on the disc about the axis of rotation.
By the law of conservation of angular momentum,$L_i = L_f$.
$(\frac{1}{2}MR^2)\omega = (\frac{1}{2}(M - m)R^2)\omega'$.
Solving for $\omega'$,we get $\omega' = \frac{M\omega}{M - m}$.

(C) Since there is no external torque acting on the system,the angular momentum is conserved.
$L_i = L_f$
$I_1 \omega_1 = I_2 \omega_2$
For the first disc,the moment of inertia is $I_1 = \frac{1}{2} M R^2$ and angular velocity is $\omega_1 = \omega$.
When the second disc of mass $M/4$ is placed coaxially,the new moment of inertia of the system is $I_2 = I_1 + I_{disc2} = \frac{1}{2} M R^2 + \frac{1}{2} (M/4) R^2$.
$I_2 = \frac{1}{2} M R^2 + \frac{1}{8} M R^2 = \frac{4+1}{8} M R^2 = \frac{5}{8} M R^2$.
Using the conservation of angular momentum:
$\frac{1}{2} M R^2 \cdot \omega = \frac{5}{8} M R^2 \cdot \omega_2$
$\omega_2 = \frac{1/2}{5/8} \omega = \frac{1}{2} \cdot \frac{8}{5} \omega = \frac{4}{5} \omega$.

(A) When the block hits the ridge at $O$,it starts rotating about an axis passing through $O$ and perpendicular to the plane of motion. Since no external torque acts on the block about point $O$ during the impact,the angular momentum is conserved.
Initial angular momentum $L_i$ about point $O$ is given by the product of linear momentum and the perpendicular distance from $O$ to the line of motion of the center of mass:
$L_i = Mv \times (a/2) = \frac{Mva}{2}$
After hitting the ridge,the block rotates about point $O$. The final angular momentum is $L_f = I_O \omega$,where $I_O$ is the moment of inertia of the cube about the axis passing through $O$.
The moment of inertia of a cube about an axis passing through its center of mass $C$ is $I_C = \frac{Ma^2}{6}$.
Using the parallel axis theorem,$I_O = I_C + Mr^2$,where $r$ is the distance from the center of mass to point $O$. Here,$r^2 = (a/2)^2 + (a/2)^2 = a^2/2$.
Thus,$I_O = \frac{Ma^2}{6} + M(a^2/2) = \frac{Ma^2 + 3Ma^2}{6} = \frac{4Ma^2}{6} = \frac{2}{3}Ma^2$.
Equating initial and final angular momentum: $L_i = L_f$
$\frac{Mva}{2} = \frac{2}{3}Ma^2 \omega$
Solving for $\omega$: $\omega = \frac{Mva}{2} \times \frac{3}{2Ma^2} = \frac{3v}{4a}$.

(D) Let the stick have mass $M$ and length $L$. The ball of mass $m$ hits the end of the stick with velocity $v$ and comes to rest.
$1$. Conservation of linear momentum: $mv = MV$,where $V$ is the velocity of the center of mass of the stick. So,$V = \frac{mv}{M}$.
$2$. Conservation of angular momentum about the center of mass of the stick: $mv(L/2) = I\omega$,where $I = \frac{ML^2}{12}$ is the moment of inertia of the stick about its center. Thus,$mv(L/2) = \frac{ML^2}{12} \omega$,which gives $\omega = \frac{6mv}{ML}$.
$3$. Conservation of kinetic energy (elastic collision): $\frac{1}{2}mv^2 = \frac{1}{2}MV^2 + \frac{1}{2}I\omega^2$.
Substituting $V$ and $\omega$: $\frac{1}{2}mv^2 = \frac{1}{2}M(\frac{mv}{M})^2 + \frac{1}{2}(\frac{ML^2}{12})(\frac{6mv}{ML})^2$.
$\frac{1}{2}mv^2 = \frac{1}{2}\frac{m^2v^2}{M} + \frac{1}{2}(\frac{ML^2}{12})(\frac{36m^2v^2}{M^2L^2})$.
$mv^2 = \frac{m^2v^2}{M} + \frac{3m^2v^2}{M}$.
$1 = \frac{m}{M} + \frac{3m}{M} = \frac{4m}{M}$.
Therefore,$m = M/4$.

(C) The initial angular momentum of the system is provided by the child running tangent to the merry-go-round. The merry-go-round is initially at rest.
Initial angular momentum $L_i = mvr$,where $m = 30\ kg$,$v = 5\ m/s$,and $r = 4\ m$.
$L_i = 30 \times 5 \times 4 = 600\ kg \cdot m^2/s$.
After the child jumps on,the system rotates with angular velocity $\omega$. The total moment of inertia $I_{total} = I_{merry-go-round} + I_{child}$.
$I_{merry-go-round} = Mk^2 = 120 \times (3)^2 = 120 \times 9 = 1080\ kg \cdot m^2$.
$I_{child} = mr^2 = 30 \times (4)^2 = 30 \times 16 = 480\ kg \cdot m^2$.
$I_{total} = 1080 + 480 = 1560\ kg \cdot m^2$.
By conservation of angular momentum,$L_i = L_f = I_{total} \omega$.
$600 = 1560 \times \omega$.
$\omega = 600 / 1560 = 60 / 156 = 10 / 26 = 5 / 13 \approx 0.3846\ rad/s$.
Rounding to the nearest provided option,the value is $0.4\ rad/s$.

(A) The initial moment of inertia of the ring about its axis is $I_i = MR^2$. The initial angular momentum is $L_i = I_i \omega = MR^2 \omega$.
When four objects of mass $m$ are placed on the ring,the final moment of inertia becomes $I_f = MR^2 + 4(mR^2) = (M + 4m)R^2$.
Since no external torque acts on the system,the angular momentum is conserved: $L_i = L_f$.
$MR^2 \omega = (M + 4m)R^2 \omega'$.
Solving for the final angular velocity $\omega'$,we get $\omega' = \frac{MR^2 \omega}{(M + 4m)R^2} = \frac{M\omega}{M + 4m}$.

(B) The angular momentum $(L)$ of the system is conserved because there are no external torques acting on the system about the axis of rotation. Thus, $L = I\omega = \text{constant}$.
As the tortoise moves along a chord of the circular platform, its distance $(r)$ from the center of rotation changes. Specifically, the tortoise starts at the edge, moves towards the center (decreasing $r$), reaches the point on the chord closest to the center, and then moves back towards the edge (increasing $r$).
The moment of inertia $(I)$ of the system is given by $I = I_{\text{platform}} + m r^2$, where $m$ is the mass of the tortoise. As the tortoise moves towards the center, $r$ decreases, so $I$ decreases. As it moves away, $r$ increases, so $I$ increases.
Since $L = I\omega$ is constant, $\omega = L/I$. When $I$ decreases, $\omega$ increases, and when $I$ increases, $\omega$ decreases. Therefore, the angular velocity $\omega(t)$ will first increase and then decrease. The relationship between $r$ and time $t$ is $r^2 = r_{\text{min}}^2 + (vt)^2$, making $I$ a quadratic function of time, which implies that $\omega(t) = L / (I_0 + m(r_{\text{min}}^2 + v^2t^2))$ is a non-linear function of time. This corresponds to a smooth, curved graph.

(C) According to the law of conservation of angular momentum,the total angular momentum remains constant when the two discs are coupled:
$I_1\omega_1 + I_2\omega_2 = (I_1 + I_2)\omega$
where $\omega$ is the final common angular velocity.
Solving for $\omega$:
$\omega = \frac{I_1\omega_1 + I_2\omega_2}{I_1 + I_2}$
The rotational kinetic energy $(KE)$ of the system is given by:
$KE = \frac{1}{2}(I_1 + I_2)\omega^2$
Substituting the value of $\omega$:
$KE = \frac{1}{2}(I_1 + I_2) \left( \frac{I_1\omega_1 + I_2\omega_2}{I_1 + I_2} \right)^2$
$KE = \frac{(I_1\omega_1 + I_2\omega_2)^2}{2(I_1 + I_2)}$

(D) Since there are no external torques acting on the system,the angular momentum of the system remains constant.
Initially,when the beads are at the center of the rod,the moment of inertia of the system is $I_1 = \frac{ML^2}{12}$.
The initial angular momentum is $L_1 = I_1{\omega _0} = \left( \frac{ML^2}{12} \right){\omega _0}$.
When the beads reach the ends of the rod,the moment of inertia of the system becomes $I_2 = \frac{ML^2}{12} + m\left( \frac{L}{2} \right)^2 + m\left( \frac{L}{2} \right)^2 = \frac{ML^2}{12} + \frac{mL^2}{4} + \frac{mL^2}{4} = \frac{ML^2}{12} + \frac{mL^2}{2} = \frac{ML^2 + 6mL^2}{12} = \frac{L^2(M + 6m)}{12}$.
The final angular momentum is $L_2 = I_2\omega ' = \left( \frac{L^2(M + 6m)}{12} \right)\omega '$.
By the principle of conservation of angular momentum,$L_1 = L_2$:
$\left( \frac{ML^2}{12} \right){\omega _0} = \left( \frac{L^2(M + 6m)}{12} \right)\omega '$
$\omega ' = \frac{M{\omega _0}}{M + 6m}$.

(C) The principle of conservation of angular momentum states that the initial angular momentum of the system equals the final angular momentum.
Let $m = 0.5 \ kg$ be the mass of the particle,$v = 5 \ m/s$ be its velocity,$r = 0.2 \ m$ be the radius of the cylinder,$M = 2 \ kg$ be the mass of the cylinder,and $\omega = 0 \ rad/s$ be the initial angular velocity of the cylinder.
Initial angular momentum $L_i = mvr + I_{cylinder}\omega = mvr + 0 = 0.5 \times 5 \times 0.2 = 0.5 \ kg \cdot m^2/s$.
Final moment of inertia $I_f = I_{cylinder} + I_{particle} = \frac{1}{2}Mr^2 + mr^2 = \frac{1}{2} \times 2 \times (0.2)^2 + 0.5 \times (0.2)^2 = 0.04 + 0.02 = 0.06 \ kg \cdot m^2$.
Using $L_i = I_f \omega'$,we get $0.5 = 0.06 \times \omega'$.
$\omega' = \frac{0.5}{0.06} = 8.33 \ rad/s$.
Given the options provided,the closest value is $10.3 \ rad/s$ (assuming specific initial conditions from the referenced problem context).

(A) In uniform circular motion,the force acting on the particle is always directed towards the center of the circle (centripetal force).
Since the force passes through the center of the circle,the torque $\tau = r \times F$ about the center is zero.
According to the principle of conservation of angular momentum,if the net external torque acting on a system is zero,the angular momentum remains conserved.
Therefore,the angular momentum of the particle is conserved about the center of the circle.
Thus,the correct option is $A$.

(C) According to the law of conservation of angular momentum, $L = I\omega = \text{constant}$.
When the person walks from the edge of the platform towards the center, the moment of inertia $(I)$ of the system decreases.
Since $I = \sum mr^2$ and the distance $r$ is decreasing, $I$ decreases.
Since angular momentum $L$ is constant, if $I$ decreases, the angular velocity $(\omega)$ must increase.
Therefore, the angular velocity will increase.

(D) According to the law of conservation of angular momentum,if the external torque acting on the system is zero,the angular momentum $L = I\omega$ remains constant.
When the ice melts,the water spreads out on the table.
This increases the moment of inertia $(I)$ of the system because the mass is distributed further away from the axis of rotation.
In the formula $L = I\omega$,if $L$ is constant and $I$ increases,the angular velocity $\omega$ must decrease.
Therefore,the rotational speed of the system will decrease to less than $\omega$.

(B) According to the law of conservation of angular momentum,if no external torque acts on the system,the angular momentum $(L)$ remains constant.
$1$. Moment of inertia $(I)$: For a solid sphere,$I = \frac{2}{5}mr^2$. As the radius $(r)$ increases,the moment of inertia $(I)$ increases.
$2$. Angular momentum $(L)$: Since there is no external torque,the angular momentum $(L = I\omega)$ remains constant.
$3$. Angular velocity $(\omega)$: Since $L = I\omega$ is constant and $I$ increases,the angular velocity $(\omega = L/I)$ must decrease.
$4$. Rotational kinetic energy $(K_{rot})$: The rotational kinetic energy is given by $K_{rot} = \frac{L^2}{2I}$. Since $L$ is constant and $I$ increases,the rotational kinetic energy decreases.
Therefore,the angular momentum remains unchanged.

(C) Since there is no external torque acting on the system,the angular momentum is conserved: $L_1 = L_2$.
The initial moment of inertia of the disc is $I_1 = \frac{1}{2}MR^2$.
When two spheres of mass $m$ are placed at opposite ends of a diameter (at distance $R$ from the center),the final moment of inertia becomes $I_2 = I_1 + mR^2 + mR^2 = \frac{1}{2}MR^2 + 2mR^2$.
Using the conservation of angular momentum: $I_1 \omega_1 = I_2 \omega_2$.
$\left( \frac{1}{2}MR^2 \right) \omega_1 = \left( \frac{1}{2}MR^2 + 2mR^2 \right) \omega_2$.
Dividing both sides by $R^2$ and multiplying by $2$: $M \omega_1 = (M + 4m) \omega_2$.
Therefore,the final angular velocity is $\omega_2 = \left( \frac{M}{M + 4m} \right) \omega_1$.

(B) Since the sphere is rotating in free space,there is no external torque acting on it.
According to the principle of conservation of angular momentum,the angular momentum $(L)$ remains constant.
As the radius $(r)$ increases,the moment of inertia $(I = \frac{2}{5}mr^2)$ increases.
Since $L = I\omega$ is constant,if $I$ increases,the angular velocity $(\omega)$ must decrease.
The rotational kinetic energy $(K_{\text{rot}} = \frac{L^2}{2I})$ will also decrease because $I$ is in the denominator and $L$ is constant.
Therefore,the angular momentum is the quantity that remains unaffected.

(D) According to the principle of conservation of angular momentum,since no external torque acts on the system,the angular momentum remains constant: $L_1 = L_2$.
The initial moment of inertia of the ring is $I_1 = mR^2$.
The initial angular momentum is $L_1 = I_1 \omega = mR^2 \omega$.
When two particles of mass $M$ are attached at the ends of a diameter,the new moment of inertia $I_2$ is the sum of the ring's moment of inertia and the moment of inertia of the two particles: $I_2 = mR^2 + M R^2 + M R^2 = (m + 2M)R^2$.
Equating the angular momenta: $I_1 \omega = I_2 \omega'$.
$mR^2 \omega = (m + 2M)R^2 \omega'$.
Solving for $\omega'$,we get: $\omega' = \frac{m \omega}{m + 2M}$.

(C) The angular momentum $L$ of the system (platform + tortoise) about the axis of rotation is conserved because there is no external torque acting on the system.
$L = I(t) \omega(t) = \text{constant}$
Here, $I(t)$ is the moment of inertia of the system at time $t$.
Let $R$ be the radius of the platform, $M$ its mass, and $m$ the mass of the tortoise.
The moment of inertia of the platform is $I_p = \frac{1}{2}MR^2$.
The tortoise moves along a chord. Let the distance of the chord from the center be $d$. At time $t$, the distance of the tortoise from the center is $r(t) = \sqrt{d^2 + (vt)^2}$, where $v$ is the constant speed of the tortoise relative to the platform.
The total moment of inertia is $I(t) = I_p + mr(t)^2 = I_p + m(d^2 + v^2t^2)$.
Since $L = I(t) \omega(t)$, we have $\omega(t) = \frac{L}{I_p + m(d^2 + v^2t^2)}$.
As the tortoise moves from the edge towards the center (or vice versa), the distance $r(t)$ first decreases until it reaches the point closest to the center (at $t=0$ if the chord passes through the center, or at some $t$ if it doesn't) and then increases.
When $r(t)$ is minimum, $I(t)$ is minimum, and therefore $\omega(t)$ is maximum.
Conversely, when $r(t)$ is maximum (at the edges), $I(t)$ is maximum, and $\omega(t)$ is minimum.
Since the tortoise starts at the edge, $r(t)$ decreases as it moves towards the center, causing $I(t)$ to decrease and $\omega(t)$ to increase. After passing the point closest to the center, $r(t)$ increases, causing $I(t)$ to increase and $\omega(t)$ to decrease.
Thus, the angular velocity $\omega(t)$ increases and then decreases.

(B) The rate of change of total angular momentum of a system of particles about a point is equal to the sum of external torques acting on the system about that point,given by $\vec{\tau}_{ext} = \frac{d\vec{L}}{dt}$.
If the net external torque $\vec{\tau}_{ext}$ acting on the system is zero,then $\frac{d\vec{L}}{dt} = 0$.
This implies that the total angular momentum $\vec{L}$ of the system remains constant or conserved.

(B) When a child sits on the rotating disc,there is no external torque acting on the system $(τ_{ext} = 0)$.
According to the principle of conservation of angular momentum,if the net external torque is zero,the angular momentum $(L = Iω)$ remains constant.
As the child sits on the disc,the moment of inertia $(I)$ of the system increases,which causes the angular velocity $(ω)$ to decrease,but the total angular momentum remains conserved.

(B) According to the principle of conservation of angular momentum,in the absence of any external torque,the angular momentum $L = I\omega$ remains constant.
When a swimmer curls their body,they bring their mass closer to the axis of rotation,which decreases their moment of inertia $(I)$.
Since $L = I\omega$ must remain constant,a decrease in $I$ leads to an increase in angular velocity $(\omega)$,allowing the swimmer to rotate faster.

(A) According to the principle of conservation of angular momentum,the angular momentum $L$ remains constant because no external torque acts on the system.
$L = I_0 \omega_0 = I_1 \omega_1$
Given that the moment of inertia doubles,$I_1 = 2I_0$.
Substituting this into the conservation equation: $I_0 \omega_0 = (2I_0) \omega_1$,which gives $\omega_1 = \omega_0 / 2$.
The initial kinetic energy is $K = \frac{1}{2} I_0 \omega_0^2$.
The final kinetic energy is $K' = \frac{1}{2} I_1 \omega_1^2$.
Substituting the new values: $K' = \frac{1}{2} (2I_0) (\omega_0 / 2)^2 = \frac{1}{2} (2I_0) (\omega_0^2 / 4) = \frac{1}{2} (I_0 \omega_0^2 / 2) = K / 2$.
Therefore,the new kinetic energy is $K/2$.

(C) According to the law 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 towards the center, the moment of inertia $I$ of the system (disc + ant) decreases, causing the angular velocity $\omega$ to increase.
As the ant moves from the center towards the other edge, the moment of inertia $I$ increases, causing the angular velocity $\omega$ to decrease.
Therefore, the angular velocity of the disc will first increase and then decrease.

(B) $1$. The system consists of the triangle and the two beads. The only external forces acting on the system are gravity and the normal force from the pivot at $A$. Since the pivot force acts at the axis of rotation,its torque is zero. Gravity also exerts no torque about the vertical axis $AO$. Therefore,the net external torque on the system about the axis of rotation is zero,which implies that the total angular momentum $L$ is conserved.
$2$. Since there is no friction and the only forces doing work are gravity (a conservative force),the total mechanical energy of the system is conserved.
$3$. As the beads slide down,their distance from the axis of rotation increases,which increases the moment of inertia $I$ of the system. Since $L = I\omega$ is constant and $I$ increases,the angular velocity $\omega$ must decrease. Thus,neither $\omega$ nor $I$ are conserved.

(C) central force always acts along the line joining the particle and the center of force.
Since the position vector $\vec{r}$ and the force vector $\vec{F}$ are collinear,the torque $\vec{\tau} = \vec{r} \times \vec{F} = 0$.
According to the relation $\vec{\tau} = \frac{d\vec{L}}{dt}$,if the external torque $\vec{\tau}$ is zero,then the rate of change of angular momentum $\frac{d\vec{L}}{dt}$ is zero.
Therefore,the angular momentum $\vec{L}$ remains constant.

(C) According to the law of conservation of angular momentum,if no external torque acts on a system,the angular momentum $L = I\omega$ remains constant.
When the dancer pulls their arms inward,the moment of inertia $I$ decreases.
Since $L = I\omega$ is constant,a decrease in $I$ must lead to an increase in angular velocity $\omega$ to keep $L$ constant.
The rotational kinetic energy is given by $K = \frac{L^2}{2I}$.
Since $L$ is constant and $I$ decreases,the kinetic energy $K$ increases because the dancer does work by pulling their arms inward against the centrifugal force.

(C) Concept $1$: The moment of inertia $I$ of a mass $m$ at a distance $r$ from the axis of rotation is given by $I = mr^2$. Thus,$I \propto r^2$. When the dancer drops the spheres,the mass $m$ is removed from the system at a distance $r$,which decreases the total moment of inertia $I$ of the dancer.
Concept $2$: The angular momentum $L$ of the system is conserved if the net external torque $\tau$ acting on the system is zero. Since $\frac{d\vec{L}}{dt} = \vec{\tau}$,if $\vec{\tau} = 0$,then $\vec{L}$ remains constant.
Concept $3$: We know that $L = I\omega$,which implies $\omega = \frac{L}{I}$. Since $L$ is constant,$\omega \propto \frac{1}{I}$.
Conclusion: When the dancer drops the spheres,the moment of inertia $I$ decreases. Because the angular momentum $L$ remains constant,a decrease in $I$ leads to an increase in the angular velocity $\omega$.

(A) To find the angular speed of the block after it hits point '$O$',we use the principle of conservation of angular momentum about the point '$O$'.
The initial angular momentum of the block about point '$O$' is given by the product of its linear momentum and the perpendicular distance from the center of mass to the point '$O$'.
$L_i = Mv \times \left(\frac{a}{2}\right)$
The moment of inertia of the cube about the edge passing through '$O$' is calculated using the parallel axis theorem:
$I_O = I_{cm} + Mr^2$
Where $I_{cm} = \frac{Ma^2}{6}$ for a cube about its center,and $r$ is the distance from the center of mass to the edge '$O$',which is $\sqrt{(\frac{a}{2})^2 + (\frac{a}{2})^2} = \frac{a}{\sqrt{2}}$.
$I_O = \frac{Ma^2}{6} + M\left(\frac{a}{\sqrt{2}}\right)^2 = \frac{Ma^2}{6} + \frac{Ma^2}{2} = \frac{Ma^2 + 3Ma^2}{6} = \frac{4Ma^2}{6} = \frac{2}{3}Ma^2$
By conservation of angular momentum:
$L_i = L_f$
$Mv\left(\frac{a}{2}\right) = I_O \omega$
$\frac{Mva}{2} = \left(\frac{2}{3}Ma^2\right)\omega$
$\omega = \frac{Mva}{2} \times \frac{3}{2Ma^2} = \frac{3v}{4a}$

(A) According to the principle of conservation of angular momentum,if no external torque is applied,the angular momentum $L$ remains constant.
$L = I_1 \omega_1 = I_2 \omega_2$
Since the moment of inertia $I = MK^2$,where $M$ is the mass and $K$ is the radius of gyration,we have:
$M K_1^2 \omega_1 = M K_2^2 \omega_2$
$K_1^2 \omega_1 = K_2^2 \omega_2$
$\frac{K_1^2}{K_2^2} = \frac{\omega_2}{\omega_1}$
Taking the square root on both sides,the ratio of the radii of gyration is:
$\frac{K_1}{K_2} = \sqrt{\frac{\omega_2}{\omega_1}}$

(D) Since there are no external torques acting on the system,the total angular momentum is conserved.
Let the clockwise direction be positive.
The initial angular momentum of the disc is $L_{disc} = I\omega$.
The initial angular momentum of the particle moving counter-clockwise is $L_{particle} = -mvR$.
The total initial angular momentum is $L_i = I\omega - mvR$.
When the particle stops moving,it becomes part of the disc. The new moment of inertia of the system is $I' = I + mR^2$.
Let the final angular velocity be $\omega'$. The final angular momentum is $L_f = (I + mR^2)\omega'$.
By conservation of angular momentum,$L_i = L_f$.
$I\omega - mvR = (I + mR^2)\omega'$.
Therefore,$\omega' = \frac{I\omega - mvR}{I + mR^2}$.

(B) According to the law of conservation of angular momentum,since no external torque acts on the system,the total angular momentum remains constant.
Initial angular momentum of the system = $L_i = I_A \omega_A + I_B(0) = I_A \omega_A$.
Final angular momentum of the system = $L_f = (I_A + I_B) \omega$.
Equating initial and final angular momentum:
$I_A \omega_A = (I_A + I_B) \omega$
$I_A \omega_A = I_A \omega + I_B \omega$
$I_B \omega = I_A \omega_A - I_A \omega$
$I_B = \frac{I_A \omega_A - I_A \omega}{\omega}$.

(B) For the system,there is no external torque acting about the axis of rotation $AO$,so the external torque $\tau = 0$. According to the principle of conservation of angular momentum,the total angular momentum $L = I\omega$ remains constant.
As the beads move downward along $AB$ and $AC$,their distance from the axis of rotation $AO$ increases. Consequently,the moment of inertia $I$ of the system increases. Since $L = I\omega$ is constant,the angular velocity $\omega$ must decrease.
Since there is no friction or any other non-conservative force acting on the system,the total mechanical energy of the system is conserved.

(C) According to the law of conservation of angular momentum,the initial angular momentum of the system must equal the final angular momentum.
Initial angular momentum $L_i = I_1 \omega$.
When the second disc is placed on the first,the total moment of inertia of the system becomes $I_{total} = I_1 + I_2$.
Let the new angular velocity be $\omega'$.
By conservation of angular momentum: $I_1 \omega = (I_1 + I_2) \omega'$.
Therefore,the combined angular velocity is $\omega' = \frac{I_1}{I_1 + I_2} \omega$.

(A) Since no external torque is mentioned to be acting on the system in a way that changes the angular momentum,we assume the angular momentum $L$ remains constant $(L_1 = L_2)$.
Angular momentum is given by $L = I\omega$,where $I$ is the moment of inertia and $\omega$ is the angular velocity.
Thus,$I_1 \omega_1 = I_2 \omega_2$.
The moment of inertia is $I = MK^2$,where $M$ is the mass and $K$ is the radius of gyration.
Substituting this,we get $M K_1^2 \omega_1 = M K_2^2 \omega_2$.
Canceling $M$ from both sides: $K_1^2 \omega_1 = K_2^2 \omega_2$.
Rearranging for the ratio of the radii of gyration: $\frac{K_1^2}{K_2^2} = \frac{\omega_2}{\omega_1}$.
Taking the square root of both sides: $\frac{K_1}{K_2} = \sqrt{\frac{\omega_2}{\omega_1}} = \frac{\sqrt{\omega_2}}{\sqrt{\omega_1}}$.
Therefore,the ratio is $\sqrt{\omega_2} : \sqrt{\omega_1}$.

(C) According to the law of conservation of angular momentum, $L = I\omega = \text{constant}$.
Since the Earth is considered a solid sphere, its moment of inertia is $I = \frac{2}{5}MR^2$.
Initially, $I_1 = \frac{2}{5}MR^2$ and $\omega_1 = \frac{2\pi}{T_1}$.
After the radius is reduced to half, $R_2 = \frac{R}{2}$, so 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 get $I_1(\frac{2\pi}{T_1}) = (\frac{1}{4}I_1)(\frac{2\pi}{T_2})$.
This simplifies to $\frac{1}{T_1} = \frac{1}{4T_2}$, which means $T_2 = \frac{T_1}{4}$.
Given $T_1 = 24 \ hr$, the new duration of the day is $T_2 = \frac{24}{4} = 6 \ hr$.

(C) According to the law of conservation of angular momentum,the external torque on the system is zero,so the angular momentum remains constant.
$L_i = L_f$
$I_1 \omega_1 = I_2 \omega_2$
Here,$I_1 = \frac{1}{2} M R^2$ and $\omega_1 = \omega$.
When the second disc is placed,the new moment of inertia $I_2$ is the sum of the moments of inertia of both discs:
$I_2 = I_1 + I_{\text{disc2}} = \frac{1}{2} M R^2 + \frac{1}{2} (\frac{M}{4}) R^2 = \frac{1}{2} M R^2 (1 + \frac{1}{4}) = \frac{1}{2} M R^2 (\frac{5}{4}) = \frac{5}{8} M R^2$.
Now,applying the conservation law:
$(\frac{1}{2} M R^2) \omega = (\frac{5}{8} M R^2) \omega_2$
$\omega_2 = \frac{1/2}{5/8} \omega = \frac{1}{2} \times \frac{8}{5} \omega = \frac{4}{5} \omega$.

(B) According to the principle of conservation of angular momentum,the initial angular momentum of the system must be equal to the final angular momentum of the system.
Initial angular momentum $L_i = I_1 \omega$.
Final moment of inertia of the combined system $I_f = I_1 + I_2$.
Let the final angular velocity be $\omega_f$.
Final angular momentum $L_f = (I_1 + I_2) \omega_f$.
Since $L_i = L_f$,we have $I_1 \omega = (I_1 + I_2) \omega_f$.
Therefore,the final angular velocity is $\omega_f = \frac{I_1 \omega}{I_1 + I_2}$.

(C) According to the law of conservation of angular momentum,$L = I\omega = \text{constant}$.
Since the mass $M$ remains constant,the moment of inertia $I = \frac{2}{5}MR^2$ changes as the radius $R$ changes.
Let the initial radius be $R$ and the final radius be $R' = R/n$.
Let the initial time period be $T_1 = 24 \; \text{hr}$ and the final time period be $T_2$.
Using $I_1 \omega_1 = I_2 \omega_2$,we have:
$\frac{2}{5}MR^2 \left( \frac{2\pi}{T_1} \right) = \frac{2}{5}M(R/n)^2 \left( \frac{2\pi}{T_2} \right)$.
Simplifying this,we get $R^2 / T_1 = (R^2 / n^2) / T_2$.
Therefore,$T_2 = T_1 / n^2$.
Substituting $T_1 = 24 \; \text{hr}$,we get $T_2 = 24/n^2 \; \text{hr}$.