Q.(a) A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of 40 rev/min. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to $2/5$ times the initial value? Assume that the turntable rotates without friction.

Concept and Intuition

For a planet orbiting under a central (gravitational) force, angular momentum about the sun is conserved, which is equivalent to saying the line joining planet and sun sweeps out equal areas in equal times — Kepler's Second Law. Mathematically, $\dfrac{dA}{dt} = \dfrac{L}{2m}$, a constant for the whole orbit.

At the two special points of the elliptical orbit — perihelion (closest) and aphelion (farthest) — the planet's velocity vector is exactly perpendicular to the radius vector (the radial component of velocity is momentarily zero there, since $r$ is at a turning point). At these two points only, the instantaneous rate of area sweep simplifies exactly to $\dfrac{dA}{dt} = \dfrac{1}{2} r v$, with no need for a general dot-product treatment. Since $r$ and $v$ are inversely related ($rv = \text{const}$ at the apsides, from angular momentum conservation), the minimum speed occurs at the maximum distance (aphelion) — a very intuitive coupling of "far means slow, near means fast."

Step-by-Step Solution

1. Areal velocity is constant: $\dfrac{dA}{dt} = 4\times10^{16}\ m^2s^{-1}$. …

Concept and Intuition

Angular momentum about the origin is $\vec L = \vec r\times\vec p$, where $\vec p = m\vec v$. We only need the z-component of the resulting cross product, which (using the standard determinant expansion) comes only from the x- and y- components of $\vec r$ and $\vec p$ (the z-components of $\vec r$ and $\vec p$ do not affect $L_z$).

Step-by-Step Solution

1. Momentum: $\vec p = m\vec v = 3\,(2\hat i+\hat j+2\hat k) = 6\hat i+3\hat j+6\hat k$, i.e. $(p_x,p_y,p_z)=(6,3,6)$.
2. Position vector: $\vec r = \hat i+2\hat j$, i.e. $(r_x,r_y,r_z)=(1,2,0)$.
3. Compute $\vec L=\vec r\times\vec p$ via the determinant:

$=\hat i(12)-\hat j(6)+\hat k(3-12) = 12\hat i-6\hat j-9\hat k$ …

Concept and Intuition

Kepler's second law says the line joining the sun to the planet sweeps out equal areas in equal times — this is a direct consequence of angular momentum conservation about the focus. The chord $BD$ (the minor axis) passes through the geometric centre $O$ of the ellipse, so it splits the ellipse into two geometrically equal halves. But the area swept from the focus $S$ is not the same as the geometric half-ellipse area, because $S$ is not at $O$ — it's offset toward $A$.

Step-by-Step Solution

1. Set up coordinates: ellipse centred at $O$, major axis along $A$–$C$, minor axis along $B$–$D$ (through $O$). Focus $S$ lies on the major axis, between $O$ and $A$ (closer to $A$, consistent with $A$ being the near/perihelion point).
2. The chord $BD$ divides the ellipse into two geometrically equal halves (each area $= \tfrac12\pi ab$), since any line through the centre bisects an ellipse's area.
3. However, the area swept by the radius vector from $S$ as the planet moves along arc $B\to A\to D$ is bounded not by the chord $BD$ but by the two segments $SB$ and $SD$. Since $S$ lies inside the $A$-side half (between $O$ and $A$, i.e. on the same side as $A$ relative to chord $BD$), the swept region is the half-ellipse area minus the triangle $DSB$: $\text{Area}_{swept}(BAD) = \tfrac12\pi ab - T$, where $T=\text{area}(\triangle DSB) > 0$. …

Concept and Intuition

Angular momentum $\vec L=\vec r\times\vec p$ changes according to $\dfrac{d\vec L}{dt}=\vec\tau=\vec r\times\vec F$. If $\vec L$ is constant, the torque about the origin due to the applied force must be zero. This happens precisely when $\vec r$ and $\vec F$ are parallel (or anti-parallel), i.e. the force line passes through the origin (a central force situation).

Step-by-Step Solution

1. Given $\vec F=2\hat i+3\hat j+4\hat k$ and $\vec r=6\hat i+b\hat j+12\hat k$.
2. Torque $\vec\tau=\vec r\times\vec F=\begin{vmatrix}\hat i&\hat j&\hat k\\6&b&12\\2&3&4\end{vmatrix}$.
3. $\hat i$ component: $b(4)-12(3)=4b-36$.
4. $\hat j$ component: $-\big(6(4)-12(2)\big)=-(24-24)=0$ — automatically zero.
5. $\hat k$ component: $6(3)-b(2)=18-2b$. …

Concept and Intuition

When an object is struck by another object and they stick together (perfectly inelastic collision) with no external force acting during the (very short) collision, two conservation laws apply for the combined system:

1. Linear momentum is conserved — this fixes the velocity of the new centre of mass.
2. Angular momentum about any fixed point in space is conserved — this fixes the angular velocity of the combined body about its own (new) centre of mass.

The velocity of any point on the rigid body afterward is then the vector sum of the centre-of-mass translational velocity and the rotational velocity ($\vec\omega \times \vec r$) of that point relative to the new centre of mass.

Step-by-Step Solution

1. Centre of mass velocity. Before collision only the particle has momentum $mv$ (rod at rest). So

2. Location of the new centre of mass. Taking the rod's own centre $O$ ($x=0$) and end $B$ at $x=L$: the particle (mass $m$) sticks at $B$, so

The distance from the new CM to $B$ is then $L-d = \dfrac{ML}{M+m}$.

3. Moment of inertia about the new CM. The rod's own moment of inertia about $O$ is $\tfrac13 ML^2$ (rod of length $2L$ about its centre). Using the parallel-axis theorem for the rod's shift, plus the particle's contribution at its own distance from the new CM:

(the last two terms combine to the standard two-body reduced-mass form $\frac{Mm}{M+m}L^2$).

4. Angular momentum conservation. Taking angular momentum about the fixed point $O$ (the rod's original centre, an inertial point unaffected by the collision):

Equating this to the final angular momentum (rotation about new CM + orbital contribution of the CM's motion about $O$) and solving gives the angular speed:

$\omega = \frac{mv}{L(M+m)\left(\frac13+\frac{m}{M+m}\right)}$ …

Concept and Intuition

This is exactly like a figure skater pulling in their arms: shrinking the radius reduces the moment of inertia, and with no external torque acting, angular momentum $L=I\omega$ stays constant, so $\omega$ must increase to compensate.

Step-by-Step Solution

1. Volume $V=\frac43\pi R^3$. If $V_{new}=V_{old}/64$, then $R_{new}^3=R_{old}^3/64\Rightarrow R_{new}=R_{old}/4$.
2. Moment of inertia of a uniform sphere: $I=\frac25MR^2$. Mass is unchanged, so $I_{new}/I_{old}=(R_{new}/R_{old})^2=(1/4)^2=1/16$. …

Concept and Intuition

When two rotating bodies are suddenly coupled (like an inelastic collision but rotational), angular momentum is conserved because the coupling forces are internal, but kinetic energy is not conserved — some is dissipated as heat/sound in the coupling process, just like inelastic linear collisions.

Step-by-Step Solution

1. Let first wheel: inertia $I_1=I$, initial angular speed $\omega$. Second wheel: inertia $I_2=3I$, at rest.
2. Conservation of angular momentum: $I\omega+3I(0)=(I+3I)\omega'\Rightarrow \omega'=\dfrac{I\omega}{4I}=\dfrac{\omega}{4}$.
3. Initial KE: $KE_i=\dfrac12 I\omega^2$.
4. Final KE: $KE_f=\dfrac12(4I)\left(\dfrac{\omega}{4}\right)^2=\dfrac12(4I)\dfrac{\omega^2}{16}=\dfrac{I\omega^2}{8}$. …

Concept and Intuition

Gravitational force on a planet is always directed toward the Sun (a central force). Torque $\vec\tau = \vec r \times \vec F$ vanishes whenever $\vec F$ is parallel (or anti-parallel) to $\vec r$, which is exactly the case here. Since $\vec\tau = d\vec L/dt$, zero torque means angular momentum $\vec L$ is strictly constant — this is precisely why planets sweep equal areas in equal times (Kepler's 2nd law), speeding up near perihelion and slowing down near aphelion, so their linear/angular speed and KE do change even though $L$ doesn't.

Step-by-Step Solution

1. Assertion states: angular speed, linear speed, KE change with time, but angular momentum is constant. This is a correct description of planetary motion (true, due to varying orbital radius in an ellipse).
2. Reason states: angular momentum is constant because no torque acts on the planet. This is the correct physical cause (true). …

Concept and Intuition

Kepler's second law says a planet sweeps out equal areas in equal times. This is really a statement about conservation of angular momentum: for a particle at position $\vec r$ moving with momentum $\vec p = m\vec v$, the infinitesimal area swept in time $dt$ is $dA = \tfrac12 |\vec r \times \vec v|\,dt$, and $|\vec r \times m\vec v| = L$ is exactly the angular momentum. So the constant areal velocity is nothing but angular momentum divided by twice the mass.

Step-by-Step Solution

1. Areal velocity: $A = \dfrac{dA_{swept}}{dt} = \dfrac12 |\vec r \times \vec v|$.
2. Angular momentum: $L = |\vec r \times \vec p| = M|\vec r \times \vec v|$ (mass of the planet is $M$).
3. Comparing the two: $|\vec r \times \vec v| = \dfrac{L}{M}$, so …