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 & Intuition

This is a classic angular momentum conservation problem. The bullet is moving linearly before impact, but after it sticks to the door, the system rotates as a rigid body. No external torque acts about the hinge (frictionless pivot), so the total angular momentum just before impact equals the total angular momentum just after. The trick: the bullet’s linear momentum must be converted to angular momentum about the hinge, using the perpendicular distance from the hinge to the bullet’s path.

Step-by-step solution

1. Identify the system and the conserved quantity The bullet + door form an isolated system for rotation about the hinge (no external torque). Therefore, angular momentum about the hinge is conserved:

2. Angular momentum of the bullet before impact The bullet moves horizontally, perpendicular to the door’s face, and strikes at the centre of the door. The distance from the hinge to the centre is half the door’s width:

The bullet’s linear momentum is $m_b v$. Its angular momentum about the hinge is

So

3. Angular momentum after the bullet embeds After embedding, the bullet and door rotate together with angular velocity $\omega = 0.625\ \text{rad s}^{-1}$. The total moment of inertia about the hinge is the sum of the door’s and the bullet’s:
   • Door: $I_{\text{door}} = \frac{M L^2}{3}$ (given).
   • Bullet: treated as a point mass at distance $r = L/2$, so $I_{\text{bullet}} = m_b r^2 = m_b (L/2)^2$. Hence

The angular momentum after is

4. Plug in the numbers Mass of door $M = 12\ \text{kg}$, width $L = 1.0\ \text{m}$, bullet mass $m_b = 10\ \text{g} = 0.010\ \text{kg}$, $\omega = 0.625\ \text{rad s}^{-1}$.

The man walks with speed $v=1.5\ \text{m s}^{-1}$ relative to the ground along a circle of radius $R=3\ \text{m}$. His angular speed relative to the ground is

$\omega=\frac{v}{R}=\frac{1.5}{3}=0.5\ \text{rad s}^{-1}$ …

Step 1 — Fix the vocabulary

For an elliptical orbit with the Sun at one focus:

• Perihelion (peri- = near, helios = Sun) — the point of closest approach.
• Aphelion (apo- = away) — the farthest point.

Therefore, immediately:

This single fact already eliminates options (A) and (C), both of which claim $r_p > r_a$.

Step 2 — Angular momentum is conserved (the central-force argument)

The gravitational force on the planet points along the line joining it to the Sun — it is a central force, i.e. $\vec{F} \parallel \vec{r}$. The torque about the Sun is therefore

And since

So the angular momentum has the same value at every point of the orbit — in particular:

This eliminates option (D), which claims $L_a < L_p$.

This conservation is Kepler's Second Law (the areal-velocity law): $\dfrac{dA}{dt} = \dfrac{L}{2m} = \text{constant}$ — equal areas swept in equal times.

Step 3 — Deduce the speeds from conserved $L$ …

Concept and intuition:

When the bullet embeds in the door, it’s an inelastic collision, but angular momentum about the hinge is conserved because there’s no external torque (frictionless hinge). The bullet’s linear momentum $p$ is converted into angular momentum about the hinge. The door rotates as a rigid body, so we need its moment of inertia about the hinge. The key is to treat the bullet as a point mass at the center of the door, and the door as a uniform rod rotating about one end.

Step-by-step solution:

1. Set up the angular momentum conservation. Before impact, only the bullet has angular momentum about the hinge. The bullet strikes at the center of the door, which is at a distance $r = 0.5\,\text{m}$ from the hinge (since the door is 1.0 m wide). Its linear momentum is $p$, so its angular momentum about the hinge is

(Direction: into the page, say positive.)

2. Moment of inertia of the door about the hinge. The door is a uniform rectangular slab, width 1.0 m, mass $M = 12\,\text{kg}$. For a rod of length $L$ rotating about one end, the moment of inertia is

Here $L = 1.0\,\text{m}$, so

3. Moment of inertia of the embedded bullet. The bullet becomes a point mass $m$ at distance $r = 0.5\,\text{m}$ from the hinge. Its contribution is

We don’t know $m$ directly, but we know its momentum $p = m v$. We’ll keep it symbolic.

4. Total moment of inertia after embedding.

5. Apply conservation of angular momentum. Initial angular momentum = final angular momentum:

So

6. Eliminate $m$ using $p = m v$. We need another relation. The bullet’s speed $v$ is not given, but we can express $m$ in terms of $p$ and $v$. However, we don’t have $v$. Wait — is there a hidden assumption? Actually, the problem gives only $p$ and the door’s properties. This suggests the answer should be independent of $m$ and $v$ individually. Let’s check:

This still contains $m$. But note: the bullet’s mass is not specified, so the answer must be in terms of $p$ only. That means the denominator must simplify to a pure number. How?

The bullet’s momentum $p$ is given, but its mass is not. However, the bullet’s mass is typically much smaller than the door’s mass, but here it’s not stated. The only way the answer is a pure number times $p$ is if the bullet’s mass is negligible compared to the door? But then $m \approx 0$ gives $\omega \approx 0.5p/4 = p/8$. That matches option (C). …

The collision has no external torque about the cylinder's axis, so angular momentum about that axis is conserved.

Moment of inertia of the solid cylinder:

Initial angular momentum of the cylinder:

Angular momentum brought in by the particle (moving tangentially at the rim, radius $R$):

$L_{\text{p}} = m v R = 1 \times 5 \times 0.2 = 1.0\ \text{kg·m}^2\text{s}^{-1}.$ …

The key concept here is areal velocity — the rate at which a planet sweeps out area as it orbits. Kepler’s second law tells us that a planet’s radius vector sweeps equal areas in equal times, which is a direct consequence of the conservation of angular momentum. The connection is geometric: the area swept per unit time is directly proportional to the angular momentum per unit mass.

Let’s derive it step by step.

1. Define areal velocity. Areal velocity, $A$, is the area swept out by the planet’s radius vector per unit time. In polar coordinates $(r, \theta)$, the infinitesimal area swept in time $dt$ is

Hence

2. Recall angular momentum. For a planet of mass $M$ moving in a plane, the angular momentum about the central body is

This is because the velocity component perpendicular to the radius is $r\dot{\theta}$, so $L = M r (r\dot{\theta}) = M r^2 \dot{\theta}$.

3. Relate the two. Compare the expressions: $$$$

Step 1 — Why angular momentum is conserved.

The gravitational pull on the planet is directed along the line joining it to the Sun, i.e. antiparallel to $\vec r$. The torque about the Sun is

And $\vec\tau = \dfrac{d\vec L}{dt}$, so

This is the defining property of any central force.

Step 2 — Areal velocity follows immediately.

In time $dt$ the radius vector sweeps the triangle of area $dA=\frac12|\vec r \times d\vec r|$, so

With $L$ constant, the areal velocity is constant — which is exactly Kepler's second law (equal areas in equal times).

Step 3 — Rule out the others. …

No external torque acts, so angular momentum is conserved:

Ring about its central axis: $I_i=MR^2$. Four point masses $m$ at radius $R$ add $4mR^2$:

Hence: …

Concept: Conservation of angular momentum for two discs coupled about a common axis (no external torque):

I₁ω₁ + I₂ω₂ = (I₁ + I₂) ω.

(4)(16) + (8)(4) = (4 + 8) ω …

Concept: Kepler's second law - the radius vector from the Sun to the planet sweeps equal areas in equal times. …

In the absence of external torque, angular momentum is conserved:

Here $\omega_2 = \omega_1/2$, so …