Two discs of moments of inertia $I_1$ and $I_2$ about their respective axes (normal to the disc and passing through the centre), and rotating with angular speed $\omega_1$ and $\omega_2$ are brought into contact face to face with their axes of rotation coincident. (a) Does the law of conservation of angular momentum apply to the situation? why? (b) Find the angular speed of the two-disc system. (c) Calculate the loss in kinetic energy of the system in the process. (d) Account for this loss.

Two discs of moments of inertia I₁ and I₂ about their respective axes (normal to the disc and passing through

NCERT Exemplar · Q24

I_1

(a) Does the law of conservation of angular momentum apply to the situation? why?

(b) Find the angular speed of the two-disc system.

(d) Account for this loss.

Sikkim CbseLong· 5mImportance★★★★★est

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Concept understanding — Conservation of Angular Momentum

Conservation of Angular Momentum

The Intuition First

Imagine you're sitting on a spinning office chair with your arms stretched out. Someone gives you a gentle push, and you start rotating slowly. Now pull your arms in tight against your chest. What happens? You spin faster. Push your arms back out — you slow down again.

Nothing external pushed you to go faster or slower. The change came from inside — from how you arranged your mass relative to the axis of rotation.

That's the core idea: Angular momentum is a quantity that stays constant for a rotating system unless an external torque acts on it. When you pulled your arms in, you didn't change your angular momentum — you changed your distribution of mass, and your rotation speed had to adjust to keep the total constant.

The Precise Statement

\vec{L} = \text{constant} \quad \text{if} \quad \vec{\tau}_{\text{ext}} = 0

Where:

$\vec{L}$ is the angular momentum of the system
$\vec{\tau}_{\text{ext}}$ is the net external torque acting on the system

In words: The total angular momentum of an isolated system (no external torque) remains constant in both magnitude and direction.

What Is Angular Momentum?

For a point mass $m$ moving with velocity $\vec{v}$ at position $\vec{r}$ from a reference point:

\vec{L} = \vec{r} \times m\vec{v}

For a rigid body rotating about a fixed axis:

L = I\omega

Where:

$I$ = moment of inertia (how mass is distributed relative to the axis)
$\omega$ = angular velocity (how fast it spins)

Note

Moment of inertia $I$ depends on where the mass is, not just how much. Mass far from the axis gives larger $I$ ; mass close to the axis gives smaller $I$ .

Why It Works: The Physics

Newton's second law for rotation says:

\tau_{\text{ext}} = \frac{dL}{dt}

If $\tau_{\text{ext}} = 0$ , then $\frac{dL}{dt} = 0$ , so $L$ is constant.

Since $L = I\omega$ , if $I$ changes (you pull arms in), $\omega$ must change in the opposite way to keep $L$ the same:

I_1 \omega_1 = I_2 \omega_2

Smaller $I$ → larger $\omega$ (spin faster). Larger $I$ → smaller $\omega$ (spin slower).

Real-World Examples

Situation	What happens	Why
Ice skater pulling arms in	Spins faster	$I$ decreases, $\omega$ increases to keep $L$ constant
Diver tucking into a ball	Rotates faster in midair	Same principle — no external torque during flight
Cat falling upside-down	Twists body to land on feet	Changes $I$ of different body parts to rotate without external torque
Planet orbiting the Sun	Speeds up when closer, slows down when farther	Gravitational force is central (torque = 0), so $L$ is constant

Watch out

Angular momentum is a vector. Its direction matters too. If no external torque acts, the axis of rotation stays fixed in space. This is why a spinning gyroscope or a bicycle wheel resists being tilted.

Common Mistake to Avoid …

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When two discs are brought coaxially into contact, angular momentum is conserved (no external torque), but kinetic energy is not — the final common angular speed is $\omega = \frac{I_1\omega_1 + I_2\omega_2}{I_1 + I_2}$ , and the kinetic energy loss is $\frac{1}{2} \frac{I_1 I_2}{I_1 + I_2} (\omega_1 - \omega_2)^2$ , dissipated as heat due to friction.

Why this approach works

The problem is a classic example of rotational inelastic collision. When the two discs are pressed together, friction between their faces does work to bring them to a common angular speed. No external torque acts about the common axis (the forces are internal to the two-disc system), so angular momentum is conserved. But kinetic energy is not conserved because friction is a non-conservative force — the "lost" energy appears as heat.

The mathematics follows directly from these two principles: conservation of angular momentum gives the final speed, and the difference in kinetic energies gives the loss.

Step-by-step solution

1. Does conservation of angular momentum apply?

Yes. The axes of both discs coincide, and they are brought into contact by forces that act along the axis (pressing them together) or are internal frictional forces between the discs. No external torque acts about the common axis of rotation. Therefore the total angular momentum of the system about that axis remains constant.

Watch out

A common mistake is to think friction is an external torque. But friction acts between the two discs — it is an internal force for the two-disc system. Internal torques cancel in pairs, so they cannot change the total angular momentum.

2. Find the final common angular speed

Let the final angular speed be $\omega$ (same for both discs once they stop slipping relative to each other).

Initial angular momentum:

L_i = I_1\omega_1 + I_2\omega_2

Final angular momentum:

L_f = (I_1 + I_2)\omega

By conservation of angular momentum:

I_1\omega_1 + I_2\omega_2 = (I_1 + I_2)\omega

Therefore:

\omega = \frac{I_1\omega_1 + I_2\omega_2}{I_1 + I_2}

Tip

This is exactly analogous to a perfectly inelastic collision in linear motion: $m_1v_1 + m_2v_2 = (m_1+m_2)v$ . The moment of inertia plays the role of mass, and angular speed plays the role of linear velocity.

3. Calculate the loss in kinetic energy

Initial rotational kinetic energy:

K_i = \frac{1}{2}I_1\omega_1^2 + \frac{1}{2}I_2\omega_2^2

Final rotational kinetic energy:

K_f = \frac{1}{2}(I_1 + I_2)\omega^2

Substitute $\omega$ from step 2:

K_f = \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)}

The loss $\Delta K = K_i - K_f$ :

\Delta K = \frac{1}{2}I_1\omega_1^2 + \frac{1}{2}I_2\omega_2^2 - \frac{(I_1\omega_1 + I_2\omega_2)^2}{2(I_1 + I_2)}

›Proof

Combine the terms over a common denominator $2(I_1 + I_2)$ :

$\Delta K = \frac{(I_1 + I_2)(I_1\omega_1^2 + I_2\omega_2^2) - (I_1\omega_1 + I_2\omega_2)^2}{2(I_1 + I_2)}$

Expand the numerator:

$= \frac{I_1^2\omega_1^2 + I_1I_2\omega_2^2 + I_1I_2\omega_1^2 + I_2^2\omega_2^2 - (I_1^2\omega_1^2 + 2I_1I_2\omega_1\omega_2 + I_2^2\omega_2^2)}{2(I_1 + I_2)}$

Cancel $I_1^2\omega_1^2$ and $I_2^2\omega_2^2$ : …

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Q.Two discs of moments of inertia $I_1$ and $I_2$ about their respective axes (normal to the disc and passing through the centre), and rotating with angular speed $\omega_1$ and $\omega_2$ are brought into contact face to face with their axes of rotation coincident.

Concept understanding — Conservation of Angular Momentum

Conservation of Angular Momentum

The Intuition First

The Precise Statement

What Is Angular Momentum?

Why It Works: The Physics

Real-World Examples

Common Mistake to Avoid …

Why this approach works

Step-by-step solution

1. Does conservation of angular momentum apply?

2. Find the final common angular speed

3. Calculate the loss in kinetic energy

Concept: Rotational Inelastic Collision — Angular Momentum Conserved, Kinetic Energy Is Not

Q.Two discs of moments of inertia I1I_1I1​ and I2I_2I2​ about their respective axes (normal to the disc and passing through the centre), and rotating with angular speed ω1\omega_1ω1​ and ω2\omega_2ω2​ are brought into contact face to face with their axes of rotation coincident.

Why this approach works

Step-by-step solution

1. Does conservation of angular momentum apply?

2. Find the final common angular speed

3. Calculate the loss in kinetic energy

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Q.Two discs of moments of inertia $I_1$ and $I_2$ about their respective axes (normal to the disc and passing through the centre), and rotating with angular speed $\omega_1$ and $\omega_2$ are brought into contact face to face with their axes of rotation coincident.

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