Two coaxial discs, having moments of inertia I₁ and I₁/2 are rotating with respective angular veloci

Q: Why is angular momentum conserved when the discs rub against each other?

The friction between the discs is an internal force within the two-disc system. Since there is no external torque acting on the system from the outside, the total angular momentum remains constant.

Q: Why isn't kinetic energy conserved during the contact?

When the discs are brought into contact, they slip against each other until they reach a common angular velocity. This slipping involves kinetic friction, which does negative work and dissipates some of the rotational kinetic energy as heat.

Q: Can we use the formula for perfectly inelastic collisions here?

Yes! This is the rotational equivalent of a perfectly inelastic collision. Just like two masses sticking together and moving with a common velocity, the two discs 'stick' (in terms of rotation) and rotate with a common angular velocity.

Q: What does the negative sign in the final answer mean?

The negative sign indicates that the final energy is less than the initial energy. It represents the loss of kinetic energy, which has been converted into heat and sound due to friction.

Q: Does it matter which disc is dropped on which?

Not at all. The conservation of angular momentum depends only on the total initial momentum and the total final moment of inertia. The order or method of contact doesn't change the final outcome as long as no external torque is applied.

Question

Two coaxial discs, having moments of inertia $I_1$ and $\frac{I_1}{2}$ are rotating with respective angular velocities $\omega_1$ and $\frac{\omega_1}{2}$, about their common axis. They are brought in contact with each other and thereafter they rotate with a common angular velocity. If $E_f$ and $E_i$ are the final and initial total energies, then $(E_f - E_i)$ is

Options:

(A) $-\frac{I_1 \omega_1^2}{24}$

(B) $-\frac{I_1 \omega_1^2}{12}$

(C) $\frac{3}{8} I_1 \omega_1^2$

(D) $\frac{I_1 \omega_1^2}{6}$

Sigma Percentile · Accepted Answer

### The Setup: Two Discs, One Axis

Imagine you are observing two discs rotating independently on the exact same axis.

The first disc is our heavy hitter, possessing a moment of inertia $I_1$ and spinning with an angular velocity $\omega_1$.

The second disc is smaller and more sluggish, with exactly half the moment of inertia $\frac{I_1}{2}$ and half the angular velocity $\frac{\omega_1}{2}$.

This is our initial state, a system of two independent rotators full of kinetic energy.

### The Initial Energy: Before the Clash

Before these discs interact, we need to know exactly how much energy they possess collectively.

The total initial kinetic energy, $E_i$, is simply the sum of their individual rotational kinetic energies.

$$E_i = \frac{1}{2} I_1 \omega_1^2 + \frac{1}{2} \left(\frac{I_1}{2}\right) \left(\frac{\omega_1}{2}\right)^2$$

Let's carefully expand the terms for the second disc.

Squaring the angular velocity gives us a factor of $\frac{1}{4}$, which multiplies with the $\frac{1}{2}$ from the moment of inertia and the $\frac{1}{2}$ from the formula.

$$E_i = \frac{1}{2} I_1 \omega_1^2 + \frac{1}{16} I_1 \omega_1^2$$

Adding these fractions together, we find our starting energy.

$$E_i = \frac{9}{16} I_1 \omega_1^2$$

### The Collision: Conservation of Angular Momentum

Now comes the critical moment: the discs are brought into contact.

As they touch, kinetic friction violently acts between their surfaces, causing the faster disc to slow down and the slower disc to speed up.

However, here is the beautiful catch—this friction is an **internal force** within our two-disc system.

Because there is absolutely no **external torque** acting from the outside, the universe demands that the total angular momentum remains perfectly conserved.

Let's equate the initial angular momentum to the final angular momentum.

$$I_1 \omega_1 + \left(\frac{I_1}{2}\right) \left(\frac{\omega_1}{2}\right) = \left(I_1 + \frac{I_1}{2}\right) \omega$$

The left side represents the sum of their individual momenta, while the right side represents them rotating together as a single rigid body with a new common angular velocity, $\omega$.

$$\frac{5}{4} I_1 \omega_1 = \frac{3}{2} I_1 \omega$$

Solving this elegant equation, we discover their final shared speed.

$$\omega = \frac{5}{6} \omega_1$$

### The Aftermath: Final Energy and the Cost of Friction

With the discs now locked in a synchronized spin, we can calculate their final kinetic energy, $E_f$.

We use their combined moment of inertia, which is $\frac{3}{2} I_1$, and our newly found common angular velocity.

$$E_f = \frac{1}{2} \left(\frac{3}{2} I_1\right) \left(\frac{5}{6} \omega_1\right)^2$$

Squaring the velocity term gives us $\frac{25}{36}$.

$$E_f = \frac{1}{2} \cdot \frac{3}{2} I_1 \cdot \frac{25}{36} \omega_1^2$$

Multiplying these fractions out reveals the final energy state of our system.

$$E_f = \frac{25}{48} I_1 \omega_1^2$$

Finally, the question asks for the change in energy, $\Delta E = E_f - E_i$.

$$\Delta E = \frac{25}{48} I_1 \omega_1^2 - \frac{9}{16} I_1 \omega_1^2$$

To subtract these, we find the common denominator, which is $48$.

$$\Delta E = \left(\frac{25 - 27}{48}\right) I_1 \omega_1^2$$

$$\Delta E = -\frac{2}{48} I_1 \omega_1^2 = -\frac{1}{24} I_1 \omega_1^2$$

The result is **negative**, which is a profound physical truth.

This negative sign proves that kinetic energy was lost, permanently dissipated as heat and sound due to the friction during their contact.

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Visualized Solution

$Initial Setup$

$E_{i} = \frac{1}{2} I_{A} ω_{A}^{2} + \frac{1}{2} I_{B} ω_{B}^{2}$

$Conservation of Angular Momentum$

$Finding Common Angular Velocity ω$

$E_{f} = \frac{1}{2} (I_{A} + I_{B}) ω^{2}$

$Δ E = E_{f} - E_{i}$

$The Way Forward$

The Sigma Insight: Conservation of Angular Momentum

The Setup

The Initial Energy

The Collision

The Aftermath

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Initial Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ei​=21​IA​ωA2​+21​IB​ωB2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conservation of Angular Momentum

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding Common Angular Velocity ω

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ef​=21​(IA​+IB​)ω2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔE=Ef​−Ei​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward