A circular disc of mass M and radius R is rotating about its axis with angular speed omega_1. If ano

Q: Why is angular momentum conserved if there is friction between the discs?

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

Q: Why isn't kinetic energy conserved in this process?

When the top disc is dropped, it initially slips against the bottom disc. This slipping involves kinetic friction, which does negative work and dissipates some of the rotational kinetic energy into heat.

Q: Does the mass of the top disc affect the percentage of energy lost?

Yes, the percentage of energy lost depends on the ratio of the moments of inertia of the two discs, which in turn depends on their masses and radii. If the top disc had a different mass, the fraction $I_2/I_1$ would change, altering the percentage loss.

Q: What would happen if the top disc was dropped off-center?

If dropped off-center, the top disc would introduce an unbalanced force and torque, causing the system to wobble. The moment of inertia would have to be calculated using the parallel axis theorem, making the math much more complex.

Q: Is there a shortcut formula for energy loss in such rotational collisions?

Yes! Just like in linear perfectly inelastic collisions, the energy loss can be directly calculated using $\Delta K = \frac{1}{2} \frac{I_1 I_2}{I_1 + I_2} (\omega_1 - \omega_2)^2$. Since the second disc is initially stationary, it simplifies to $\Delta K = \frac{1}{2} \frac{I_1 I_2}{I_1 + I_2} \omega_1^2$.

Question

A circular disc of mass $M$ and radius $R$ is rotating about its axis with angular speed $\omega_1$. If another stationary disc having radius $\frac{R}{2}$ and same mass $M$ is dropped co-axially on to the rotating disc. Gradually, both discs attain constant angular speed $\omega_2$. The energy lost in the process is $p\%$ of the initial energy. Value of $p$ is ...........

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine a large circular disc of mass $M$ and radius $R$ spinning smoothly on its central axis with an initial angular speed $\omega_1$. Now, a second disc, identical in mass $M$ but with half the radius $\frac{R}{2}$, is gently dropped co-axially onto the first one.

Initially, the second disc is completely stationary. But as soon as it makes contact, friction acts between the two surfaces. The bottom disc tries to drag the top disc along, while the top disc tries to slow the bottom one down. Eventually, the slipping stops, and both discs rotate together as a single rigid body with a new, constant angular speed $\omega_2$.

### The Master Equation: Conservation of Angular Momentum
Because the top disc is dropped vertically, there are no external torques acting on the system about the axis of rotation. The friction between the discs is an *internal* force. Therefore, the total angular momentum of the system must remain perfectly conserved.

$$L_i = L_f$$

Before we plug in the values, let's determine the moment of inertia for both discs. For the large bottom disc, the moment of inertia is standard:
$$I_1 = \frac{1}{2} M R^2$$

For the smaller top disc, we substitute its radius $\frac{R}{2}$ into the formula:
$$I_2 = \frac{1}{2} M \left(\frac{R}{2}ight)^2 = \frac{1}{8} M R^2$$

Now, applying the conservation of angular momentum:
$$I_1 \omega_1 + I_2 (0) = (I_1 + I_2) \omega_2$$

Substituting the inertias:
$$\frac{1}{2} M R^2 \omega_1 = \left(\frac{1}{2} M R^2 + \frac{1}{8} M R^2ight) \omega_2$$

$$\frac{1}{2} \omega_1 = \frac{5}{8} \omega_2$$

Solving for the final angular speed, we get:
$$\omega_2 = \frac{4}{5} \omega_1$$

### Calculating the Energy Loss
In perfectly inelastic rotational collisions like this one, kinetic energy is never conserved. Let's find out exactly how much energy was lost.

The initial kinetic energy of the system is just the energy of the bottom disc:
$$K_i = \frac{1}{2} I_1 \omega_1^2 = \frac{1}{2} \left(\frac{1}{2} M R^2ight) \omega_1^2 = \frac{1}{4} M R^2 \omega_1^2$$

The final kinetic energy of the combined system is:
$$K_f = \frac{1}{2} (I_1 + I_2) \omega_2^2$$

Substitute the total inertia and the new angular speed:
$$K_f = \frac{1}{2} \left(\frac{5}{8} M R^2ight) \left(\frac{4}{5} \omega_1ight)^2$$

$$K_f = \frac{1}{2} \cdot \frac{5}{8} \cdot \frac{16}{25} M R^2 \omega_1^2 = \frac{1}{5} M R^2 \omega_1^2$$

The energy lost during the slipping phase is the difference between the initial and final kinetic energies:
$$\Delta K = K_i - K_f = \left(\frac{1}{4} - \frac{1}{5}ight) M R^2 \omega_1^2 = \frac{1}{20} M R^2 \omega_1^2$$

### Final Calculation
The problem asks for the percentage loss of the initial energy, denoted as $p\%$.
$$p = \frac{\Delta K}{K_i} 	imes 100$$

$$p = \frac{\frac{1}{20} M R^2 \omega_1^2}{\frac{1}{4} M R^2 \omega_1^2} 	imes 100$$

$$p = \frac{4}{20} 	imes 100 = 20$$

**The value of $p$ is 20.** This $20\%$ of the initial kinetic energy didn't just vanish; it was dissipated as heat and sound due to the kinetic friction between the two discs before they achieved a common angular speed.

Enter Numerical Value:

Visualized Solution

$System Setup$

$Conservation of Angular Momentum$

$Moments of Inertia$

$Applying Conservation Law$

$Final Angular Speed$

$Initial Kinetic Energy$

$Final Kinetic Energy$

$Energy Loss$

$Percentage Loss$

$The Way Forward$

The Sigma Insight: Conservation of Angular Momentum

Analyzing the Setup Imagine a large circular disc of mass $M$ and radius $R$ spinning smoothly on its central axis with an initial angular speed $ω_{1}$

The Master Equation

Calculating the Energy Loss In perfectly inelastic rotational collisions like this one, kinetic energy is never conserved

Final Calculation

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Enter Numerical Value:

Visualized Solution

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

.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; } Moments of Inertia

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

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

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

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

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

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

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