Two uniform circular discs are rotating independently in the same direction around their common axis

Q: Why is angular momentum conserved but not kinetic energy?

When the discs come into contact, kinetic friction acts between them until they reach a common angular velocity. This internal friction does negative work, dissipating some kinetic energy as heat. However, since these frictional torques are internal to the two-disc system and cancel each other out, the net external torque is zero, conserving angular momentum.

Q: Can we use conservation of energy to find the final angular velocity?

No! As explained above, this is an inelastic collision. Kinetic energy is strictly not conserved. Using $K_i = K_f$ will lead to a completely wrong answer.

Q: What if the discs were rotating in opposite directions initially?

If they were rotating in opposite directions, you would assign a positive sign to one angular velocity and a negative sign to the other. The total initial angular momentum would be the difference of their magnitudes.

Q: Does the order in which they are dropped matter?

No, the final state only depends on the total initial angular momentum and the total final moment of inertia. The sequence of events leading to the sticking does not change the final common angular velocity.

Q: Is this similar to a perfectly inelastic collision in linear kinematics?

Yes, it is the exact rotational analog. In linear kinematics, $m_1v_1 + m_2v_2 = (m_1+m_2)v$. Here, $I_1\omega_1 + I_2\omega_2 = (I_1+I_2)\omega$. In both cases, the objects stick together and move with a common velocity, resulting in a maximum loss of kinetic energy.

Question

Two uniform circular discs are rotating independently in the same direction around their common axis passing through their centres. The moment of inertia and angular velocity of the first disc are $0.1 	ext{ kg-m}^2$ and $10 	ext{ rad s}^{-1}$ respectively, while those for the second one are $0.2 	ext{ kg-m}^2$ and $5 	ext{ rad s}^{-1}$, respectively. At some instant they get stuck together and start rotating as a single system about their common axis with some angular speed. The kinetic energy of the combined system is

Options:

(A) $\frac{10}{3} 	ext{ J}$

(B) $\frac{20}{3} 	ext{ J}$

(C) $\frac{5}{3} 	ext{ J}$

(D) $\frac{2}{3} 	ext{ J}$

Sigma Percentile · Accepted Answer

The problem of two rotating discs sticking together is a classic demonstration of the **Conservation of Angular Momentum**. It is the rotational equivalent of a perfectly inelastic collision in linear mechanics. Let's break down the physics step-by-step.

### The Setup: Two Discs, One Axis

Imagine two uniform circular discs, both spinning independently around the same vertical axis.

The first disc is lighter, with a moment of inertia $I_1 = 0.1 	ext{ kg-m}^2$, and it's spinning quite fast at an angular velocity $\omega_1 = 10 	ext{ rad s}^{-1}$.

The second disc is heavier, with a moment of inertia $I_2 = 0.2 	ext{ kg-m}^2$, and it's spinning slower at $\omega_2 = 5 	ext{ rad s}^{-1}$.

Suddenly, they are brought into contact. Friction acts between their surfaces, causing the faster one to slow down and the slower one to speed up until they lock together and rotate as a single rigid body.

### The Master Principle: Conservation of Angular Momentum

When the discs rub against each other, they exert equal and opposite frictional torques on one another. However, if we consider both discs together as our **system**, these frictional torques are purely *internal*.

Since there is absolutely no *external* torque acting on the system from the outside world, the total angular momentum of the system must remain perfectly conserved.

$$L_{	ext{initial}} = L_{	ext{final}}$$

The initial angular momentum is simply the sum of the individual angular momenta of the two discs:

$$L_{	ext{initial}} = I_1\omega_1 + I_2\omega_2$$

When they stick together, they form a single object with a combined moment of inertia $(I_1 + I_2)$ rotating at a new, common angular velocity $\omega$.

$$L_{	ext{final}} = (I_1 + I_2)\omega$$

Equating the two gives us our master equation:

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

### Finding the Common Angular Velocity

Now, we just need to plug in the given values to find the final angular velocity $\omega$.

$$(0.1)(10) + (0.2)(5) = (0.1 + 0.2)\omega$$

Calculating the left side:

$$1 + 1 = 0.3\omega$$

$$2 = 0.3\omega$$

Solving for $\omega$:

$$\omega = \frac{2}{0.3} = \frac{20}{3} 	ext{ rad s}^{-1}$$

So, the combined system rotates at $\frac{20}{3} 	ext{ rad s}^{-1}$.

### The Final Kinetic Energy

The question asks for the kinetic energy of the combined system. The formula for rotational kinetic energy is:

$$K = \frac{1}{2}I\omega^2$$

For our combined system, the total moment of inertia is $I_{	ext{sys}} = I_1 + I_2 = 0.3 	ext{ kg-m}^2$, and the angular velocity is $\omega = \frac{20}{3} 	ext{ rad s}^{-1}$.

Substituting these values:

$$K_f = \frac{1}{2} (0.3) \left(\frac{20}{3}ight)^2$$

Let's simplify the math:

$$K_f = \frac{1}{2} \left(\frac{3}{10}ight) \left(\frac{400}{9}ight)$$

$$K_f = \frac{1}{2} 	imes \frac{1200}{90} = \frac{1}{2} 	imes \frac{40}{3} = \frac{20}{3} 	ext{ J}$$

The final kinetic energy of the combined system is **$\frac{20}{3} 	ext{ J}$**.

### The Hidden Physics: Where Did the Energy Go?

You might wonder, is kinetic energy conserved in this process? Let's check!

The initial kinetic energy was:
$$K_i = \frac{1}{2}I_1\omega_1^2 + \frac{1}{2}I_2\omega_2^2 = \frac{1}{2}(0.1)(100) + \frac{1}{2}(0.2)(25) = 5 + 2.5 = 7.5 	ext{ J}$$

The final kinetic energy is $\frac{20}{3} 	ext{ J} \approx 6.67 	ext{ J}$.

Notice that $K_f < K_i$. Kinetic energy was **lost**! This is because the collision is perfectly inelastic. The "lost" energy was dissipated as heat and sound due to the kinetic friction between the discs as they slipped against each other before finally locking together. This is a beautiful rotational analog to perfectly inelastic linear collisions!

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Visualized Solution

$Initial and Final States$

$Conservation of Angular Momentum$

$Substituting Values$

$Calculating Momenta$

$Final Angular Velocity$

$Rotational Kinetic Energy$

$Substituting Values for KE$

$Simplifying Expression$

$Final Kinetic Energy$

$Energy Loss in Collision$

The Sigma Insight: Conservation of Angular Momentum

The Setup

The Master Principle

Finding the Common Angular Velocity

The Final Kinetic Energy

The Hidden Physics

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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 and Final States

.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; } Substituting Values

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values for KE

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

.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 in Collision