A uniform circular disc of mass 50 kg and radius 0.4 m is rotating with an angular velocity of 10 ra

Q: Why doesn't the friction between the rings and the disc change the angular momentum?

Friction here acts as an *internal* force within the disc-rings system. While it causes the rings to spin up and the disc to slow down until they move together, the *net* torque on the entire system from this internal friction is exactly zero. Hence, total angular momentum is conserved.

Q: Why do we use the parallel axis theorem for the rings? Can't we just use $mr^2$?

The formula $mr^2$ is the moment of inertia of a ring about its *own* central axis. However, the rings are revolving around the *disc's* central axis. Since their centers are shifted by a distance $r$ from the axis of rotation, we must use the parallel axis theorem: $I = I_{\text{cm}} + md^2 = mr^2 + mr^2 = 2mr^2$.

Q: What does 'gently placed' mean in physics problems?

'Gently placed' is a classic physics code phrase! It means the objects are added without imparting any initial velocity or external impulsive forces. This guarantees that no external torque is applied to the system, making it a perfect candidate for the Conservation of Angular Momentum.

Q: Why did the angular velocity decrease from 10 to 8 rad/s?

Think of an ice skater spinning. When they throw their arms out, they slow down. By adding the rings, we increased the total moment of inertia of the system (from 4 to 5 $\text{kg}\cdot\text{m}^2$). To keep the total angular momentum ($L = I\omega$) constant, the angular velocity $\omega$ must decrease proportionally.

Q: Would the final angular velocity be different if the rings were placed at the edge of the disc?

Absolutely! If placed at the edge, the distance $d$ in the parallel axis theorem would be $R - r$ or $R$ depending on how they are placed. This would drastically increase the final moment of inertia, causing the final angular velocity to drop much lower. The further the mass is from the axis, the harder it is to spin!

Question

A uniform circular disc of mass $50	ext{ kg}$ and radius $0.4	ext{ m}$ is rotating with an angular velocity of $10	ext{ rad/s}$ about its own axis, which is vertical. Two uniform circular rings, each of mass $6.25	ext{ kg}$ and radius $0.2	ext{ m}$, are gently placed symmetrically on the disc in such a manner that they are touching each other along the axis of the disc and are horizontal. Assume that the friction is large enough such that the rings are at rest relative to the disc and the system rotates about the original axis. The new angular velocity (in $	ext{rad s}^{-1}$) of the system is

Sigma Percentile · Accepted Answer

### The Beauty of Rotational Dynamics

Imagine a large, heavy circular disc spinning smoothly like a playground merry-go-round. It possesses a certain amount of 'rotational momentum'—a stubbornness to keep spinning at its current rate. This problem is a beautiful exploration of what happens when we disturb this system, not by pushing or pulling it, but simply by adding mass to it. It's a classic demonstration of one of the most profound laws in physics: the Conservation of Angular Momentum.

### Visualizing the Setup

We start with a uniform circular disc of mass $M = 50	ext{ kg}$ and radius $R = 0.4	ext{ m}$. It is rotating freely about its vertical central axis with an initial angular velocity $\omega_1 = 10	ext{ rad/s}$.

Then comes the intervention. Two uniform circular rings, each of mass $m = 6.25	ext{ kg}$ and radius $r = 0.2	ext{ m}$, are 'gently placed' symmetrically on the disc. The phrase 'gently placed' is the golden key here. It means the rings are dropped onto the disc without any twisting force or external torque. They are placed such that they touch each other exactly at the axis of the disc. This geometric constraint is crucial: it tells us exactly how far the center of each ring is from the main axis of rotation.

### The Principle of Conservation

Because the rings are placed gently, the net external torque acting on the entire system (disc + rings) is zero ($	au_{	ext{ext}} = 0$). According to Newton's laws applied to rotation, when the net external torque is zero, the total angular momentum of the system must remain constant.

Mathematically, this is expressed as:

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

$I_1 \omega_1 = I_2 \omega_2$

Where $I_1$ is the initial moment of inertia, $\omega_1$ is the initial angular velocity, $I_2$ is the final moment of inertia, and $\omega_2$ is the final angular velocity we want to find.

### Calculating the Initial State

Before the rings are added, the only object rotating is the uniform disc. The moment of inertia of a uniform disc about its central axis is given by the standard formula:

$I_1 = I_{	ext{disc}} = \frac{1}{2}MR^2$

Let's plug in the given values to see what we are working with:

$I_1 = \frac{1}{2}(50	ext{ kg})(0.4	ext{ m})^2 = \frac{1}{2}(50)(0.16) = 4	ext{ kg}\cdot	ext{m}^2$

### The Parallel Axis Theorem Challenge

Now, we need to find the final moment of inertia, $I_2$, which includes the disc and the two rings. The moment of inertia of the disc remains the same, but we must carefully calculate the moment of inertia of the rings.

Here is where many students fall into a trap. The moment of inertia of a ring is $mr^2$, right? Yes, but *only* about its own central axis! Our rings are not rotating about their own centers; they are revolving around the central axis of the disc.

Because the rings touch each other at the disc's axis, the center of each ring is shifted by a distance $d = r$ from the axis of rotation. To find their moment of inertia about the disc's axis, we must invoke the **Parallel Axis Theorem** ($I = I_{	ext{cm}} + md^2$):

$I_{	ext{ring}} = I_{	ext{cm}} + md^2 = mr^2 + m(r)^2 = 2mr^2$

Since there are two identical rings, their total contribution to the moment of inertia is:

$I_{	ext{added}} = 2 	imes (2mr^2) = 4mr^2$

Let's calculate this numerical value:

$I_{	ext{added}} = 4(6.25	ext{ kg})(0.2	ext{ m})^2 = 4(6.25)(0.04) = 25(0.04) = 1	ext{ kg}\cdot	ext{m}^2$

So, the total final moment of inertia is:

$I_2 = I_{	ext{disc}} + I_{	ext{added}} = 4 + 1 = 5	ext{ kg}\cdot	ext{m}^2$

### The Final Calculation

We now have all the pieces of the puzzle. We return to our conservation equation:

$I_1 \omega_1 = I_2 \omega_2$

$(4)(10) = (5)\omega_2$

$40 = 5\omega_2$

$\omega_2 = \frac{40}{5} = 8	ext{ rad/s}$

### Conclusion and Physical Intuition

The final angular velocity is **$8	ext{ rad/s}$**.

Does this make physical sense? Absolutely. We added mass to the system, and more importantly, we added mass at a distance from the axis of rotation. This increased the system's rotational inertia (its resistance to spinning). To conserve the total angular momentum, the system had to slow down its rotation rate. It's the exact same physics that causes an ice skater to spin slower when they extend their arms outward!

Enter Numerical Value:

Visualized Solution

$Initial State$

$τ_{ext} = 0 ⟹ L_{i} = L_{f}$

$I_{1} = \frac{1}{2} M R^{2}$

$Placing the Rings$

$I_{ring} = I_{cm} + m d^{2}$

$I_{2} = I_{disc} + 2 I_{ring}$

$ω_{2} = \frac{I _{1}}{I _{2}} ω_{1}$

$Substituting Values$

$ω_{2} = 8 rad/s$

$Conclusion$

The Sigma Insight: Conservation of Angular Momentum

The Beauty of Rotational Dynamics

Visualizing the Setup

The Principle of Conservation

Calculating the Initial State

The Parallel Axis Theorem Challenge

The Final Calculation

Conclusion and Physical Intuition

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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; } Initial State

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } τext​=0⟹Li​=Lf​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I1​=21​MR2

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Iring​=Icm​+md2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I2​=Idisc​+2Iring​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ω2​=I2​I1​​ω1​

.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; } ω2​=8 rad/s

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