A disc of mass M and radius R is free to rotate about its vertical axis as shown in the figure. A ba

Q: Why do we consider the orbital angular momentum of the small disc?

Even though the small disc is spinning on its own axis, its center of mass is located at a distance $R$ from the central axis. As the large disc rotates, it carries the small disc along with it, giving the small disc an orbital velocity and hence an orbital angular momentum.

Q: Is the angular velocity $\omega$ of the small disc absolute or relative to the large disc?

In this specific problem, the standard interpretation of 'rotates at a uniform angular speed $\omega$' without further qualification implies absolute angular speed. If it were relative, the spin term would be $\frac{1}{2}m(R/2)^2(\omega - \omega')$, leading to a different answer.

Q: Why is the initial angular momentum zero?

The problem states that 'Initially, both the discs are at rest'. Since nothing is moving, the initial angular momentum of the entire system is exactly zero.

Q: Why do the large and small discs rotate in opposite directions?

To conserve angular momentum! Since the initial angular momentum is zero, the positive angular momentum generated by the small disc's spin must be perfectly cancelled by the negative angular momentum of the large disc and the small disc's orbital motion.

Q: Can we apply conservation of energy here?

No! The battery-operated motor does work on the system to start the rotation. Mechanical energy is not conserved because the motor converts electrical energy into kinetic energy.

Question

A disc of mass $M$ and radius $R$ is free to rotate about its vertical axis as shown in the figure. A battery operated motor of negligible mass is fixed to this disc at a point on its circumference. Another disc of the same mass $M$ and radius $R/2$ is fixed to the motor's thin shaft. Initially, both the discs are at rest. The motor is switched on so that the smaller disc rotates at a uniform angular speed $\omega$. If the angular speed at which the large disc rotates is $\omega/n$, then the value of $n$ is_____.

Sigma Percentile · Accepted Answer

The problem of the rotating discs is a classic test of your understanding of the **Conservation of Angular Momentum**. It might look intimidating with a motor and two different axes of rotation, but the underlying physics is beautifully simple.

### The Setup: A System Free to Rotate

Imagine you are looking at this system from above. We have a large disc of mass $M$ and radius $R$, and mounted on its edge is a smaller disc of mass $M$ and radius $R/2$. Initially, everything is perfectly still.

When the motor is switched on, it forces the small disc to spin. But here is the crucial observation: the motor is internal to the system. There are absolutely no external twisting forces, or **torques**, acting on the entire setup about the central vertical axis.

According to Newton's laws applied to rotational motion, if the net external torque is zero, the total angular momentum of the system must remain constant. Since it started at zero, it must stay zero throughout the motion.

### Unpacking the Angular Momentum

When the small disc starts spinning with an absolute angular velocity $\omega$, it gains spin angular momentum. To keep the total angular momentum at zero, the large disc must react by rotating in the opposite direction. Let's call its angular speed $\omega'$.

But we must be very careful! The small disc is not just spinning on its own axis; it is physically attached to the edge of the large disc. As the large disc rotates, it carries the small disc along with it in a circle of radius $R$.

Therefore, the small disc has *two* types of angular momentum about the central axis:
1. **Spin Angular Momentum:** Due to its rotation about its own axis.
2. **Orbital Angular Momentum:** Due to its center of mass revolving around the central axis.

### The Master Equation

Let's write down the mathematical expression for each component.

The spin angular momentum of the small disc is its moment of inertia multiplied by its angular velocity:
$$L_{	ext{spin, small}} = \frac{1}{2} M \left(\frac{R}{2}ight)^2 \omega$$

The orbital angular momentum of the small disc is its mass times the radius squared times the angular velocity of the large disc:
$$L_{	ext{orbit, small}} = M R^2 \omega'$$

The spin angular momentum of the large disc is:
$$L_{	ext{spin, large}} = \frac{1}{2} M R^2 \omega'$$

Since the total angular momentum must be zero, the spin of the small disc must perfectly balance the combined orbital motion and the spin of the large disc:
$$\frac{1}{2} M \left(\frac{R}{2}ight)^2 \omega - M R^2 \omega' - \frac{1}{2} M R^2 \omega' = 0$$

### The Final Calculation

Now, we just need to perform the algebra. Let's simplify the equation by expanding the squared term:
$$\frac{1}{8} M R^2 \omega = M R^2 \omega' + \frac{1}{2} M R^2 \omega'$$

Notice how the mass $M$ and the radius squared $R^2$ appear in every single term? We can cancel them out completely! This shows that the final ratio of speeds is independent of the actual mass and radius of the discs.
$$\frac{1}{8} \omega = \frac{3}{2} \omega'$$

Solving for $\omega'$, we get:
$$\omega' = \frac{\omega}{12}$$

The problem states that the large disc rotates at a speed of $\omega/n$. By comparing our result, we can clearly see that **$n = 12$**.

This elegant cancellation is a hallmark of well-designed physics problems, rewarding you for setting up the conceptual framework correctly before diving into the numbers!

Enter Numerical Value:

Visualized Solution

$L_{ini t ia l} = 0$

$τ_{e x t} = 0 ⟹ Δ L = 0$

$L_{s ma l l}$

$L_{l a r g e}$

$L_{f ina l} = 0$

$Simplification$

$Final Answer$

$Food for Thought$

The Sigma Insight: Conservation of Angular Momentum

The Setup

Unpacking the Angular Momentum

The Master Equation

The 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; } Linitial​=0

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

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

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

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

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

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

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