A flat surface of a thin uniform disk A of radius R is glued to a horizontal table. Another thin uni

Q: Why is the distance of the center of mass of B taken as $2R$?

Disk A has radius $R$ and disk B has radius $R$. Since B rolls on the circumference of A, the distance between their centers is $R + R = 2R$.

Q: Why is the velocity of the contact point zero?

The problem states that disk B rolls *without slipping* on disk A. Since disk A is glued and stationary, the point on disk B in contact with A must also be momentarily at rest.

Q: How did we get $\omega_0 = 2\omega$?

The velocity of the center of B is $v_c = \omega(2R)$. The velocity of the contact point is $v_c - \omega_0 R = 0$. Substituting $v_c$, we get $2\omega R = \omega_0 R$, which gives $\omega_0 = 2\omega$.

Q: Why do we add the two angular momentum terms?

The total angular momentum of a rigid body about any point is the sum of its spin angular momentum about its own center of mass ($I_{cm}\omega_0$) and the orbital angular momentum of its center of mass about that point ($\vec{r} \times \vec{p}$).

Q: Does the direction of $\omega$ and $\omega_0$ matter?

Yes! Since disk B rolls on the outside of A, as its center moves counterclockwise, it must also spin counterclockwise to have zero velocity at the inner contact point. Thus, both angular momentum vectors point in the same direction (out of the plane) and simply add up.

Question

A flat surface of a thin uniform disk $A$ of radius $R$ is glued to a horizontal table. Another thin uniform disk $B$ of mass $M$ and with the same radius $R$ rolls without slipping on the circumference of $A$, as shown in the figure. A flat surface of $B$ also lies on the plane of the table. The center of mass of $B$ has fixed angular speed $\omega$ about the vertical axis passing through the center of $A$. The angular momentum of $B$ is $nM\omega R^2$ with respect to the center of $A$. Which of the following is the value of $n$?

Options:

(A) 2

(B) 5

(C) $\frac{7}{2}$

(D) $\frac{9}{2}$

Sigma Percentile · Accepted Answer

The problem of a disk rolling on another fixed disk is a classic test of your understanding of rigid body kinematics and angular momentum. It beautifully combines the concepts of pure rolling, orbital motion, and spin. Let's break it down step by step.

### Analyzing the Setup

Imagine you are looking at the table from above. We have two identical disks, $A$ and $B$, both of radius $R$. Disk $A$ is firmly glued to the table, meaning it cannot move or rotate. Disk $B$, on the other hand, is rolling on the circumference of disk $A$.

The center of mass of disk $B$ is revolving around the center of disk $A$ with a constant angular speed $\omega$. Because both disks have a radius $R$, the distance between their centers is simply $R + R = 2R$.

This means the center of disk $B$ is moving in a circular path of radius $2R$. Therefore, the linear velocity of the center of mass of disk $B$, let's call it $v_c$, is given by:
$$v_c = \omega(2R)$$

### The Condition for Pure Rolling

Here is where many students make a mistake. The question explicitly states that disk $B$ rolls **without slipping** on disk $A$. What does this physically mean? It means that at the exact point where the two disks touch, there is no relative motion between them. Since disk $A$ is stationary, the velocity of the contact point on disk $B$ must be exactly zero.

Let's assume disk $B$ is spinning about its own center of mass with an angular velocity $\omega_0$. The net velocity of the contact point is the vector sum of the translational velocity of the center of mass ($v_c$) and the tangential velocity due to its spin ($\omega_0 R$).

For the contact point to be at rest, these two velocities must perfectly cancel each other out:
$$v_c - \omega_0 R = 0$$

Substituting our earlier expression for $v_c$, we get:
$$2\omega R - \omega_0 R = 0$$

Solving this gives us the spin angular velocity of disk $B$:
$$\omega_0 = 2\omega$$

This is a crucial insight! Disk $B$ is spinning twice as fast about its own axis as it is revolving around disk $A$.

### The Master Equation of Angular Momentum

Now, we need to find the total angular momentum of disk $B$ with respect to the center of disk $A$. For a rigid body undergoing both translation and rotation, the total angular momentum about a point is the sum of two parts:
1. The spin angular momentum about its own center of mass ($\vec{L}_{cm}$).
2. The orbital angular momentum of its center of mass about the reference point ($\vec{r}_{cm} 	imes \vec{p}_{cm}$).

Mathematically, this is expressed as:
$$\vec{L}_A = \vec{L}_{cm} + \vec{r}_{cm} 	imes \vec{p}_{cm}$$

Let's calculate each part carefully. The spin angular momentum is the moment of inertia of the disk about its center multiplied by its spin angular velocity:
$$L_{cm} = I_{cm}\omega_0 = \left(\frac{1}{2}MR^2ight)(2\omega) = MR^2\omega$$

Next, the orbital angular momentum is the cross product of the position vector and the linear momentum. Since the velocity is always perpendicular to the position vector in circular motion, the magnitude is simply $r_{cm} M v_c$:
$$L_{orbital} = (2R)M(2\omega R) = 4MR^2\omega$$

### Final Calculation

Since both the spin and the orbital motion are in the same direction (counterclockwise), their angular momentum vectors point in the same direction (out of the page). We can simply add their magnitudes:
$$L_A = MR^2\omega + 4MR^2\omega = 5MR^2\omega$$

The problem states that the angular momentum is $nM\omega R^2$. By comparing our result with this expression, we can immediately see that:
$$n = 5$$

This elegant result shows how the different components of motion contribute to the total angular momentum. The orbital motion actually contributes four times as much angular momentum as the spin!

Select Answer:

Visualized Solution

Setup and Kinematics

Velocity of Center of Mass

Condition for Pure Rolling

Angular Velocity of Disk B

Solving for $ω_{0}$

Total Angular Momentum

Calculating Components

Summing Angular Momentum

Final Answer

Food for Thought

The Sigma Insight: Dynamics of Rigid Body Rotation

Analyzing the Setup

The Condition for Pure Rolling

The Master Equation of Angular Momentum

Final Calculation

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