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 $n$?

Step 1: Determine the distance and velocity of the center of mass of disk $B$.

Since both disks have radius $R$ and touch externally, the distance from the center of disk $A$ (point $O$) to the center of disk $B$ is:

$$d = R + R = 2R$$

The center of mass of $B$ moves in a circle of radius $2R$ with angular speed $\omega$ about $O$, giving a tangential speed of:

$$v = 2R\omega$$

Step 2: Find the spin angular speed of disk $B$.

Since disk $B$ rolls without slipping on the fixed disk $A$, the contact point is instantaneously at rest.

$$v = \omega_B R$$

Substitute $v = 2R\omega$ to find the spin angular speed $\omega_B$ of disk $B$ about its own center:

$$2R\omega = \omega_B R$$ $$\omega_B = 2\omega$$

Step 3: Calculate the spin angular momentum of disk $B$.

The moment of inertia of the uniform disk $B$ about its center of mass is:

$$I_{\text{cm}} = \frac{1}{2}MR^2$$

The spin angular momentum about its center of mass is:

$$L_{\text{spin}} = I_{\text{cm}}\omega_B$$ $$L_{\text{spin}} = \left(\frac{1}{2}MR^2\right)(2\omega)$$ $$L_{\text{spin}} = MR^2\omega$$

Step 4: Calculate the orbital angular momentum of disk $B$.

The orbital angular momentum of the center of mass of $B$ about the center of $A$ is:

$$L_{\text{orb}} = M(2R)^2\omega$$ $$L_{\text{orb}} = 4MR^2\omega$$

Step 5: Find the total angular momentum and determine $n$.

The total angular momentum $L_O$ of disk $B$ with respect to the center of $A$ is the sum of its spin and orbital angular momenta:

$$L_O = L_{\text{spin}} + L_{\text{orb}}$$ $$L_O = MR^2\omega + 4MR^2\omega$$ $$L_O = 5MR^2\omega$$

Comparing this with the given expression $L_O = n M \omega R^2$:

$$n = 5$$

Answer: $n = 5$