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 ω about the vertical axis passing through the center of A. The angular momentum of B is $nM{R}^2$ with respect to the center of A. Which of the following is the value of n ?

The correct answer is 5.

Step 1: Determine the distance to the center of mass of disk B

Disk A has a radius $R$ and its center is fixed at point $O$. Disk B also has a radius $R$ and rolls around the circumference of disk A.

The position vector of the center of mass $C$ of disk B relative to the center $O$ of disk A is at a distance:

$r_{\text{cm}}=R+R=2R$

Step 2: Calculate the linear speed of the center of mass of B

The center of mass of disk B rotates about the vertical axis passing through $O$ with an angular speed $\omega$.

Therefore, the linear speed of the center of mass $v_{\text{cm}}$ is:

$v_{\text{cm}}=r_{\text{cm}}\omega =2\omega R$

Step 3: Determine the spin angular speed of disk B

Let ${\omega }_B$ be the spin angular speed of disk B about its own center of mass $C$.

Since disk B rolls without slipping on the glued (stationary) disk A, the point of contact $P$ on disk B must have a instantaneous velocity of zero ($v_P=0$).

The velocity of the contact point $P$ is related to the center of mass velocity and spin rotation by:

$v_P=v_{\text{cm}}-{\omega }_{B}R=0$

Substituting $v_{\text{cm}}=2\omega R$ into the equation:

$2\omega R-{\omega }_{B}R=0$

$\Rightarrow {\omega }_B=2\omega$

Step 4: Calculate total angular momentum of disk B about the center of A

By the parallel axis theorem for angular momentum (Chasles' theorem for angular momentum), the total angular momentum $L_O$ of a rotating rigid body about an arbitrary point $O$ is the sum of its intrinsic angular momentum about its center of mass ($L_{\text{cm}}$) and its orbital angular momentum of the center of mass about $O$ ($L_{\text{orbital}}$):

$L_O=L_{\text{cm}}+L_{\text{orbital}}$

1. Intrinsic Angular Momentum ($L_{\text{cm}}$):

For a thin uniform disk of mass $M$ and radius $R$, the moment of inertia about its center of mass is $I_{\text{cm}}=\frac{1}{2}M{R}^2$.

$L_{\text{cm}}=I_{\text{cm}}{\omega }_B=\left(\frac{1}{2}M{R}^2\right)\left(2\omega \right)=M{R}^2\omega$

2. Orbital Angular Momentum ($L_{\text{orbital}}$):

$L_{\text{orbital}}=M{v}_{\text{cm}}{r}_{\text{cm}}=M\left(2\omega R\right)\left(2R\right)=4M{R}^2\omega$

Combining both terms:

$L_O=M{R}^2\omega +4M{R}^2\omega =5M{R}^2\omega$

Comparing $L_O=5M{R}^2\omega$ to the given expression $nM{R}^2\omega$, we find:

$n=5$