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 $$n$$?

1. Translational (Orbital) Angular Momentum

• Disk A is fixed at the origin. Since both disks have a radius of $$R$$, the center of mass (CM) of disk B moves in a circular path of radius $$r = R + R = 2R$$.
• The center of mass of disk B moves with a fixed angular speed $\omega$ around the center of A. Therefore, the linear velocity of its center of mass ($$v_{\text{cm}}$$) is: $$v_{\text{cm}} = \omega(2R) = 2\omega R$$
• The orbital angular momentum ($$L_{\text{orbital}}$$) of disk B relative to the center of A is given by: $$L_{\text{orbital}} = M \cdot v_{\text{cm}} \cdot (2R) = M(2\omega R)(2R) = 4M\omega R^2$$

2. Rotational (Spin) Angular Momentum

• Because disk B rolls without slipping on the stationary circumference of disk A, its contact point with disk A must have an instantaneous velocity of zero.
• Let $$\omega_B$$ be the spinning angular velocity of disk B about its own center of mass. The non-slip condition dictates: $$v_{\text{cm}} = \omega_B R \implies 2\omega R = \omega_B R \implies \omega_B = 2\omega$$
• The moment of inertia of a uniform disk about its central axis is $$I_{\text{cm}} = \frac{1}{2}MR^2$$.
• The intrinsic spin angular momentum ($$L_{\text{spin}}$$) about its own center is: $$L_{\text{spin}} = I_{\text{cm}}\omega_B = \left(\frac{1}{2}MR^2\right)(2\omega) = M\omega R^2$$

3. Total Angular Momentum

Since both the orbital revolution and the intrinsic spin occur in the same direction (counter-clockwise), their magnitudes add up directly:

$$L_{\text{total}} = L_{\text{orbital}} + L_{\text{spin}}$$

$$L_{\text{total}} = 4M\omega R^2 + M\omega R^2 = 5M\omega R^2$$

Conclusion

Comparing our result with the expression given in the problem, $$n M \omega R^2$$:

$$n M \omega R^2 = 5 M \omega R^2 \implies n = 5$$