A thin horizontal disc is rotating about a vertical axis passing...

The angular momentum of a rigid body about any arbitrary point $P$ is given by the relation:

$\vec{L}_P = \vec{L}_{cm} + \vec{r}_{cm/P} \times M\vec{v}_{cm}$

where $\vec{L}_{cm}$ is the angular momentum of the body about its center of mass, $\vec{r}_{cm/P}$ is the position vector of the center of mass with respect to point $P$, $M$ is the total mass of the body, and $\vec{v}_{cm}$ is the velocity of the center of mass.

In the given problem, the disc is rotating about a vertical axis passing through its fixed center $O$. Since the center of mass of the disc is at $O$ and is fixed, its velocity is zero, i.e., $\vec{v}_{cm} = 0$.

Substituting $\vec{v}_{cm} = 0$ into the angular momentum equation, we get:

$\vec{L}_P = \vec{L}_{cm} + \vec{r}_{cm/P} \times 0$

$\vec{L}_P = \vec{L}_{cm}$

This shows that the angular momentum of the disc is independent of the choice of the reference point and is equal to its angular momentum about the center of mass.

Therefore, the angular momentum about point $A$ is equal to the angular momentum about point $B$:

$L_A = L_B = L_{cm}$

Hence, the ratio is:

$\dfrac{L_A}{L_B} = 1$

Answer: $1$