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Question

(C) The angular momentum $L$ of a rigid body rotating about a fixed axis passing through its center of mass $O$ with angular velocity $\omega$ is give

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(C) The angular momentum $L$ of a rigid body rotating about a fixed axis passing through its center of mass $O$ with angular velocity $\omega$ is given by $L_O = I_O \omega$, where $I_O$ is the moment of inertia about the axis passing through $O$.For any other point $P$ in the plane of the disc, the angular momentum $L_P$ is given by $L_P = L_O + \vec{r}_{OP} 	imes \vec{P}_{total}$, where $\vec{P}_{total}$ is the total linear momentum of the disc.Since the disc is rotating about its center of mass $O$, the total linear momentum $\vec{P}_{total} = M \vec{v}_{cm} = 0$, because the velocity of the center of mass $\vec{v}_{cm} = 0$.Therefore, the angular momentum about any point $P$ in the plane of the disc is equal to the angular momentum about the center of mass $O$, i.e.,$L_P = L_O$.Thus, $L_A = L_O$ and $L_B = L_O$.Consequently, $\frac{L_A}{L_B} = \frac{L_O}{L_O} = 1$.

$A$ thin horizontal disc is rotating about a vertical axis passing through its fixed centre $O$. Its angular momentum is $L_A$ and $L_B$ computed about points $A$ and $B$, respectively, with $OB = 2 \times OA$. The value of $\frac{L_A}{L_B}$ is :

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