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, where \(OB=2\times OA\).
The value of
\[
\frac{L_A}{L_B}
\]
is:
Show Hint
Angular momentum depends on axis.
For a fixed rotation axis, shifting origin along axis does not change angular momentum.
Always identify the rotation axis first.
Remember rigid body rotation properties.
Step 1: Picture the situation. A thin disc spins about a fixed vertical axis through its centre $O$. We are asked to compare its angular momentum about two points $A$ and $B$ on that axis, with $OB = 2\,OA$. Step 2: Recall the key idea. For a rigid body rotating about a fixed axis, the spin angular momentum is $L = I\omega$, where $I$ is the moment of inertia about that axis. Step 3: Identify the axis. Both $A$ and $B$ lie on the very same vertical rotation axis. The body still rotates about that one axis. Step 4: Why position along the axis does not matter. The angular momentum of spin about a fixed axis is set by $I$ and $\omega$, both of which describe rotation about that axis, not by how far up or down the axis the reference point sits. Step 5: Compare the two values. Since $A$ and $B$ are on the same axis, the spin angular momentum is identical. \[ L_A = L_B \] Step 6: Form the ratio. Dividing equal quantities gives unity. \[ \frac{L_A}{L_B} = 1 \] \[ \boxed{1} \]
