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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 × OA. The value of $\frac{L_A}{L_B}$ is:
Detailed Solution
For a body in pure rotation, the angular momentum is the same about every point: $L=I\omega$. So $L_A=\frac{MR^2}{2}\omega$ and $L_B=\frac{MR^2}{2}\omega$, giving $\frac{L_A}{L_B}=1$.