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 :
The question asks for the ratio of angular momenta computed about two different points, A and B, for a rotating disc with center O. Given that \( OB = 2 \times OA \), we need to find the value of \(\frac{L_A}{L_B}\).
Angular momentum \( L \) for a rotating body is given by the formula:
\(L = I \omega + m \mathbf{r} \times \mathbf{v}\),
where:
- \(I\) is the moment of inertia.
- \(\omega\) is the angular velocity.
- \(\mathbf{r} \times \mathbf{v}\) is the cross product of the position vector with the velocity vector.
Since the disc is rotating about the axis through its center O, the contribution to angular momentum due to the off-center point is only because of the term \( m \mathbf{r} \times \mathbf{v} \).
For point A:
\(L_A = I_O \omega + m r_A^2 \omega\)
For point B:
\(L_B = I_O \omega + m r_B^2 \omega\)
Given \(r_B = 2 r_A\), we can substitute and rewrite:
\(L_A = I_O \omega + m r_A^2 \omega\)
\(L_B = I_O \omega + m (2 r_A)^2 \omega = I_O \omega + 4 m r_A^2 \omega\)
Taking the ratio:
\(\frac{L_A}{L_B} = \frac{I_O \omega + m r_A^2 \omega}{I_O \omega + 4 m r_A^2 \omega}\)
This simplifies to:
\(\frac{I_O \omega + m r_A^2 \omega}{I_O \omega + 4 m r_A^2 \omega} = 1\) (since the contributions of \(m r_A^2 \omega\) cancel out in the context given with \(I_O = 0\))
Therefore, the correct answer is 1. The angular momentum values are equivalent, reflecting the symmetric rotational dynamics about these points.
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