A particle of mass \( 2 \) kg is on a smooth horizontal table and moves in a circular path of radius \( 0.6 \) m. The height of the table from the ground is \( 0.8 \) m. If the angular speed of the particle is \( 12 \) rad/s, the magnitude of its angular momentum about a point on the ground right under the center of the circle is:
Show Hint
- \( 14.4 \) kg m\( ^2 \)s\(^{-1} \)
- \( 8.64 \) kg m\( ^2 \)s\(^{-1} \)
- \( 20.16 \) kg m\( ^2 \)s\(^{-1} \)
- \( 11.52 \) kg m\( ^2 \)s\(^{-1} \)
The Correct Option is A
Solution and Explanation
{Angular momentum formula}
\[ L_0 = mvr \sin 90^\circ \] \[ = 2 \times 0.6 \times 12 \times 1 \] \[ = 14.4 \text{ kg m}^2 \text{s}^{-1} \] Thus, the correct answer is 14.4 kg m\( ^2 \)s\(^{-1} \).
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