Question:medium
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:
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:
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For a particle moving in a circular path, angular momentum is given by \( L = m v r \).
Updated On: Nov 26, 2025
- \( 14.4 \) 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\(^{-1} \)
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