PhysicsA particle of mass 2 kg is on a smooth horizontal table and moves in a circular path of radius 0.6m. The height of the table from the ground is 0.8 m. If the angular speed of the particle is 12 rad s–1, the magnitude of its angular momentum about a point on the ground right under the centre of the circle is

A particle of mass 2 kg is on a smooth horizontal table and moves in a circular path of radius 0.6m. The height of the table from the ground is 0.8 m. If the angular speed of the particle is 12 rad s–1, the magnitude of its angular momentum about a point on the ground right under the centre of the circle is

  1. A

    11.52 kg m2s–1

  2. B

    20.16 kg m2s–1

  3. C

    14.4 kg m2s–1

  4. D

    8.64 kg m2s–1

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    Solution:

    The magnitude of the angular momentum is 

    L = mVR 

    We know V = rω = 0.6 × 12 = 7.2m/s

    Also R=r2+h2= 0.62+0.82=1

    L = 2 x 7.2 x 1 = 14.4 kg m2s1

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