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Question:

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 centre of the circle is?

14.4kgm2s𕒵

8.64kgm2s𕒵

11.52kgm2s𕒵

20.16kgm2s𕒵

Solution:

→r=x^i+y^j+z^k=rcosθ^i+rsinθ^j+z^k=0.6(cosθ^i+sinθ^j)+0.8^kwhere θ is the angle the radius makes with the x-axis as the particle moves in the circle.