A particle of mass 2 kg is on a smooth horizontal table and moves...

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 \mathrm{rad} \mathrms^{-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 0.6 m. The height of the

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

Options

A $14.4 \mathrm{ kg} \mathrm{ m}^2 \mathrm{ s}^{-1}$
B $8.64 \mathrm{ kg} \mathrm{ m}^2 \mathrm{ s}^{-1}$
C $20.16 \mathrm{ kg} \mathrm{ m}^2 \mathrm{ s}^{-1}$
D $11.52 \mathrm{ kg} \mathrm{ m}^2 \mathrm{ s}^{-1}$

Correct answer

A. $14.4 \mathrm{ kg} \mathrm{ m}^2 \mathrm{ s}^{-1}$

Step-by-step solution

Angular momentum

L_0=\mathrm{mvr} \sin 90^{{}^{\circ}}=2 \times 0.6 \times 12 \times 1 \times 1 \left[\right.

V=r \omega, \left.\operatorname{Sin}90^{{}^{\circ}}=1\right]

So,

L_0=14.4 \mathrm{kgm}^2

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