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A particle of mass $2 \mathrm{~kg}$ is on a smooth horizontal table and moves in a circular path of radius $0.6 \mathrm{~m}$. The height of the table from the ground is 0.8 $\mathrm{m}$. If the angular speed of the particle is $12 \mathrm{rad} \mathrm{s}^{-1}$ the magnitude of its angular momentum about a point on the ground right under the centre of the circle is
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The correct answer is:
$14.4 \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1}$
Angular momentum, $L_0=m v r \sin 90^{\circ}$ 
$=2 \times 0.6 \times 12 \times 1 \times 1 \mid\left[\right.$ As $\left.V=r \omega, \operatorname{Sin} 90^{\circ}=1\right]$
So, $L_0=14.4 \mathrm{kgm}^2 / \mathrm{s}$
$=2 \times 0.6 \times 12 \times 1 \times 1 \mid\left[\right.$ As $\left.V=r \omega, \operatorname{Sin} 90^{\circ}=1\right]$
So, $L_0=14.4 \mathrm{kgm}^2 / \mathrm{s}$
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