A particle of mass $2 \mathrm{~kg}$ is on a smooth horizontal table and moves in a circular path of radius…

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
  1. $14.4 \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-1}$
  2. $8.64 \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-1}$
  3. $20.16 \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-1}$
  4. $11.52 \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-1}$

Solution

Angular momentum, $L_{0}=\operatorname{mvr} \sin 90^{\circ}$
$=2 \times 0.6 \times 12 \times 1 \times 1$
$\left[\right.$ As $\left.V=r \omega, \operatorname{Sin} 90^{\circ}=1\right] \quad 0.8 \mathrm{~m}$
So, $L_{0}=14.4 \mathrm{kgm}^{2} / \mathrm{s}$
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Asked in: JEE Mains - Rotational Motion - Test 4

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