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 ⁻¹, the magnitude of its angular momentum about a point on the ground right under the center of the circle is: (a) 8.64 kg m2s ⁻¹ (b) 11.52 kg m2s ⁻¹ (c) 14.4 kg m2s ⁻¹ (d) 20.16 kg m2s ⁻¹

Answer: (c). 0.6 m0.8 mr = 1.0 mground point v = ω r = 12 × 0.6 = 7.2 m/s, p = mv = 2 × 7.2 = 14.4 kg m/s. Position vector from the ground point: 0.6 m horizontal and 0.8 m vertical, so | r| = √(0.6² + 0.8²) = 1.0 m. p is tangential, perpendicular to both components of r, so L = r p = 1.0 × 14.4. So the answer is (c) 14.4 kg m ² s ⁻¹.

A particle of mass 2 kg is on a smooth horizontal table and moves in… | JEE Main 2015 (11 Apr, Shift 1)

Q67JEE Main 2015 · 11 APR · SHIFT 1Medium

Rotational Kinematics & Dynamics

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 ${}^{-1}$ , the magnitude of its angular momentum about a point on the ground right under the center of the circle is:

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Answer: (c)

$v = \omega r = 12 \times 0.6 = 7.2$ m/s, $p = mv = 2 \times 7.2 = 14.4$ kg m/s.
Position vector from the ground point: $0.6$ m horizontal and $0.8$ m vertical, so $|\vec r| = \sqrt{0.6^2 + 0.8^2} = 1.0$ m.
$\vec p$ is tangential, perpendicular to both components of $\vec r$ , so $L = r p = 1.0 \times 14.4$ .
So the answer is (c) $14.4$ kg m $^2$ s $^{-1}$ .

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