First slide

Angular momentum

Question

A particle of mass 2 kg is on a smooth horizontal table and moves in a circular path of radius 0.6m. 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 centre of the circle is

Moderate

A

11.52 kg m²s^–1
B

20.16 kg m²s^–1
C

14.4 kg m²s^–1
D

8.64 kg m²s^–1

Solution

L = mVR ; m = 2kg
V = rω = 0.6 × 12 = 7.2m/s
$\large R = \sqrt {{r^2} + {h^2}} = 1m$
$\large \therefore L = 2 \times 7.2 \times 1 = 14.4kg{m^2}{s^{ - 1}}$

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