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)` . Find the magnitude of its angular momentum about a point on the ground right under the centre of the circle .

Explanation

Here , m = 2 kg , `omega = 12 rad s^(-1)`
`r = sqrt((0.8)^(2) + (0.6)^(2)) = 1` m
Angular momentum of the particle about point O ,
`L = mvr sin 90^(@)`
`= m xx (0.6 omega) r`
`=2 xx 0.6 xx 12 xx 1 = 14.4 kg m^(2) s^(-1)` .