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 centre of the circle is

Options

A. `14.4kgm^(2)s^(-1)`
B. `8.64kgm^(2)s^(-1)`
C. `20.16kgm^(2)s^(-1)`
D. `11.52kgm^(2)s^(-1)`

Answer

A

Explanation

Angular momentum,
`L_(0)=mvr sin 90^(@)`
`=2xx0.6xx12xx1xx1`
[As `V=romega,Sin90^(@)=1`]
So, `L_(0)=14.4kgm^(2)//s`