Why the angular momentum perpendicular to the axis ${L_ \bot }$ in a rotational motion about a fixed axis ? 

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 ........ $kg\, m^2\,s^{-1}$

A disc of mass  $M$  and radius  $R$  is rolling with angular speed $\omega $ on a horizontal plane as shown. The magnitude of angular momentum of the disc about the origin $O$ is

$A$ particle of mass $0.5\, kg$ is rotating in a circular path of radius $2m$ and centrepetal force on it is $9$ Newtons. Its angular momentum (in $J·sec$) is:

The time dependence of the position of a particle of mass $m = 2$ is given by $\vec r\,(t)\, = \,2t\,\hat i\, - 3{t^2}\hat j$ Its angular momentum with respect to the origin at time $t = 2$ is

Angular momentum of a single particle moving with constant speed along circular path: