When a particle is moving in a circular path around a fixed point in a plane,the direction of its angular momentum is ........

$A$ particle moves along a circular path with decreasing speed. Hence,

$A$ ball of mass $1 \,kg$ is projected with a velocity of $20 \sqrt{2} \,m/s$ from the origin of an $xy$ coordinate axis system at an angle of $45^{\circ}$ with the $x$-axis (horizontal). The angular momentum [in $SI$ units] of the ball about the point of projection after $2 \,s$ of projection is [take $g = 10 \,m/s^2$] ($y$-axis is taken as vertical).

Find the components along the $x, y, z$ axes of the angular momentum $\vec{l}$ of a particle whose position vector is $\vec{r}$ with components $x, y, z$ and momentum is $\vec{p}$ with components $p_x, p_y, p_z$. Show that if the particle moves only in the $x-y$ plane,the angular momentum has only a $z$-component.

The angular momentum of a particle is:

$A$ thin rod of mass $M$ and length $L$ is struck at one end by a ball of clay of mass $m$,moving with speed $v$ as shown in the figure. The ball sticks to the rod. After the collision,the angular momentum of the clay-rod system about $A$,the midpoint of the rod,is: