Two particles $A$ and $B$ are moving as shown in the figure. Their total angular momentum about the point $O$ is

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

The position vector of a particle of mass $m$ moving with a constant velocity $\vec{v} = v \hat{i}$ is given by $\vec{r} = x(t) \hat{i} + b \hat{j}$,where $b$ is a constant. At an instant,$\vec{r}$ makes an angle $\theta$ with the $x$-axis. The variation of the magnitude of the angular momentum $|\vec{L}|$ of the particle about the origin with $\theta$ will be:

Two rigid bodies $A$ and $B$ rotate with rotational kinetic energies $E_A$ and $E_B$ respectively. The moments of inertia of $A$ and $B$ about the axis of rotation are $I_A$ and $I_B$ respectively. If $I_A = I_B/4$ and $E_A = 100 E_B$,the ratio of angular momentum $(L_A)$ of $A$ to the angular momentum $(L_B)$ of $B$ is

$A$ small particle of mass $m$ is projected at an angle $\theta$ with the $x$-axis with an initial velocity $v_0$ in the $x-y$ plane as shown in the figure. At a time $t < \frac{v_0 \sin \theta}{g}$,the angular momentum of the particle is

$A$ particle of mass $M$ moves with a constant velocity parallel to the $X$-axis. What happens to its angular momentum with respect to the origin?