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

The moment of inertia of a uniform rod of mass $M$ and length $L$ about an axis through its centre and perpendicular to its length is given by $\frac{ML^2}{12}$. Now,consider one such rod pivoted at its centre,free to rotate in a vertical plane. The rod is at rest in the vertical position. $A$ bullet of mass $M$ moving horizontally at a speed $v$ strikes and embeds itself in one end of the rod. The angular velocity of the rod just after the collision will be

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:

If a body has a moment of inertia $I$ and an angular velocity $\omega \, rad/s$,then what is the value of its angular momentum $L$?

$A$ uniform rod of mass $8m$ and length $6a$ is placed on a horizontal table. Two point masses $m$ and $2m$ are moving with speeds $2v$ and $v$ respectively,as shown in the figure. They strike the rod and stick to it after the collision. Find the angular velocity of the rod about an axis passing through its center of mass.

Write the general formula of total angular momentum of rotational motion about a fixed axis.