$A$ ring of mass $10 \ kg$ and diameter $0.4 \ m$ is rotating about its axis. If it completes $2100$ revolutions per minute,what is its angular momentum in $kg \cdot m^2/s$?

The position of a particle is given by $\vec r = (\hat i + 2\hat j - \hat k)$ and its momentum is $\vec p = (3\hat i + 4\hat j - 2\hat k)$. The angular momentum is perpendicular to:

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:

$A$ mass $m$ is moving with a constant velocity $v$ along a line parallel to the $X$-axis. Its angular momentum with respect to the origin or the $Z$-axis is:

$A$ particle starts from the point $(0, 8) \, m$ and moves with a uniform velocity of $\vec{v} = 3 \hat{i} \, m/s$. What is the angular momentum of the particle about the origin after $5 \, s$ (mass of the particle is $1 \, kg$)?

$A$ uniform rod $AB$ of mass $m$ and length $l$ is at rest on a smooth horizontal surface. An impulse $P$ is applied to the end $B$. The time taken by the rod to turn through a right angle is