$A$ ring of mass $10 \ kg$ and diameter $0.4 \ m$ is rotated about its axis. If it rotates at $2100 \ rpm$,its angular momentum will be ........... $kg \cdot m^2/s$.

$A$ particle of mass $m$ moves with a constant velocity $\vec{v}$ in the $X-Y$ plane. Its angular momentum with respect to the origin is:

Obtain the relation between torque of a system of particles and angular momentum.

If the Earth is considered a sphere of radius $R$ and mass $M$,then its angular momentum about its axis of rotation in terms of the time period $T$ will be:

$A$ thin uniform rod of length $L$ and mass $m$ is kept on a frictionless horizontal table with a massless string of length $L$ fixed to one end (top view is shown in the figure). The other end of the string is pivoted to a point $O$. If a horizontal impulse $P$ is imparted to the rod at a distance $x = L/n$ from the mid-point of the rod (see figure), then the rod and string revolve together around the point $O$, with the rod remaining aligned with the string. In such a case, the value of $n$ is:

$A$ particle of mass $m$ is moving along a straight line parallel to the $x$-axis with a constant velocity $v$ at a distance $b$ from the origin. Find the angular momentum about the origin in vector form.