$A$ body of mass $5 \,kg$ moving with a uniform speed $3 \sqrt{2} \,m/s$ in the $X-Y$ plane along the line $y=x+4$. The angular momentum of the particle about the origin will be . . . . . . $kg \,m^2/s$.

$A$ constant torque of $200 \, N-m$ turns a flywheel, which is at rest, about an axis through its centre and perpendicular to its plane. If its moment of inertia is $50 \, kg-m^{2}$, then in $4 \, s$, what will be the change in its angular momentum?

$A$ small mass $m$ is attached to a massless string whose other end is fixed at $P$ as shown in the figure. The mass is undergoing circular motion in the $x-y$ plane with centre at $O$ and constant angular speed $\omega$. If the angular momentum of the system,calculated about $O$ and $P$ are denoted by $\vec{L}_O$ and $\vec{L}_P$ respectively,then

$A$ particle is moving in a circular path of radius $a$ with a constant speed $v$ as shown in the figure. The centre of the circle is marked by $C$. The angular momentum about the origin $O$ can be written as:

Two particles,each of mass $m$ and speed $v$,travel in opposite directions along parallel lines separated by a distance $d$. Show that the angular momentum vector of the two-particle system is the same whatever be the point about which the angular momentum is taken.

Explain the Cartesian components of the angular momentum of a particle.