$A$ particle of mass $m = 5$ is moving with a uniform speed $v = 3\sqrt{2}$ in the $XOY$ plane along the line $Y = X + 4$. The magnitude of the angular momentum of the particle about the origin is .......

Angular momentum is a:

$A$ 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

$A$ particle of mass $5 \ kg$ moves along the straight line $y = 2x + 4$ with a velocity of $3\sqrt{5} \ m/s$. Find its angular momentum about the origin in $kg \ m^2 \ s^{-1}$.

$A$ 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

$A$ mass tied to a string is whirled in a horizontal circular path with a constant angular velocity and its angular momentum is $L$. If the string is now halved,keeping the angular velocity the same,then the new angular momentum will be: