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

$A$ 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?

$A$ particle moves along a circular path with decreasing speed. Hence,

$A$ small particle of mass $m$ is projected at an angle $\theta$ with the $x$-axis with an initial velocity $v_0$ in the $x-y$ plane as shown in the figure. At a time $t < \frac{v_0 \sin \theta}{g}$,the angular momentum of the particle is

$A$ uniform rod of mass $m$ and length $l$ is moving with velocity $u$ in a direction perpendicular to its length. $A$ blow of impulse $J$ is applied in the direction of $u$,perpendicular to its length at a distance $l/4$ from its center at point $P$,such that the instantaneous velocity of point $P$ is $2u$. Then the velocity of its center of mass will be:

$A$ 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 length of the string is now halved, keeping the angular velocity the same, then the angular momentum will be: