$A$ 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 along a straight line $AB$ in the $XY$ plane. Let $L_A$ and $L_B$ be the angular momentum of the particle about the origin $O$ when it is at points $A$ and $B$ respectively. Then:

$A$ thin rod of mass $M$ and length $L$ is struck at one end by a ball of clay of mass $m$,moving with speed $v$ as shown in the figure. The ball sticks to the rod. After the collision,the angular momentum of the clay-rod system about $A$,the midpoint of the rod,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 $1 \, kg$)?

Two bodies rotate with kinetic energies $E_1$ and $E_2$. Their moments of inertia about their axes of rotation are $I_1$ and $I_2$. If $I_1 = \frac{I_2}{3}$ and $E_1 = 27 E_2$,then the ratio of their angular momenta $L_1$ to $L_2$ is:

The diameter of a flywheel is $1 \,m$. It has a mass of $20 \,kg$. It is rotating about its axis with a speed of $120$ rotations in one minute. Its angular momentum in $kg-m^2/s$ is (in $4$)