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.

A particle of mass $m$ is projected with a velocity $v$ making an angle of $30^{\circ}$ with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height $h$ is

$A$ particle of mass $m$ is rotating in a plane is $a$ circular path of radius $r$, its angular momentum is $L$. The centripital force acting on the particle is :

Two rigid bodies $A$ and $B$ rotate with rotational kinetic energies $E_A$ and $E_B$ respectively.  The moments of inertia of $A$ and $B$ about the axis of rotation are $I_A$ and $I_B$ respectively. If $I_A = I_B/4 \,$and$ \, E_A = 100\ E_B$ the ratio of angular momentum $(L_A)$ of $A$ to the angular momentum $(L_B)$ of $B$ is

A thin rod of mass $M$ and length $a$ is free to rotate in horizontal plane about a fixed vertical axis passing through point $O$. A thin circular disc of mass $M$ and of radius $a / 4$ is pivoted on this rod with its center at a distance $a / 4$ from the free end so that it can rotate freely about its vertical axis, as shown in the figure. Assume that both the rod and the disc have uniform density and they remain horizontal during the motion. An outside stationary observer finds the rod rotating with an angular velocity $\Omega$ and the disc rotating about its vertical axis with angular velocity $4 \Omega$. The total angular momentum of the system about the point $O$ is $\left(\frac{ M a^2 \Omega}{48}\right) n$. The value of $n$ is. . . . .