$A$ cube of side '$a$' moves with a velocity '$v$' on a horizontal surface as shown in the figure. It hits an edge at point '$O$'. What will be the angular speed of the block after it hits point '$O$'?

If the torque applied on a particle is zero,then its angular momentum will be:

$A$ particle is moving in a circular orbit of radius $r_1$ with an angular velocity $\omega_1$. It jumps to another circular orbit of radius $r_2$ and attains an angular velocity $\omega_2$. If $r_2 = 0.5\, r_1$ and assuming that no external torque is applied to the system,then the angular velocity $\omega_2$ is given by:

The total angular momentum of all particles of a system is conserved if:

$A$ smooth uniform rod of length $L$ and mass $M$ has two identical beads of negligible size,each of mass $m$,which can slide freely along the rod. Initially,the two beads are at the center of the rod and the system is rotating with angular velocity $\omega_0$ about its axis perpendicular to the rod and passing through its mid-point (see figure). There are no external torques. When the beads reach the ends of the rod,the angular velocity of the system is:

$A$ cubical block of side $a$ is moving with velocity $v$ on a horizontal smooth plane as shown. It hits a ridge at point $O$. The angular speed of the block after it hits $O$ is