$A$ ring of mass $m$ and radius $r$ is placed on a rough surface with an initial angular velocity $\omega_0$. The initial velocity of the center of the ring is zero. What will be the velocity of its center when the ring stops slipping?

$A$ 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 an axis perpendicular to the rod and passing through the midpoint of the rod (see figure). There are no external torques. When the beads reach the ends of the rod,the angular velocity of the system is

$A$ horizontal platform is rotating with uniform angular velocity around the vertical axis passing through its centre. At some instant of time, a viscous fluid of mass $m$ is dropped at the centre and is allowed to spread out and finally fall off. The angular velocity during this period

Two discs of same moment of inertia $I$ are rotating about their central axes perpendicular to their planes with angular velocities $\omega_1$ and $\omega_2$. They are brought into contact face to face,such that their axes of rotation coincide. The expression for the loss of energy during this process is:

$A$ swimmer can easily perform somersaults in the air while jumping from a height if he:

$A$ solid cylinder of mass $2 \ kg$ and radius $0.2 \ m$ is rotating with an angular velocity of $3 \ rad/s$. $A$ particle of mass $0.5 \ kg$ moving with a velocity of $5 \ m/s$ strikes its periphery tangentially and sticks to it. Find the angular velocity of the system after the collision in $rad/s$.