An ant is sitting on the edge of a rotating disc. If the ant walks along the diameter to reach the other end, how will the angular velocity of the disc change?

Two uniform circular discs are rotating independently in the same direction around their common axis passing through their centres. The moment of inertia and angular velocity of the first disc are $0.1 \; kg \cdot m^{2}$ and $10 \; rad \cdot s^{-1}$ respectively,while those for the second one are $0.2 \; kg \cdot m^{2}$ and $5 \; rad \cdot s^{-1}$ respectively. At some instant,they get stuck together and start rotating as a single system about their common axis with some angular speed. The kinetic energy of the combined system is ........... $J$.

The angular momentum of a system of particles is not conserved:

$Assertion$ : There are very small sporadic changes in the speed of rotation of the earth.
$Reason$ : Shifting of large air masses in the earth's atmosphere produces a change in the moment of inertia of the earth, causing its speed of rotation to change.

$A$ child with mass $m$ is standing at the edge of a disc with moment of inertia $I$,radius $R$,and initial angular velocity $\omega$. See the figure given below. The child jumps off the edge of the disc with tangential velocity $v$ with respect to the ground. The new angular velocity of the disc is

$A$ rotating table completes one revolution in $10 \ s$. Its moment of inertia is $100 \ kg \cdot m^2$. $A$ man of mass $50 \ kg$ stands at the center of the rotating table. If the man moves $2 \ m$ away from the center,what will be the angular velocity of the table in $rad/s$?