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

$A$ man standing on a turn-table is rotating at a certain angular frequency with his arms outstretched. He suddenly folds his arms. If his moment of inertia with folded arms is $75 \%$ of that with outstretched arms,then his rotational kinetic energy will

$A$ horizontal circular platform of radius $R = 0.5 \ m$ and mass $M = 0.45 \ kg$ is free to rotate about its axis. Two massless spring toy-guns,each carrying a steel ball of mass $m = 0.05 \ kg$,are attached to the platform at a distance $r = 0.25 \ m$ from the centre on its either sides along its diameter (see figure). Each gun simultaneously fires the balls horizontally and perpendicular to the diameter in opposite directions. After leaving the platform,the balls have a horizontal speed of $v = 9 \ m/s$ with respect to the ground. The rotational speed of the platform in $rad/s$ after the balls leave the platform is:

$A$ horizontal disc of moment of inertia $4.25 \,kg \cdot m^2$ with respect to its axis of symmetry is spinning counter-clockwise at $15 \,rps$ about its axis,as viewed from above. $A$ second disc of moment of inertia $1.80 \,kg \cdot m^2$ with respect to its axis of symmetry is spinning clockwise at $25 \,rps$ as viewed from above about the same axis and is dropped on top of the first disc. The two discs stick together and rotate as one about their axis of symmetry. The new angular velocity of the system as viewed from above is close to

$A$ bullet of mass $10 \; g$ and speed $500 \; m/s$ is fired into a door and gets embedded exactly at the centre of the door. The door is $1.0 \; m$ wide and weighs $12 \; kg$. It is hinged at one end and rotates about a vertical axis practically without friction. Find the angular speed of the door just after the bullet embeds into it.

When a person dancing on ice pulls their arms inward,they start spinning faster. This is due to: