The angular momentum of a moving body remains constant, if

$(a)$ $A$ child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of $40\; rev/min$. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to $2/5$ times the initial value? Assume that the turntable rotates without friction.
$(b)$ Show that the child's new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy?

$A$ thin smooth rod of length $L$ and mass $M$ is rotating freely with angular speed $\omega_0$ about an axis perpendicular to the rod and passing through its centre. Two beads of mass $m$ and negligible size are at the centre of the rod initially. The beads are free to slide along the rod. The angular speed of the system,when the beads reach the opposite ends of the rod,will be:

$A$ circular ring of mass $m$ and radius $R$ is rotating about its axis with a constant angular velocity $\omega$. Two particles of mass $M$ are gently attached to the ring at the ends of a diameter. The ring will now rotate with an angular velocity $\omega'$ equal to:

$A$ particle of mass $m$ moves with a velocity $v_0$ in a circle of radius $R_0$ on a smooth horizontal plane. The mass is tied to a string passing through a hole in the smooth plane as shown in the figure. The tension in the string is gradually increased,and finally,the mass $m$ moves in a circle of radius $\frac{R_0}{2}$. What is the ratio of the final kinetic energy to the initial kinetic energy?

$A$ thin uniform circular disc of mass $M$ and radius $R$ is rotating with angular velocity $\omega$ in a horizontal plane about an axis passing through its centre and perpendicular to its plane. Another disc of the same radius but of mass $\frac{M}{3}$ is placed gently on the first disc co-axially. The new angular velocity will be: