$A$ thin circular ring of mass $M$ and radius $r$ is rotating with a constant angular velocity $\omega$. Two particles,each of mass $m$,are attached to the opposite ends of a diameter of the ring. The new angular velocity of the ring will be:

The position vector of a particle is given by $\vec{r} = 2\hat{i} - 6\hat{j} - 12\hat{k}$ units. $A$ force $\vec{F} = p\hat{i} + 3\hat{j} + 6\hat{k}$ units acts on it. For what value of $p$ is the angular momentum conserved?

Two small beads are initially at $A$ on an equilateral triangle made of uniform wires. The triangle is rotated about the vertical axis $AO$. The beads are released from rest simultaneously and slide down along $AB$ and $AC$ respectively (as shown in the figure). Which quantities are conserved as the beads slide down? (Friction is negligible)

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

$A$ uniform disc of radius $R$ is placed on a smooth horizontal surface. $A$ particle of mass $m$ is moving with velocity $v$ on the surface and hits the disc at the periphery and sticks to it. The angular velocity of the system after the collision is:

$(a)$ $A$ 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?