The angular momentum of a moving body remains constant, if

$A$ thin circular ring of mass $M$ and radius $r$ is rotating about its axis with a constant angular velocity $\omega$. Two objects,each of mass $m$,are attached gently to the opposite ends of a diameter of the ring. The ring will now rotate with an angular velocity:

$A$ particle of mass $1 \ kg$ is subjected to a force which depends on the position as $\vec{F} = -k(x \hat{i} + y \hat{j}) \ N$ with $k = 1 \ kg \ s^{-2}$. At time $t = 0$,the particle's position is $\vec{r} = (\frac{1}{\sqrt{2}} \hat{i} + \sqrt{2} \hat{j}) \ m$ and its velocity is $\vec{v} = (-\sqrt{2} \hat{i} + \sqrt{2} \hat{j} + \frac{2}{\pi} \hat{k}) \ m \ s^{-1}$. Let $v_x$ and $v_y$ denote the $x$ and $y$ components of the particle's velocity,respectively. Ignore gravity. When $z = 0.5 \ m$,the value of $(x v_y - y v_x)$ is . . . . . $m^2 \ s^{-1}$.

$A$ child is standing at the center of a rotating platform with hands folded. The kinetic energy of the system is $K$. The child now stretches his hands,doubling the moment of inertia. The kinetic energy of the system will now become ........

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

$A$ thin circular disc of mass $M$ and radius $R$ is rotating in a horizontal plane with an angular velocity $\omega$ about an axis passing through its center and perpendicular to its plane. If another disc of the same radius but one-fourth of its mass is placed gently on top of it,what will be the angular velocity of the new system?