$A$ 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:

If the value of torque is zero,then

$A$ thin circular ring of mass $M$ and radius $R$ is rotating about its axis with a constant angular velocity $\omega$. Four objects,each of mass $m$,are kept gently on the opposite ends of two perpendicular diameters of the ring. The new angular velocity of the ring will be:

$A$ 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 ........

$A$ 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

$A$ 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 . . . . . $m^2 \ s^{-1}$.