An ant is sitting at the edge of a rotating disc. If the ant reaches the other end, after moving along the diameter, the angular velocity of the disc will

$A$ circular platform is mounted on a frictionless vertical axle. Its radius $R = 2\, m$ and its moment of inertia about the axle is $200\, kg\, m^2$. It is initially at rest. $A$ $50\, kg$ man stands on the edge of the platform and begins to walk along the edge at the speed of $1\, m/s$ relative to the ground. Time taken by the man to complete one revolution is

If the radius of a rotating disc is suddenly halved while its mass remains the same,its angular velocity will become:

$A$ circular disk of moment of inertia $I_t$ is rotating in a horizontal plane,about its symmetry axis,with a constant angular speed $\omega_i$. Another disk of moment of inertia $I_b$ is dropped coaxially onto the rotating disk. Initially,the second disk has zero angular speed. Eventually,both the disks rotate with a constant angular speed $\omega_f$. The energy lost by the initially rotating disk to friction is

$A$ 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 gently placed at opposite ends of a diameter of the ring. The new angular velocity of the ring will be

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