$A$ body changes its angular speed from $\omega_1$ to $\omega_2$ without any external torque applied. This change is due to a change in its moment of inertia. What is the ratio of the radii of gyration in the two cases?

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

$A$ ring of mass $m$ and radius $r$ is placed on a rough surface with an initial angular velocity $\omega_0$. The initial velocity of the center of the ring is zero. What will be the velocity of its center when the ring stops slipping?

$A$ disc of mass $1 \, kg$ and radius $0.1 \, m$ is rotating with angular velocity $20 \, rad/s$. What is the angular velocity (in $rad/s$) if a mass of $0.5 \, kg$ is placed on the circumference of the disc?

$A$ circular platform of moment of inertia $100 \ kg\ m^2$ with a boy of mass $20 \ kg$ standing at the centre completes one rotation in $10 \ s$. If he walks $2 \ m$ away from the centre along the radius,then one rotation (with respect to the ground) will be completed in $.... \ s$.