$A$ solid sphere is rolling on a horizontal plane without slipping. If the ratio of angular momentum about the axis of rotation of the sphere to the total energy of the moving sphere is $\pi: 22$,then the value of its angular speed will be $...........\,rad/s$.

$A$ circular hoop of radius $50 \ cm$ and mass $1 \ kg$ rotating with an angular velocity $\omega_0$ is placed on a rough horizontal surface. The initial velocity of the centre of the hoop is zero. Let $v$ be the velocity of the centre of the hoop when it ceases to slip. The ratio $v / \omega_0$ will be (in $cm$)

$A$ solid sphere of mass $1\,kg$ rolls without slipping on a plane surface. Its kinetic energy is $7 \times 10^{-3}\,J$. The speed of the centre of mass of the sphere is $.........\,cm\,s^{-1}$.

$A$ force $F$ is applied at the centre of a disc of mass $M$. The minimum value of the coefficient of friction of the surface for pure rolling is:

$A$ sphere rolls without slipping. The radius of gyration about an axis passing through its center of mass is $K$. If the radius of the sphere is $R$,what fraction of the total energy is in the form of rotational kinetic energy?

$A$ uniform disc of mass $0.5\,kg$ and radius $r$ is projected with velocity $18\,m/s$ at $t = 0\,s$ on a rough horizontal surface. It starts off with a purely sliding motion at $t = 0\,s$. After $2\,s$ it acquires a purely rolling motion (see figure). The total kinetic energy of the disc after $2\,s$ will be $..............J$ (given,coefficient of friction is $0.3$ and $g = 10\,m/s^2$).