Consider a point $P$ on the circumference of a disc rolling along a horizontal surface. If $R$ is the radius of the disc,the distance through which $P$ moves in one full rotation of the disc is:

$A$ disc of mass $M$ and radius $R$ is rolling on a horizontal surface with an angular speed $\omega$. The angular momentum of the disc about the origin $O$ is:

In the case of pure rolling,what will be the velocity of point $A$ of the ring of radius $R$?

$A$ disc rolls without slipping with a constant velocity. What fraction of its total kinetic energy is in the form of rotational kinetic energy?

$A$ solid sphere is moving on a horizontal plane. The ratio of its translational kinetic energy to its rotational kinetic energy is:

$A$ thin circular ring and a circular disc of equal mass are rolling without sliding. If their linear velocities are equal and the total kinetic energy of the disc is $6 \ J$,then the total kinetic energy of the ring is (in $J$)