$A$ 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?

$A$ disc of radius $1\,m$ and mass $4\,kg$ rolls on a horizontal plane without slipping in such a way that its centre of mass moves with a speed of $10\,cm/s$. Its rotational kinetic energy is (in $,J$)

$Assertion$: $A$ rigid disc rolls without slipping on a fixed rough horizontal surface with uniform angular velocity. Then the acceleration of the lowest point on the disc is zero.
$Reason$: For a rigid disc rolling without slipping on a fixed rough horizontal surface,the velocity of the lowest point on the disc is always zero.

$A$ solid sphere of mass $2 \ kg$ is rolling on a frictionless horizontal surface with velocity $6 \ m \ s^{-1}$. It collides with the free end of an ideal spring whose other end is fixed. The maximum compression produced in the spring will be (Force constant of the spring $= 36 \ N \ m^{-1}$). (in $m$)

$A$ thin hollow cylinder,open at both ends,slides without rotating and then rolls without slipping with the same speed. The ratio of kinetic energies in the two cases will be:

$A$ spool consists of a cylinder of radius $R_1$ wrapped with thread,and its end caps have radius $R_2$ as shown in the end view illustrated in the figure. The mass of the spool,including the thread,is $m$,and its moment of inertia about an axis through its center is $I$. The spool is placed on a rough horizontal surface so that it rolls without slipping. When a force $\vec{T}$ acting to the right is applied to the free end of the thread,the friction force exerted by the surface on the spool is given by: