$A$ 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:

$A$ 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 is rolling (without slipping) on a horizontal surface. $C$ is its centre and $Q$ and $P$ are two points on the same horizontal line passing through $C$,such that $Q$ is at a distance $r$ from $C$ and $P$ is at a distance $r$ from $C$ on the opposite side. Let $V_P, V_Q$ and $V_C$ be the magnitudes of velocities of points $P, Q$ and $C$ respectively,then:

When a body is rolling without slipping on a rough horizontal surface,the work done by friction is ........

$A$ solid uniform sphere resting on a rough horizontal plane is given a horizontal impulse directed through its centre so that it starts sliding with an initial velocity $v_{0}$. When it finally starts rolling without slipping, the speed of its centre is

$A$ circular disc of mass $0.41 \ kg$ and radius $10 \ m$ rolls without slipping with a velocity of $2 \ m/s$. The total kinetic energy of the disc is ....... $J$.