$A$ solid sphere spinning about a horizontal axis with an angular velocity $\omega$ is placed on a horizontal surface. Subsequently,it rolls without slipping with an angular velocity of

The speed of rolling of a ring of mass $M$ changes from $V$ to $3\ V$. What is the change in its kinetic energy (in $,MV^2$)?

$A$ solid sphere is in rolling motion. In rolling motion,a body possesses translational kinetic energy $(K_t)$ as well as rotational kinetic energy $(K_r)$ simultaneously. The ratio $K_t : (K_t + K_r)$ for the sphere is

$A$ small roller of diameter $20 \ cm$ has an axle of diameter $10 \ cm$. It is on a horizontal floor and a meter scale is positioned horizontally on its axle with one edge of the scale on top of the axle. The scale is now pushed slowly on the axle so that it moves without slipping on the axle,and the roller starts rolling without slipping. After the roller has moved $50 \ cm$,the position of the scale will look like (figures are schematic and not drawn to scale)-

$A$ 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:

$A$ 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: