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 $,MV^2$)?

If a solid sphere of mass $1\, kg$ and radius $0.1\, m$ rolls without slipping at a uniform velocity of $1\, m/s$ along a straight line on a horizontal floor,the kinetic energy is (in $, J$)

$A$ string is wound around a hollow cylinder of mass $5\, kg$ and radius $0.5\, m$. If the string is now pulled with a horizontal force of $40\, N$,and the cylinder is rolling without slipping on a horizontal surface (see figure),then the angular acceleration of the cylinder will be ......... $rad/s^2$. (Neglect the mass and thickness of the string)

The ratio of kinetic energies of two spheres rolling with equal centre of mass velocities is $2 : 1$. If their radii are in the ratio $2 : 1$,then the ratio of their masses will be

Consider a point $P$ on the circumference of a disc of radius $R$ rolling along a horizontal surface. The distance traveled by point $P$ in one full rotation of the disc is:

Find the displacement of the point of contact with the ground when a ring of radius $R$ completes a half-rotation.