The ratio of the total kinetic energy to the rotational kinetic energy for a rolling disc is:

$A$ thin uniform circular disc rolls with a constant velocity without slipping on a horizontal surface. Its total kinetic energy is

$A$ solid homogeneous sphere is rolling with sliding on a horizontal rough surface. For this type of motion of the sphere:

$A$ circular ring of mass $10 \text{ kg}$ rolls along a horizontal floor. The center of mass of the ring has a speed of $1.5 \text{ m/s}$. The work required to stop the ring is: (in $\text{ J}$)

$A$ sphere cannot roll on a ..........

$A$ circular hoop of radius $50 \ cm$ and mass $1 \ kg$ rotating with an angular velocity $\omega_0$ is placed on a rough horizontal surface. The initial velocity of the centre of the hoop is zero. Let $v$ be the velocity of the centre of the hoop when it ceases to slip. The ratio $v / \omega_0$ will be (in $cm$)