$A$ solid spherical ball is rolling on a frictionless horizontal plane surface about its axis of symmetry. The ratio of rotational kinetic energy of the ball to its total kinetic energy is ................

$A$ uniform solid cylinder of mass $M = 3 \ kg$ and radius $R = 10 \ cm$ is connected about an axis through the centre of the cylinder to a horizontal spring with spring constant $k = 8 \ N/m$. The cylinder is pulled back,stretching the spring $x = 1 \ m$ from equilibrium. When released,the cylinder rolls without slipping. What is the speed of the center of the cylinder when it returns to equilibrium? .................. $m/s$

$A$ solid sphere is moving on a horizontal plane. The ratio of its translational kinetic energy to its rotational kinetic energy is:

$A$ sphere is rolling without slipping on a fixed horizontal plane surface. $A$ is the point of contact,$B$ is the centre of the sphere,and $C$ is the topmost point. Then:

$A$ wheel is rolling straight on ground without slipping. If the axis of the wheel has speed $v$,the instantaneous velocity of a point $P$ on the rim,defined by angle $\theta$ relative to the vertical,relative to the ground will be:

$A$ 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