$A$ disc of mass $3 \, kg$ rolls down an inclined plane of height $5 \, m$. The translational kinetic energy of the disc on reaching the bottom of the inclined plane is .......... $J$. (Take $g = 10 \, m/s^2$)

$A$ ball is rolling without slipping. The radius of gyration of the ball about an axis passing through its center of mass is $K$. If the radius of the ball is $R$,what fraction of the total energy is rotational kinetic energy?

$A$ uniform disk of mass $m$ and radius $R$ rolls without slipping down an inclined plane of length $l$ and inclination $\theta$. Initially,the disk was at rest at the top of the inclined plane. Its angular momentum about the point of contact with the inclined plane when it reaches the bottom will be equal to:

$A$ solid sphere of mass $1 kg$ and radius $1 m$ rolls without slipping on a fixed inclined plane with an angle of inclination $\theta = 30^{\circ}$ from the horizontal. Two forces of magnitude $1 N$ each,parallel to the incline,act on the sphere,both at a distance $r = 0.5 m$ from the center of the sphere,as shown in the figure. The acceleration of the sphere down the plane is . . . $m s^{-2}$. (Take $g = 10 m s^{-2}$.)

$A$ solid cylinder is released from rest from the top of an inclined plane of inclination $30^{\circ}$ and length $60 \,cm$. If the cylinder rolls without slipping, then the speed when it reaches the bottom is (in $\,m/s$)

$A$ ring takes times $t_1$ and $t_2$ to reach the bottom of an inclined plane of length $L$ by sliding and rolling,respectively. What is the ratio of $t_1$ to $t_2$?