$A$ 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:

$A$ cylinder is pure rolling up an incline plane. It stops momentarily and then rolls back. The force of friction

$A$ solid sphere of mass $m$ and radius $r$ is allowed to roll without slipping from the highest point of an inclined plane of length $L$ that makes an angle $30^{\circ}$ with the horizontal. The speed of the sphere at the bottom of the plane is $v_1$. If the angle of inclination is increased to $45^{\circ}$ while keeping $L$ constant,the new speed of the sphere at the bottom of the plane is $v_2$. The ratio of $v_1^2 : v_2^2$ is

$A$ 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?

$A$ solid sphere and a solid cylinder of identical radii approach an incline with the same linear velocity (see figure). Both roll without slipping throughout the motion. The two climb maximum heights $h_{sph}$ and $h_{cyl}$ on the incline. The ratio $\frac{h_{sph}}{h_{cyl}}$ is given by

$A$ body is rolling down an inclined plane. If the kinetic energy of rotation is $40\%$ of the translational kinetic energy,then the body is a