$A$ thin metal disc of radius $0.25\,m$ and mass $2\,kg$ starts from rest and rolls down an inclined plane. If its rotational kinetic energy is $4\,J$ at the foot of the inclined plane,then the linear velocity (in $m/s$) at the same point is:

$A$ solid cylinder rolls down an inclined plane from a height $h$. At any moment,the ratio of rotational kinetic energy to the total kinetic energy is:

$A$ ring of radius $4a$ is rigidly fixed in a vertical position on a table. $A$ small disc of mass $m$ and radius $a$ is released as shown in the figure. When the disc rolls down,without slipping,to the lowest point of the ring,then its speed will be

$A$ cylinder of mass $m$ and radius $R$ rolls without slipping down an inclined plane of length $L$ and height $h$. What will be the velocity of its center of mass when the cylinder reaches the bottom?

$A$ solid cylinder and a hollow cylinder,both of the same mass and same external diameter,are released from the same height at the same time on an inclined plane. Both roll down without slipping. Which one will reach the bottom first?

Prove that the velocity $v$ of translation of a rolling body (like a ring,disc,cylinder,or sphere) at the bottom of an inclined plane of height $h$ is given by $v^{2} = \frac{2gh}{1 + k^{2}/R^{2}}$ using dynamical considerations (i.e.,by considering forces and torques). Note: $k$ is the radius of gyration of the body about its symmetry axis,and $R$ is the radius of the body. The body starts from rest at the top of the plane.