$A$ body rolls down an inclined plane without slipping. The kinetic energy of rotation is $50 \%$ of its translational kinetic energy. The body is :

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.

$A$ sphere of mass $m$ and radius $r$ rolls on a horizontal plane without slipping with the speed $u$. Now if it rolls up an incline, the maximum height it would attain will be

$A$ solid cylinder is suspended symmetrically by two massless strings,as shown in the figure. The distance the cylinder should fall by unbinding the strings to achieve a speed of $4\,ms^{-1}$ is $........cm$. (Take $g=10\,ms^{-2}$)

$A$ solid cylinder of mass $M$ and radius $R$ rolls down an inclined plane of height $h$. When it reaches the foot of the plane,its rotational kinetic energy is ($g=$ acceleration due to gravity).

$A$ uniform solid cylinder of mass $m$ and radius $r$ rolls along an inclined rough plane of inclination $45^{\circ}$. If it starts to roll from rest from the top of the plane,then the linear acceleration of the cylinder axis will be: