$A$ body is in pure rotation. The linear speed $v$ of a particle,the distance $r$ of the particle from the axis,and the angular velocity $\omega$ of the body are related as $\omega = \frac{v}{r}$. Thus,

$A$ homogeneous solid cylindrical roller of radius $R$ and mass $M$ is pulled on a cricket pitch by a horizontal force $F$ applied at its center. Assuming rolling without slipping,the angular acceleration of the cylinder is

$A$ bob of mass $m$ is tied by a massless string whose other end is wound on a flywheel (disc) of radius $R$ and mass $m$. When released from rest,the bob starts falling vertically downwards. If the bob has covered a vertical distance $h$,then the angular speed of the wheel will be (There is no slipping between the string and the wheel,$g$ is the acceleration due to gravity).

$A$ uniform cylinder of mass $m$ can rotate freely about its own axis,which is horizontal. $A$ particle of mass $m_0$ hangs from the end of a light string wound around the cylinder,which does not slip over it. When the system is allowed to move,the acceleration of the descending mass will be

$A$ mass $m$ is supported by a massless string wound around a uniform cylinder of mass $M$ and radius $R$. On releasing the mass from rest,it will fall with an acceleration of:

$A$ block of mass $2\,kg$ hangs from the rim of a wheel of radius $0.5\,m$. On releasing from rest,the block falls through $5\,m$ height in $2\,s$. The moment of inertia of the wheel will be ...... $kg \cdot m^2$.