$A$ flywheel has a moment of inertia of $4 \ kg \cdot m^2$ and a kinetic energy of $200 \ J$. Calculate the number of revolutions it makes before coming to rest if a constant opposing couple of $5 \ N \cdot m$ is applied to the flywheel.

The $M.I.$ of a body about a given axis is $1.2 \ kg \cdot m^2$ and initially the body is at rest. In order to produce a rotational kinetic energy of $1500 \ J$,an angular acceleration of $25 \ rad/s^2$ must be applied about that axis for a time duration of $....... \ s$.

The power $(P)$ is supplied to a rotating body having moment of inertia '$I$' and angular acceleration '$\alpha$'. Its instantaneous angular velocity is

$A$ tangential force $F$ is applied on a disc of radius $R$,due to which it deflects through an angle $\theta$ from its initial position. The work done by this force would be

Three objects,$A$ (a solid sphere),$B$ (a thin circular disk),and $C$ (a circular ring),each have the same mass $M$ and radius $R$. They all spin with the same angular speed $\omega$ about their own symmetry axes. The amounts of work $(W)$ required to bring them to rest would satisfy the relation:

$A$ uniform solid cylinder of mass $M$ and radius $R$ can freely rotate around its axis $O$. There is an elastic string of relaxed length $L$ and stiffness $K$ attached to the cylinder and a static wall. Initially,the string is relaxed. As the cylinder starts rotating,the string will wind around the cylinder. The surface of the cylinder is very rough,so the string does not slip on the cylinder's surface. The minimum initial angular speed of the cylinder,${\omega _0}$,so that it can rotate through an angle $2\pi$ is (Assume Hooke's law to be valid.)