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:

Write the formula for power and angular momentum in rotational motion.

$A$ light rod carries three equal masses $A$,$B$,and $C$ as shown in the figure. If the rod is released from the horizontal position,what will be the velocity of mass $B$ when the rod reaches the vertical position?

$A$ disc of mass $1\,kg$ and radius $R$ is free to rotate about a horizontal axis passing through its centre and perpendicular to the plane of the disc. $A$ body of the same mass as that of the disc is fixed at the highest point of the disc. Now the system is released. When the body comes to the lowest position,its angular speed will be $4 \sqrt{\frac{x}{3 R}} \text{ rad s}^{-1}$ where $x=$ (Given $g = 10 \text{ m s}^{-2}$)

Explain the work done by torque.

$A$ 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.)