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

A disc of mass $1\,kg$ and radius $R$ is free of rotate about a horizontal axis passing through its centre and perpendicular to the plane of disc. A body of same mass as that of 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 $4 \sqrt{\frac{x}{3 R}}$ rad s$^{-1}$ where $x=$

$\left( g =10\,ms ^{-2}\right)$

The moment of inertia of a body about a given axis is $1.2 \;kg m^{2}$. Initially, the body is at rest. In order to produce a rotational kinetic energy of $1500\; joule$, an angular acceleration of $25 \;rad s^{-2}$ must be applied about that axis for a duration of

A rod of length $50\,cm$ is pivoted at one end. It is raised such that if makes an angle of $30^o$ fro the horizontal as shown and released from rest. Its angular speed when it passes through the horizontal (in $rad\,s^{-1}$ ) will be $(g = 10\,ms^{-2})$

$A$ thin rod $AB$ is sliding between two fixed right angled surfaces. At some instant its angular velocity is $ \omega $. If $I_x$ represent moment of inertia of the rod about an axis perpendicular to the plane and passing through the point $X$ ($A, B, C$ or $D$), the kinetic energy of the rod is