$A$ thin uniform rod of length $L$ and mass $M$ is swinging freely about a horizontal axis passing through its end. Its maximum angular speed is $\omega$. Its centre of mass rises to a maximum height of (where $g$ is the acceleration due to gravity):

The angular velocity of a body is $\vec{\omega} = 2\hat{i} + 3\hat{j} + 4\hat{k}$ and a torque $\vec{\tau} = \hat{i} + 2\hat{j} + 3\hat{k}$ acts on it. The rotational power will be .......... $W$.

$A$ 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?

$A$ thin rod $MN$,free to rotate in the vertical plane about the fixed end $N$,is held horizontal. When the end $M$ is released,the speed of this end,when the rod makes an angle $\alpha$ with the horizontal,will be proportional to (see figure):

$A$ 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.)

$A$ circular disc of radius $R$ meter and mass $M$ kg is rotating around the axis perpendicular to the disc. An external torque is applied to the disc such that $\theta(t) = 5t^2 - 8t$,where $\theta(t)$ is the angular position of the rotating disc as a function of time $t$. How much power is delivered by the applied torque,when $t = 2$ s (in $MR^2$)?