$A$ 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 $MR^2$)?

$A$ disc with a moment of inertia of $\frac{9.8}{\pi^2} \ kg \ m^2$ is rotating at $600 \ rpm$. If the rotational frequency changes from $600 \ rpm$ to $300 \ rpm$,the work done is .......... $J$.

When $12000 \ J$ of work is done on a flywheel,its frequency of rotation increases from $10 \ Hz$ to $20 \ Hz$. The moment of inertia of the flywheel about its axis of rotation is (take $\pi^2=10$):

$A$ 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.)

$A$ 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):

The moment of inertia of a body about a given axis is $2.4 \ kg \cdot m^2$. To produce a rotational kinetic energy of $750 \ J$,an angular acceleration of $5 \ rad/s^2$ must be applied about that axis for how many seconds?