$A$ circular disc rotating with frequency $f_0 = 1.3 \, rev/sec$ comes to a stop in $30 \, seconds$. The approximate angular acceleration is ....... $rad/sec^2$.

$A$ circular disc of radius $2 \ m$ is rotating at $240 \ rpm$. If a torque is applied to the disc, it slows down at a rate of $\pi \ rad/s^2$. How long will it take to stop? How many revolutions will the disc complete before coming to rest?

Torques of equal magnitude are applied to a hollow cylinder and a solid sphere,both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry,and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time?

Two circular discs of radius $10 \ \text{cm}$ each are joined at their centres by a rod of length $30 \ \text{cm}$ and mass $600 \ \text{g}$ as shown in the figure. If the mass of each disc is $600 \ \text{g}$ and the applied torque between the two discs is $43 \times 10^{5} \ \text{dyne cm}$, the angular acceleration of the discs about the given axis $AB$ is . . . . . . $\text{rad/s}^{2}$.

$A$ wheel is rotating about an axis through its centre at $720 \, rpm$. It is acted on by a constant torque opposing its motion for $8 \, s$ to bring it to rest finally. The value of torque (in $N \cdot m$) is (Given $I = \frac{24}{\pi} \, kg \cdot m^2$)

$A$ uniform rod of length $l$,hinged at the lower end,is free to rotate in the vertical plane. If the rod is held vertically in the beginning and then released,what is the angular acceleration of the rod when it makes an angle of $45^{\circ}$ with the horizontal? (Given: Moment of inertia $I = ml^2/3$)