$A$ wheel having a moment of inertia of $2 \, kg \cdot m^2$ about its vertical axis rotates at the rate of $60 \, rpm$ about the axis. The torque required to stop the wheel's rotation in one minute would be

$A$ wheel rotates about its geometric axis at a speed of $60 \ rpm$. If the moment of inertia of the wheel about this axis is $2 \ kg \ m^2$,what torque is required to stop its rotation in one minute?

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 . . . . . . $\text{rad/s}^{2}$.

$A$ thin uniform rod of mass $m$ and length $L$ is pivoted at one end so that it can rotate in a vertical plane. The free end is held vertically above the pivot and then released. The angular acceleration of the rod when it makes an angle $\theta$ with the vertical is [consider negligible friction at the pivot] ($g=$ acceleration due to gravity).

The instantaneous angular position of a point on a rotating wheel is given by the equation $\theta(t) = 2t^3 - 6t^2$. The torque on the wheel becomes zero at $t = $ ...... $s$.

$A$ string is wound around the rim of a mounted flywheel (disc) of mass $20 \, kg$ and radius $20 \, cm$. $A$ steady pull of $25 \, N$ is applied on the cord. Neglecting friction and the mass of the string,the angular acceleration of the wheel in $rad/s^2$ is