$A$ flywheel is rotating at a rate of $150 \text{ rev/minute}$. If it slows at a constant retardation of $\pi \text{ rad/s}^2$, then the time required for the wheel to come to rest is (in $\text{ s}$)

$A$ same torque is applied to a disc and a ring of equal mass and radii. Then:

$A$ rectangular solid rod of length $0.3\, m$ is held horizontally,with one of its sides on the edge of a platform of height $5\, m$. When released,it slips off the table in a very short time $\Delta t = 0.01\, s$,remaining essentially horizontal. The angle by which it would rotate when it hits the ground will be (in radians) close to:

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

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