The moment of inertia of a body about a given axis is $2.4 \, kg \cdot m^2$. For what duration of time must an angular acceleration of $5 \, rad/s^2$ be applied about this axis to generate a rotational kinetic energy of $750 \, J$? (in seconds)

If $I = 50\,kg-m^2$,then how much torque (in $N-m$) must be applied to stop it in $10\,s$,given its initial angular speed is $20\,rad/s$?

$A$ 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:

If a disc of mass $M$ and radius $R$ rotates with an angular acceleration $a$,the torque acting on the disc is

$A$ pulley of radius $2 \ m$ is rotated about its axis by a force $F = (20t - 5t^2) \ N$ (where $t$ is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is $10 \ kg \cdot m^2$,the number of rotations made by the pulley before its direction of motion is reversed is:

$A$ pulley has the shape of a uniform solid disc of mass $2 \ kg$ and radius $0.5 \ m$. $A$ string is wrapped over its rim and is pulled by a force of $2.5 \ N$. The pulley is free to rotate about its axis. Initially,the pulley is at rest. Find the angular velocity of the pulley just after $10 \ s$. (in $rad/s$)