$A$ torque of $50\, Nm$ acting on a wheel at rest rotates it through $200\, rad$ in $5\, s$. Calculate the angular acceleration produced. ........... $rad\, s^{-2}$.

$A$ 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 $10 \ kg \ m^2$,the number of rotations made by the pulley before its direction of motion is reversed is approximately equal to: (in $.5$)

$A$ disc of radius $0.4 \,m$ and mass $1 \,kg$ rotates about an axis passing through its center and perpendicular to its plane. The angular acceleration is $10 \,rad \,s^{-2}$. The tangential force applied to the rim of the disc is (in $\,N$)

$A$ uniform rod $AB$ of length $l$ and mass $m$ is free to rotate about point $A.$ The rod is released from rest in the horizontal position. Given that the moment of inertia of the rod about $A$ is $ml^2/3$,the initial angular acceleration of the rod will be:

The moment of inertia of a wheel about an axis passing through its center is $200 \, kg \cdot m^2$. $A$ constant torque of $1000 \, N \cdot m$ is applied to rotate the wheel. After $3 \, s$,the angular velocity of the wheel will be ........ $rad/s$.

Two equal and opposite forces are applied tangentially to a uniform disc of mass $M$ and radius $R$ as shown in the figure. If the disc is pivoted at its centre and free to rotate in its plane,the angular acceleration of the disc is ............