The angular momentum of a wheel changes from $2L$ to $5L$ in $3$ seconds. The magnitude of the torque acting on the wheel 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:

$A$ 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:

$A$ 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 $I = ml^2/3$)

$A$ non-uniform rod $OA$ of linear mass density $\lambda = \lambda_0 x$ (where $\lambda_0$ is a constant) is suspended from the ceiling with a hinge joint at $O$ and a light string as shown in the figure. Find the angular acceleration of the rod just after the string is cut.

$A$ slender uniform rod of mass $M$ and length $l$ is pivoted at one end so that it can rotate in a vertical plane (see figure). There is negligible friction at the pivot. The rod is released from a position where it makes an angle $\theta$ with the vertical. The angular acceleration of the rod when it makes an angle $\theta$ with the vertical is