$A$ uniform rod of length $l$ and mass $m$ is pivoted at point $A$. The rod is released from a horizontal position. If the moment of inertia of the rod about point $A$ is $ml^2/3$,then its initial angular acceleration is:

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 ........ $rad/s$.

The moment of inertia of a body about a given axis is $1.2 \; kg \cdot m^2$. Initially,the body is at rest. In order to produce a rotational kinetic energy of $1500 \; J$,an angular acceleration of $25 \; rad/s^2$ must be applied about that axis for a duration of: (in $; s$)

$A$ 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:

In the following figure,$r_1 = 5 \, cm$ and $r_2 = 30 \, cm$. If the moment of inertia of the wheel is $5100 \, kg \cdot m^2$,then its angular acceleration will be:

$A$ wheel of radius $0.4 \,m$ can rotate freely about its axis as shown in the figure. $A$ string is wrapped over its rim and a mass of $4 \,kg$ is hung. An angular acceleration of $8 \,rad \,s^{-2}$ is produced in it due to the torque. Then, the moment of inertia of the wheel is $(g = 10 \,m \,s^{-2})$.