The power $(P)$ is supplied to a rotating body having moment of inertia $I$ and angular acceleration $\alpha$. Its instantaneous angular velocity $\omega$ is

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

The angular velocity of a body changes from $6 \ rad \ s^{-1}$ to $21 \ rad \ s^{-1}$ in a time of $1.5 \ s$. If the moment of inertia of the body is $100 \ g \ m^2$,then the rate of change of angular momentum of the body is (in $N \ m$)

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

$A$ constant torque acting on a uniform circular wheel changes its angular momentum from $A_0$ to $4 \,A_0$ in $4 \,s$. The magnitude of the torque is

$A$ rod $PQ$ of mass $M$ and length $L$ is hinged at end $P$. The rod is kept horizontal by a massless string tied to point $Q$ as shown in the figure. When the string is cut,the initial angular acceleration of the rod is