$A$ rod of mass $M$ and length $L$ is placed in a horizontal plane with one end hinged about a vertical axis. $A$ horizontal force of $F = \frac{Mg}{2}$ is applied at a distance $\frac{5L}{6}$ from the hinged end. The angular acceleration of the rod will be:

$A$ string is wrapped around the rim of a wheel of moment of inertia $0.40 \ kg \cdot m^2$ and radius $10 \ cm$. The wheel is free to rotate about its axis. Initially,the wheel is at rest. The string is now pulled by a force of $40 \ N$. The angular velocity of the wheel after $10 \ s$ is $x \ rad/s$,where $x$ is $\qquad$

$A$ wheel with a moment of inertia of $5 \times 10^{-3} \ kg \cdot m^2$ is rotating at a rate of $20 \ rev/s$. The torque required to bring it to rest in $10 \ s$ is:

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

$A$ wheel is rotating about an axis through its centre at $720 \, rpm$. It is acted on by a constant torque opposing its motion for $8 \, s$ to bring it to rest finally. The value of torque (in $N \cdot m$) is (Given $I = \frac{24}{\pi} \, kg \cdot m^2$)

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 $; s$)