$A$ constant torque of $1000 \; Nm$ turns a wheel of moment of inertia $200 \; kg \cdot m^2$ about an axis through its centre. The wheel is at rest initially. Its angular velocity after $3 \; s$ 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:

Write the Newton's second law for rotational motion about a fixed axis.

$A$ 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:

$A$ 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

The moment of inertia of a wheel about its transverse axis is $2 \ kg \cdot m^2$ and it is rotating at $60 \ rpm$ about that axis. The torque required to stop the wheel in one minute is: