$A$ wheel having a moment of inertia of $5 \times 10^{-3} \ kg \ m^2$ is rotating at a rate of $20 \ rev/s$. The torque required to stop the wheel in $10 \ s$ is $... \times 10^{-2} \ N \ m$. (in $\pi$)

Obtain $\tau = I\alpha$ from the angular momentum of a rigid body.

$A$ wheel is at rest in a horizontal position. Its moment of inertia about the vertical axis passing through its centre is $I$. $A$ constant torque $\tau$ acts on it for $t$ seconds. The change in rotational kinetic energy is:

Moment of inertia of a body about an axis is $4 \,kg-m^2$. The body is initially at rest and a torque of $8 \,N-m$ starts acting on it along the same axis. Work done by the torque in $20 \,s$, in joules, is

$A$ disc of mass $25 \ kg$ and radius $0.2 \ m$ is rotating at $240 \ r.p.m.$ $A$ retarding torque brings it to rest in $20 \ s$. If the torque is due to a force applied tangentially on the rim of the disc,then the magnitude of the force is:

The angular speed of a motor wheel is increased from $1200 rpm$ to $3120 rpm$ in $16 s$. The angular acceleration of the motor wheel is