$A$ 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

$A$ 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:

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

$A$ flywheel of mass $25 \,kg$ has a radius of $0.2 \,m$. It is rotating at $240 \,rpm$. What is the torque necessary to bring it to rest in $20 \,s$?

$A$ flywheel is rotating at a rate of $150 \text{ rev/minute}$. If it slows at a constant retardation of $\pi \text{ rad/s}^2$, then the time required for the wheel to come to rest is (in $\text{ s}$)