Question

A 140kg hoop rolls along a horizontal floor so that the hoops center of mass has a speed of 0.150m/sec, how much work must be done on the hoop to stop it?

Answer 1

The moment of inertia of a hoop is

$$I=mr^2.$$

The rolling loop has kinetic energy of linear motion and rotation:

$$EK=\frac{1}{2}(mv^2+I\omega^2)=\frac{1}{2}(mv^2+mr^2\omega^2)=mv^2.$$

According to work-energy theorem, the change from this value of kinetic energy to zero is equal to the work performed to stop the loop:

$$W=mv^2=3.15\text{ J}.$$