Question

A hoop of radius 2 m weighs 100 kg. It rolls along a horizontal floor so that its centre of mass has a speed of 20 cm/s. How much work has to be done to stop it?

Answer 1

Total energy of the hoop=Translational Kinetic energy+ Rotational Kinetic energy=$\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$

For hoop, $I=m{r}^2$

Also $v=r\omega$

Thus total energy=$\frac{1}{2}m{v}^2+\frac{1}{2}m{r}^2(\frac{v}{r}{)}^2=m{v}^2$

Hence the work required to stop the hoop=Its total energy=$m{v}^2=100\times {0.2}^{2}J=4J$