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

Radius of the hoop, r = 2 m
Mass of the hoop, m = 100 kg
Velocity of the hoop, v = 20 cm/s = 0.2 m/s
Total energy of the hoop = Translational KE + Rotational KE
$E_T=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2{E}_r=\frac{1}{2}m{v}^2+\frac{1}{2}\left(m{r}^2\right){\omega }^2$
But we have the relation, $v=r\omega$
$\therefore E_r=\frac{1}{2}m{v}^2+\frac{1}{2}m{r}^2{\omega }^2=\frac{1}{2}m{v}^2+\frac{1}{2}m{v}^2=m{v}^2$
The work required to be done for stopping the hoop is equal to the total energy of the hoop.
$\therefore$ Required work to be done, $W=m{v}^2=100\times (0.2{)}^2=4J$