A hoop of radius $2 \mathrm{m}$ weighs $100 \mathrm{kg}.$ It rolls along a horizontal floor so that its center of mass has a speed of $20 \mathrm{cm}{ \mathrm{s}}^{-1}$. How much work has to be done to stop it?

Here, $R=2 \mathrm{m}, M=100 \mathrm{kg},$

$v=20 \mathrm{cm}{ \mathrm{s}}^{-1}=20\times {10}^{-2} \mathrm{m}{ \mathrm{s}}^{-1}$

Total kinetic energy of the hoop $=K_T+K_R$

$$\begin{gathered} =\frac{1}{2}M{v}^2+\frac{1}{2}I{\omega }^2=\frac{1}{2}M{v}^2+\frac{1}{2}M{R}^2{\omega }^2 \\ [\because \text{ For a hoop, ]} \\ =\frac{1}{2}M{v}^2+\frac{1}{2}M{v}^2 [\therefore v=R\omega ] \end{gathered}$$

Work required to stop the hoop = Total kinetic energy of the hoop $=M{v}^2=(100 \mathrm{kg}){\left(20\times {10}^{-2} \mathrm{m}{ \mathrm{s}}^{-1}\right)}^2=4 \mathrm{J}$