Question

A ring of radius 2m 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 ring, r = 2m
​mass of the ring,=100kg
velocity of the ring, v=20 cm/s = 0.2m/s
total energy of the ring = translational KE + rotational KE
${\mathrm{E}}_{\mathrm{r}}=\frac{1}{2}{\mathrm{mv}}^2+\frac{1}{2}{\mathrm{I\omega }}^2$
moment of inertia of the ring about its centre, I=mr^{2
${\mathrm{E}}_{\mathrm{r}}=\frac{1}{2}{\mathrm{mv}}^2+\frac{1}{2}\left({\mathrm{mr}}^2\right){\mathrm{\omega }}^2$
but we hve the relation, $v=r\omega$
$$\begin{gathered} \mathrm{therefore}, \\ {\mathrm{E}}_1=\frac{1}{2}{\mathrm{mv}}^2+\frac{1}{2}{\mathrm{mr}}^2{\mathrm{\omega }}^2 \\ =\frac{1}{2}{\mathrm{mv}}^2+\frac{1}{2}{\mathrm{mv}}^2 \\ ={\mathrm{mv}}^2 \\ \mathrm{the}\mathrm{work}\mathrm{required}\mathrm{to}\mathrm{be}\mathrm{done}\mathrm{for}\mathrm{stopping}\mathrm{the}\mathrm{ring}\mathrm{is}\mathrm{the}\mathrm{total}\mathrm{energy}\mathrm{of}\mathrm{the}\mathrm{ring}. \\ \mathrm{hence},\mathrm{required}\mathrm{work}\mathrm{to}\mathrm{be}\mathrm{done},\mathrm{W}={\mathrm{mv}}^2=100\times (0.2{)}^2=4\mathrm{J} \end{gathered}$$}