Three particles are attached to a ring of mass m and radius R as shown in the figure. The centre of mass of the ring has a speed v₀ and rolls without slipping on a horizontal surface. The kinetic energy of the system in the position shown in the figure

Given the body is pure rolling 

${\mathrm{V}}_{\circ }=\mathrm{R\omega }$

Now the kinetic energy of the ring is

${\mathrm{E}}_{\mathrm{ring}}=\frac{1}{2}\mathrm{m}{{\mathrm{V}}_{\circ }}^2+\frac{1}{2}{\mathrm{I\omega }}^2=\frac{1}{2}\mathrm{m}{{\mathrm{V}}_{\circ }}^2+\frac{1}{2}{\mathrm{mR}}^2{\mathrm{\omega }}^2$

${\mathrm{E}}_{\mathrm{ring}}=\frac{1}{2}\mathrm{m}{{\mathrm{V}}_{\circ }}^2+\frac{1}{2}\mathrm{m}{{\mathrm{V}}_{\circ }}^2=\mathrm{m}{{\mathrm{V}}_{\circ }}^2$

The left-most particle's speed is

${\mathrm{v}}_1=\sqrt{{{\mathrm{V}}_{\circ }}^2+\left({\mathrm{R\omega }}^2\right)}=\sqrt{{{\mathrm{V}}_{\circ }}^2+{{\mathrm{V}}_{\circ }}^2}={\mathrm{V}}_{\circ }\sqrt{2}$

Kinetic energy,

${\mathrm{E}}_1=\frac{1}{2}\times 2\mathrm{m}\times {{\mathrm{v}}_1}^2=2\mathrm{m}{{\mathrm{V}}_0}^2$

The right-most particle's speed is

${\mathrm{v}}_2=\sqrt{{{\mathrm{V}}_{\circ }}^2+\left({\mathrm{R\omega }}^2\right)}=\sqrt{{{\mathrm{V}}_{\circ }}^2+{{\mathrm{V}}_{\circ }}^2}={\mathrm{V}}_{\circ }\sqrt{2}$

Kinetic energy

${\mathrm{E}}_2=\frac{1}{2}\times \mathrm{m}\times {{\mathrm{v}}_2}^2=\mathrm{m}{{\mathrm{V}}_0}^2$

The speed at the top

${\mathrm{v}}_3={\mathrm{V}}_{\circ }+\mathrm{R\omega }=2{\mathrm{V}}_{\circ }$

Kinetic energy

${\mathrm{E}}_3=\frac{1}{2}\times \mathrm{m}\times {{\mathrm{v}}_3}^2=2\mathrm{m}{{\mathrm{V}}_{\circ }}^2$

Thus the total energy

$\mathrm{E}={\mathrm{E}}_{\mathrm{ring}}+{\mathrm{E}}_1+{\mathrm{E}}_2+{\mathrm{E}}_3=6\mathrm{m}{{\mathrm{V}}_{\circ }}^2$

Hence the correct answer is $6\mathrm{m}{{\mathrm{V}}_{\circ }}^2.$