The point of the ring that is touching the ground (position 1) has instant speed 0. The diametrical opposite point on the ring (the most upper point, position 2 in the figure) has speed `2*v_0` . The points in position 3 and 4 in the figure both have the absolute value of the speed `v_0` (having opposite directions).

The ring itself has a kinetic energy of translation `(m*v_0^2)/2` .

Particle at position 2 has kinetic energy `(m*(2*v_0)^2)/2 =4(m*v_0^2)/2`

Particle at position 3 has the kinetic energy `(2m*v_0^2)/2 =2(m*v_0^2)/2`

Particle at position 4 has the kinetic energy `(m*v_0^2)/2`

Because the kinetic energy is a scalar the total energy of the system is simply the sum of the energy of its constituents.

Therefore the total kinetic energy of the system (ring+3 particles) is

`E_k =(1+4+2+1)(m*v_0^2)/2 =8*(m*v_0^2)/2 =4*m*v_0^2`

**The correct answer choice is C)** `4mv_0^2`

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