Q. A ring of mass `m` and radius `R` has three particles attached to the ring as shown in the figure.The centre of the ring has a speed `v_0.` The kinetic energy of the system is:-(Slipping is absent)
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`