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A ring of mass m and radius R has three ...

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).

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Velocity of any point in the ring is the vector sum of `V_CM` and the tangential velocity `Romega` due to spin about CM. For rolling without slipping, `V_cm = Romega``V_A=V_(CM)+Romega=2V_(CM)``V_(B)=sqrt(V_(CM)^(2)+R^2omega^2)=sqrt2V_(CM)``V_D=sqrt(V_(CM)^(2)+R^2omega^2)=sqrt2V_(CM)``K_(t otal)=1/2mV_(A)^(2)+1/2mV_(D)^(2)+1/2(2m)V_(B)^(2)+K(ri ng)``=1/2m(V_(A)^(2)+V_(D)^(2)+V_(B)^(2))+(1/2mV_(CM)^(2)+1/2I_(CM)omega^2``1/2m(4+2+4)V_(CM)^(2)+1/2m(V_(CM)^(2)+R^2V_(CM)^(2)/R^2)=6mV_(CM)^(2)rArr` the correct answer is A.
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