a ring of mass M hangs from a thread, and two beads of mass m slide on it without friction. the beads are released simultaneously from the top of the ring and slide down opposite sides. show that the ring will start to rise if m>3M/2, and find the angle at which this occurs. what i did: i got the displacement x=R(1-cos(a)) by energies we have 2mgR=(2R-x)mg+mv^2/2 so v^2=2xg now, i think because it rises, the increase in the normal force equals the change of forces, because both beads have a centripatl force acting on them, then the change in normal force is (2mv^2/R)cos(a) (by the third law of newton, action reaction force), so i think to myself this increase should be greater than Mg, but from here im stuck (if im even right on this claim), can someone point where im wrong if im wrong, and how to correct this, becuase i dont know from here to conclude that m>3M/2 and the angle when this occurs. thanks in advance.