A rigid circular ring (M) of radius R is hanged from a string. Two equal mass beads (m) are released from rest simultaneously from the top of the ring and begin to slide down opposite sides.
Show that if the mass of the bead exceeds 1.5 times of the ring, the ring will jump up when the beads fall to a certain position. Ignore friction.
The Attempt at a Solution
when upward force acted on the ring exceeds its gravitational force, it experiences upward accceleration, denote that angular position as θ, measuring from the top.
Denote N as the reaction force exerted by the ring to the beads, by 3rd law, an opposite N is exerted on the ring (upward).
So 2N cosθ > Mg, as tension in string is 0 at the moment
energy conservation: 0.5mv^2 = mgR(1-cosθ)
Centripetal acceleration: N+mg cosθ = mv^2/R
N = 2mg(1-cosθ)-mg cosθ = mg(2-3cosθ)
Mg/2cosθ < mg(2-3cosθ)
so I get m/M = 1/(2cosθ)(2-3cosθ)
It seems that I still miss one equation. Where is it? Thanks in advance.