Solving Mechanics Problem: Jumping Ring When Mass Exceeds 1.5 Times

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SUMMARY

The discussion focuses on a mechanics problem involving a rigid circular ring and two equal mass beads. When the mass of each bead exceeds 1.5 times the mass of the ring, the ring experiences an upward acceleration as the beads slide down. The key equations utilized include energy conservation and centripetal acceleration, leading to the conclusion that the condition for the ring to jump is defined by the inequality m/M = 1/(2cosθ)(2-3cosθ). The problem is resolved through the application of Newton's third law and the analysis of forces acting on the system.

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Homework Statement



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.

Homework Equations





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.
 
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