Why aren't the weights of the beads considered in this equation?

Question

I was solving this problem:

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 slides down in the opposite sides.

We are asked to find the condition on $m$ such that the ring will move up during the motion of the beads.

Now I wrote down the equation $$ N + mg\cos\theta =\frac{mv^2}{r} $$ where $N$ is the normal reaction force provided by the ring (I am working in the frame of reference of the bead) and by using the work energy theorem I get $$ \frac{mv^2}{r} = 2mg(1-\cos\theta)$$ After that, by solving for $N$, I take the downward component of $N$ and multiply it by $2$ for the two beads so it becomes $2N\cos\theta$ which provides force to lift the ring up. Now differentiating and finding maximum force for corresponding $\theta$, we get $$F_\text{max} = \frac{2mg}{3}$$ Now my question is, I will get the correct answer which is $m>3M/2$ if I use $F>Mg$ where $F = 2N\cos\theta$, but shouldn't I write it as $F>(M+2m)g$ considering the weight of the other two small beads sliding upon the ring?

Answer 1

Your confusion will be removed if you consider the FBD of the ring itself.

(Mg acts from center of mass of the ring.)
I think you can now see why the weight of the beads is inconsequential while determining equilibrium condition of the ring: regardless of the various forces felt by the beads, the ring itself feels only the following forces:
1. Weight
2. Tension of the ring
3. Normal reaction from both the beads
Hence the weight of the beads is not to be used in the comparison: F>Mg
NOTE:
Do, as an additional exercise, check whether the beads will tend to "fly off" from the ring before or after the ring will tend to move upwards(though logic says it WILL happen so).