Problem on beads sliding on a ring 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 $m\lt\frac{3M}{2}$, and find the angle $\theta$ at which this occurs.

I need some help in understanding a concept of this problem. I used 2nd Newton's Law on both objects(a bead and the ring) I got :
For a single bead:
$$N+mg\cos(\theta)=\frac{mv^2}{R}$$
For the ring:
$$T-Mg+2N\cos(\theta)=Ma$$
However, I get stuck on deciding what is the condition for the ring to rise Is it $T = 0$ or will the ring rise when $Ma\gt 0$ so I then I should calculate when
$$T-Mg+2N\cos(\theta)=0$$
The concept I'm stuck on is knowing what are the conditions for a body that is at rest to start moving upwards when some weight is exerted on it. I get confused because I don't know if the tension plays any role in this movement, and I would just like to understand which things need to happen for this upwards movement (which is not very intuitive for me) to take place.
Thanks!
 A: Your equation
N+mgcos(theta) = mv^2/R
is right, you'll then need to find $v^2$
do this by considering the loss in potential energy when a bead has gone to the angle $\theta$.  This becomes the kinetic energy, equating these gets you $v^2$
Then you'll have $N$.
The $N$ acts the opposite direction on the ring, it can be upwards on the ring if $\theta$ is less than $90^\circ$.
Include another factor to do with the direction $N$ is acting and you can find the total upward force on the ring, it must be bigger than the ring's weight to make it rise.
The problem also involves finding the maximum value of $4c-6c^2$ where $c=cos\theta$
best of luck.
A: Trying here to just give a hint (since this looks like homework):
It may seem counterintuitive that the ring would rise, but it shouldn't be.  Consider, for example, the masses not on the ring but tied to two strings, that go over two pulleys, and each string is then tied to the ring.  Here the answer is simple, $m > M/2$.  That is, there's a force between the beads and the ring, and if that force is large enough, the ring will rise.
For the beads on the ring it's the same idea: there is a force between between the beads and the ring (but here it's the force which keeps the beads moving in a circle).  If the component of that force in the upward direction becomes greater than the weight of the ring, then the ring will begin to rise.
A: If the vertical upward force exerted by the beads on the ring equals the downward force of gravity exerted on the ring, the ring will have zero weight. When is that?
How to translate this in math? You know the force the beads exert on the ring due to the (always vertical and downward)pull of gravity. You know the force the beads exert on the ring due to their circular motion (this force has a vertical and horizontal component; only the vertical is required as both horizontal components cancel; sometimes the vertical components are downward, sometimes upward; when the beads move away from each other the force is upward and when they move towards each other the force is downward). You know what force gravity exerts on the ring (always vertical and downward). The only condition is that the total vertical force on the ring is zero or bigger.
This should be enough.
Compare what would happen to the same ring in outer space. If the beads have opposite velocities, but the same speed, what will happen? The ring will oscillate. Why? Because total momentum has to be conserved. The total momentum of the beads changes direction continuously (the beads collide, their velocity reverses, they collide, etc.) so the momentum of the ring will change accordingly, to keep the total momentum constant. The beads pull the ring to and fro in sync with their to and fro motion on the ring.
In T- Mg +2Ncos(theta) = 0 you wrote a T too much...
