Why is 'Force resolution method' and 'energy conservation principle method' giving two different answers in this problem? A ring of mass 'm' slides on a smooth vertical rod; attached to the ring is a light string passing over a smooth peg at a distance 'a' from the rod, and at the other end of this string is a mass M (>m). The ring is held on a level with the peg and released. Show that it first comes to rest after falling a distance 2mMa/(M^2-m^2).
I have a couple of issues in this question.

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*Why is the ring falling in the first place? If M>m, shouldn't it be staying at its initial position itself, with Tension = Mg pulling it in the horizontal direction?


*If it is falling down some way, I resolved the forces in equilibrium and found out the distance it would fall. But I got a different answer from what is given in the question. If you apply energy conservation principle, you will get what is given in the question. So where have I gone wrong?
Here is my calculation.

 A: This is a very good question. You have correctly said that M > m but you are not seeing that it is T cos(θ) which is responsible for balancing mg. At t = 0, the ring is said to be held on a level with the peg which results in θ = 90°. So the vertical component of force will be zero and there will be nothing to balance mg and resist it from falling down. But as it falls, θ decreases and cos(θ) increases until a time comes that it becomes sufficient to balance mg and stop the ring from falling which is asked in the question.
As for your next question, balancing the forces gives you the equilibrium position but it does not mean that it will stop at that position. For example, if you accelerate a car and then turn off the engine, it will not stop moving and will continue moving indefinitely unless an external force acts on it (Newton's 1st Law). So it is not giving you the correct answer. But using the work energy theorem ensures that the final kinetic energy becomes zero and the body has no motion whatsoever. Note that this position of zero kinetic energy is not always the equilibrium position.
Hope this helps.
