How does a mass that's dropped when strung by an inelastic, slack string continue its motion? [closed]

Question

Say a mass is connected to a light, inextensible string of length $l$. The other end of the string is fixed to a point O. If initially, the mass is kept at a horizontal distance of $\frac {l}{2}$to the point O (with the string slack) and is released from rest, what will be its velocity at the lowest point of the subsequent circular motion?

My head gives 2 answers.

First, simply by using the law of conservation of Energy from the start of the motion to the point where it's at the lowest point, I would say the velocity at the bottom will be: $$\frac{1}{2}mv^2 = mgl$$$$v =\sqrt{2gl}$$However, I also had the idea that as soon as the string becomes taut, we should consider the tangential velocity of the mass to the circular path at that point. That is, we should consider the component of the downward velocity of that point, which is in the tangential direction to the circular path. If so, how does the energy conservation law hold here? The velocity at the bottom will be less than the previous answer in this case.

Answer 1

There are multiple systems that could be intended by this setup.

If you intend a system that conserves mechanical energy then the mass will "bounce" when the string becomes taut. It will not simply begin to smoothly arc. You could model such a system as a sort of non-linear spring whose spring constant is 0 while it is not taut and infinite when it is taut.

If you intend it to be a system that smoothly transitions to an arc, then mechanical energy will be lost at the transition. This would be best to model as you described. Basically you will make a piecewise solution and specify the initial conditions of the second piece from the final conditions of the first piece.

If so, how does the energy conservation law hold here?

In systems where mechanical energy is lost, it must be converted to some other form. Usually heat or sound.