NTesla's latest activity

[Mentor Note: this post has been heavily edited to remove numerous insults.] You mentioned that my blind assertion is that Energy is...

Even if we consider that Energy is not conserved, then in that case also, the value of tension at the bottom of both, the static portion...

When I equate this d(P.E) with negative of d(K.E), I get an expression for velocity in terms of y: ##v^{2} = gy\frac {2L - y}{L - y}##...

Yes, that I understand. When I wrote that I meant that the length of the static portion increases by ##\frac {v}{2}## per unit of time...

Angular momentum of purely vertical falling portion of the chain = ##\lambda (\frac {L}{2} - \frac {x}{2})2rv##. since the velocity of...

In the limit ##r=0##, the static portion's length increases by ##\frac {v}{2}##, and the moving portion of the chain falls by velocity...

Just before point D in direction of point E, the acceleration is a. Just at point D, the downward acceleration is less than a, and it's...

hmm.. that's a valid point. However, in the limit that the distance between the two points O and A, tending to zero, and the assumption...

I would like to understand which part did you thought was strange ? Only then, can i attempt to figure out the rationale.

IMO, It's catenary as long as both it's end is fastened to the ceiling. It's not catenary from the moment one of it's end is released...

Only one portion is falling, i.e. portion DE. When I apply conservation of energy, I'll have to take the speed of portion DE. I'm not...

Actually, the diagram is drawn for the situation at time t > 0, when the freely falling part has fallen a distance y in time t. At time...

In case of a straight chain falling on a surface, the force required to stop an element of length ##\lambda dx## of the chain is =...

Who says I cannot be bothered to ask the question here. Did you ask ? Even once ? The answer is NO. Not even once. If you didn't...

As I've mentioned in post#93 and 96, I'm finding it not just difficult to understand the explanation given in post#92 and 95, but near...