Recent content by NTesla

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 understand the question itself, you should have asked question, or atleast let me know that you are not able to understand the question. I would have tried to frame...

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 impossible to understand as to how the explanation given in these 2 posts would help me understand the question that I had raised in post#84. And post#97 is in...

Yes. I agree. Reason: There's no point on the chain that doesn't have tension, except the ends of the chain. So, yes, there is tension at the bend. However, I'm not sure whether the tension at the inlet would be ##\lambda v^{2}## or not. This is because, ##\lambda v^{2}## is the leftward force...

Yes. I agree with this. However, I suppose you meant to write ##\lambda h \frac {dv}{dt}## instead of ##\lambda \frac {dv}{dt}##. It was just a minor typo, I think. Yes I agree. For Scenario 2, you wrote: This is the way I'm seeing this: When scenario 2 starts with Scenario 1, this means...

No retardation condition is fulfilled when no motion is happening. So, yes, when the chain is not moving at all, then Tension is more than when it is moving. So, this statement is true. I'm still trying to understand this. Let's say that before the chain starts to move, the tension at the...

(1) Although I agree with your assertion that tension at it's right hand end is only infinitesimally different from that at ##v. dt## to the left of that point, but I'm having trouble understanding what you meant when you mentioned that: "the effect of that difference on the motion over time...

I still have a question. When I am trying to solve this question by equating rate of change of momentum to the force applicable, then there is a slight conundrum that I'm facing. Here's the details: For the horizontal portion of the chain, ##\frac {dp}{dt}## = ##\lambda ((l - x - h)\frac...

@jbriggs444 , Kindly see post No. 67 and respond if and when possible.

No, that particular issue in that question still remains unresolved. No one picked up that particular point.

This post did clarify a crucial error in my calcualtion, especially the part written in edit. Much appreciated. Thank you.

It did solve the problem that I was facing. I had given up on solving the question using energy considerations, until your intervention. It allowed me to see the entire thing in a new way. So, Thank you very much. And the resulting differential equation did solve the question.

If we consider an element of length ##\Delta x## at point B, which is just about to strike the table, then it's K.E at an instant ##t## = ##\frac {1}{2}\lambda \Delta x v^{2}##. Let's suppose that at time ##t + \Delta t##, it comes to a stop. Then in time ##\Delta t##, ##\Delta K.E = -\frac...

Yes, thank you. That seems plausible. Here's my attempt using energy method: K.E of the chain, at the instant the upper end of the chain has moved a distance ##x## from point ##A## = ##\frac {1}{2} \lambda (l - x)v^{2}##. Potential Energy ##U## of the chain at the same instant = ##\lambda (l -...

Much appreciated. Thank you. However, now the equation that I'm writing for the hanging part of the chain is cancelling out all together. Here's the calculation: in the ##\hat j## direction: ##\vec {\Delta P} = \lambda h(v + \Delta v)(-\hat j) - \lambda hv(-\hat j)## = ##\lambda h\Delta v(-\hat...

There's no reason to go on resolving the equations further if the 1st equation itself seems incomplete or wrong to begin with. I expect you to atleast go through the equations once, before you jump to the conclusion that I've fallen into the classic trap or any other conclusion for that matter...