NTesla's latest activity

"I am older than she" never made sense to me. "I am older that she is" sounds correct, but with out the "is" feels wrong.

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...

I am sorry, but I am rather snowed now. I'll try to get back to this in a day or two.

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...

No retardation condition is fulfilled when no motion is happening. So, yes, when the chain is not moving at all, then Tension is more...

(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...

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...

Very good. I'm glad I could help.

@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...

Your result ##-\frac 1 2 \lambda v^3## is correct for the rate at which KE is lost to heat at point B. I’m not sure of your thought...

You may have misunderstood what @TSny was asking, which was to write an expression for ##dt## in terms of ##dK##. And you have that, in...

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##...