How Long Will It Take for the Entire Chain to Slide Off the Table?

[ehild]

I too have the same doubt.
The answer from energy conservation method is 0.55 and answer from momentum technique is 0.44.Why they are't equal?

Remember, you got 0.55 s from energy conservation assuming that the whole chain moves with the same speed.
With the assumption, that only the vertical part moves, the result is 0.44 s.

ehild

 

[haruspex]

The defect in the momentum approach is, for example, the assumption that the infinitesimal chain element acquires a finite velocity in infinitesimal time, which means infinite acceleration. The assumption that such an infinitesimal element is possible in a chain may itself be considered a flaw.

I don't believe it makes that assumption.

 

[haruspex]

Remember, you got 0.55 s from energy conservation assuming that the whole chain moves with the same speed.
With the assumption, that only the vertical part moves, the result is 0.44 s.

ehild

Sure, but applying work conservation to the bunched chain case produces a rather different equation from that based on momentum conservation.

 

[Orodruin]

Sure, but applying work conservation to the bunched chain case produces a rather different equation from that based on momentum conservation.

I got 0.40 s for the energy conserving approach for the bunched chain. Although I did it quite hastily so there may be errors, but this is similar to what I would expect (loss of work should result in the chain moving slower).

 

[ehild]

Satvick, see that video.

www.youtube.com/watch?v=tj5vm4nnZRo.

It is a simpler problem, assuming outstretched chain, and solved both with F=ma and conservation of energy.
Or read that https://www.google.com/url?sa=t&rct...Sj9th1TF7CuifoQ&bvm=bv.76802529,d.ZWU&cad=rjt

And try to play with a chain. You will see that the outstrecthed one needs more time to slide down then the folded one, shown in the picture.

 

[voko]

I don't believe it makes that assumption.

How do you interpret the term $(dm) v$?

Another interesting assumption in this approach is that the only acting force is that of gravity; specifically, no tension is assumed at the top of the chain column.

 

[Satvik Pandey]

Satvick, see that video.

www.youtube.com/watch?v=tj5vm4nnZRo.

I saw that video.It was quite useful..

Or read that https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&ved=0CB0QFjAA&url=https://wiki.brown.edu/confluence/download/attachments/2752884/Chain.pdf?version=2&modificationDate=1385496323000&ei=U9UuVKLxJcj3OvKVgPgD&usg=AFQjCNHEve_SSVfEbe6RhlWLafvz4l7uvw&sig2=6hZD4Z6Sj9th1TF7CuifoQ&bvm=bv.76802529,d.ZWU&cad=rjt

I understood it, till page(1) but I don't know how to solve that differential equation.
Thank you for helping! :)

 

[ehild]

Do you mean the differential equation $\frac{d^2x(t)}{dt^2}=\frac{g}{L}x(t)$?
It is a very simple second order linear de with constant coefficients, a homogeneous one. You will learn about them soon.
Try the solutions in the form x=e^kt. You get a quadratic equation for k, with two roots, k₁ and k₂. The general solution of the de is the linear combination of the two solutions, belonging to k₁ and k_2. X=c1 e^{k₁ t}+c2 e^k₂t.

ehild

 

[haruspex]

How do you interpret the term $(dm) v$?

ok, that was a bit glib. But it does not assume a nonzero mass undergoes an infinite acceleration. It is only infinite in the limit as dt tends to zero, and dm is also zero in that limit.
The key assumption is that the motion is homogeneous, for the part that is moving. If you want a work conseving version then that assumption must be dropped. This makes the problem far more complex, opening up the possibility of longitudinal oscilations. Thus it may be that even if work is completely conserved, the time taken is greater than the 'simple' calculation suggests.

 

[ehild]

How do you interpret the term $(dm) v$?
.

You can interpret it as an inelastic collision between the moving part and a link at the edge of the table still in rest. That link "sticks" to the moving part and they move together, with a bit less speed. Energy is not conserved.

ehild

 

[voko]

That is the equation that can be obtained as indicated by Orodruin in #35. So its solution is known.

But it does not assume a nonzero mass undergoes an infinite acceleration. It is only infinite in the limit as dt tends to zero, and dm is also zero in that limit.

That still sounds like an infinite acceleration to me. But, as I said earlier, taking care of this and other nuances moves the problem into a different league.

 

[Satvik Pandey]

Do you mean the differential equation $\frac{d^2x(t)}{dt^2}=\frac{g}{L}x(t)$?
It is a very simple second order linear de with constant coefficients, a homogeneous one. You will learn about them soon.
Try the solutions in the form x=e^kt. You get a quadratic equation for k, with two roots, k₁ and k₂. The general solution of the de is the linear combination of the two solutions, belonging to k₁ and k_2. X=c1 e^{k₁ t}+c2 e^k₂t.

ehild

Thank you ehild. You really help me a lot! :) I am just a beginner in calculus.
I looked for the solution of second differential equation in senior secondary school mathematics book but I didn't found it. It only contain the solution of linear differential equation. I think I will learn about them later.
But I will try to learn about it through internet.:w

 

[ehild]

I looked for the solution of second differential equation in senior secondary school mathematics book but I didn't found it. It only contain the solution of linear differential equation. I think I will learn about them later.
But I will try to learn about it through internet.:w

Do not run. Wait till you learn about them in school.

ehild

 

[Satvik Pandey]

Do not run. Wait till you learn about them in school.

ehild

Thanks ehild. I will follow your advice.:)

 

[ehild]

I know that you won't. :D

ehild

 

[Satvik Pandey]

I know that you won't. :D

ehild

:D
Yes I will try to learn about them but not now.

 

[Satvik Pandey]

I found equation in #post 41 by a slightly different method.

As $\frac { d(mv) }{ dt } =\frac { M }{ L } (x)g$

so $mv\frac { d(mv) }{ dt } =\frac { M }{ L } (x)g\times mv$

or $mv\frac { d(mv) }{ dt } =g{ \left( \frac { M }{ L } \right) }^{ 2 }{ x }^{ 2 }\frac { dx }{ dt\\ }$

Let $s=mv$

So $\frac { { s }^{ 2 } }{ 2 } ={ \left( \frac { M }{ L } \right) }^{ 2 }\frac { { x }^{ 3 } }{ 3 } +c$

now $c=-{ \left( \frac { M }{ L } \right) }^{ 2 }\frac { { x }_{ 0 }^{ 3 } }{ 3 }$
putting this value I got

$\frac { { v }^{ 2 } }{ 2 } =\frac { g }{ 3 } \left( x-\frac { { x }_{ 0 }^{ 3 } }{ { { x }^{ 2 } } } \right)$

or ${ v }^{ 2 }=\frac { 2g }{ 3 } \left( x-\frac { { x }_{ 0 }^{ 3 } }{ { { x }^{ 2 } } } \right)$

or $\int _{ { x }_{ 0 } }^{ x }{ { \sqrt { \frac { 2g }{ 3 } \left( x-\frac { { x }_{ 0 }^{ 3 } }{ { { x }^{ 2 } } } \right) } }^{ -1/2 } } dx=\int _{ 0 }^{ t }{ dt }$
This is what haruspex found in #post 41.:D

 

[Satvik Pandey]

I have a confusion.
If a part of a chain is piled up on a table and rest part of it hangs from the table.And suppose we are asked to find the time in which whole chain falls off the table then will it's answer be the solution of the last equation in post #107?

 

[haruspex]

I have a confusion.
If a part of a chain is piled up on a table and rest part of it hangs from the table.And suppose we are asked to find the time in which whole chain falls off the table then will it's answer be the solution of the last equation in post #107?

If you assume no elasticity, yes. If there is elasticity it becomes very difficult. There may be longitudinal oscillations.

 

[Satvik Pandey]

If you assume no elasticity, yes. If there is elasticity it becomes very difficult. There may be longitudinal oscillations.

Thank you!
Is solution in #post107 correct?

 

[Orodruin]

Thank you!
Is solution in #post107 correct?

Apart from the additional /2 in the final integral, which I assume is a typo, it is fine.

 

[Satvik Pandey]

Apart from the additional /2 in the final integral, which I assume is a typo, it is fine.

Thank you!
Oh! Sorry, that was a typo.

 

[Satvik Pandey]

Do not run. Wait till you learn about them in school.

ehild

I think I solved that.

As $\frac { { d }^{ 2 }x }{ { dt }^{ 2 } } =\frac { gx }{ l\\ }$

or $\frac { dv }{ dt } =\frac { gx }{ l }$

or $\frac { vdv }{ dt } =\frac { gx }{ l\\ } \frac { dx }{ dt }$

or $vdv=\frac { gx }{ l\\ } dx$

or $\frac { { v }^{ 2 } }{ { 2 } } =\frac { g }{ l\\ } \frac { { x }^{ 2 } }{ { 2 } } +c$

at $t=0$ $v=0$ and $x=x_{0}$

So $c=-\frac { g{ x }_{ 0 }^{ 2 } }{ 2l }$

or $\frac { { v }^{ 2 } }{ 2 } =\frac { g }{ l } \frac { { x }^{ 2 } }{ { 2 } } -\frac { g{ x }_{ 0 }^{ 2 } }{ 2l }$

or $\frac { dx }{ dt } =\sqrt { \frac { g }{ l } \left( { x }^{ 2 }-{ x }_{ 0 }^{ 2 } \right) }$

or $\sqrt { \frac { l }{ g } } \int _{ { x }_{ 0 } }^{ l }{ \frac { dx }{ \sqrt { \left( { x }^{ 2 }-{ x }_{ 0 }^{ 2 } \right) } } } =t$

or $\sqrt { \frac { l }{ g } } { ln|{ x }+\sqrt { \left( { x }^{ 2 }-{ x }_{ 0 }^{ 2 } \right) } | }_{ { x }_{ 0 } }^{ l }=t$

Is it right ehild?:D

 

[ehild]

It is correct, Satvik. Good job!
That was the hard way to do it.

The easy method is when one has studied differential equations, and recognises that it is a second order linear DE with constant coefficients.
One knows that the solutions are of the form x=e^λt.
Replacing back the trial solution into the DE, you get an equation for the constant λ.

λ²=g/l; λ1=√(g/l), λ2=-√(g/l). Let be k= √(g/l). There are two independent solutions: y1=e^kt and y2=e^-kt.

Again, when you studied differential equations, you know, that any linear combinations of the independent solutions are a solution again.

So you can write up the general solution as x(t)=c₁e^kt+c₂e^-kt.

Fit the constants to the initial conditions: at t=0 x(0)=x₀ and v(0)=0---> c₁+c₂=x₀.
v(t)=c₁ke^kt-c₂ke^-kt
v(0)=0=k(c₁-c₂).

The constants are c₁=c₂=x₀/2 . The solution is x(t)=(x₀/2) ( e^kt+e^-kt). You need to solve it for t when x=l.ehild

 

[Satvik Pandey]

It is correct, Satvik. Good job!
That was the hard way to do it.

The easy method is when one has studied differential equations, and recognises that it is a second order linear DE with constant coefficients.
One knows that the solutions are of the form x=e^λt.
Replacing back the trial solution into the DE, you get an equation for the constant λ.

λ²=g/l; λ1=√(g/l), λ2=-√(g/l). Let be k= √(g/l). There are two independent solutions: y1=e^kt and y2=e^-kt.

Again, when you studied differential equations, you know, that any linear combinations of the independent solutions are a solution again.

So you can write up the general solution as x(t)=c₁e^kt+c₂e^-kt.

Fit the constants to the initial conditions: at t=0 x(0)=x₀ and v(0)=0---> c₁+c₂=x₀.
v(t)=c₁ke^kt-c₂ke^-kt
v(0)=0=k(c₁-c₂).

The constants are c₁=c₂=x₀/2 . The solution is x(t)=(x₀/2) ( e^kt+e^-kt). You need to solve it for t when x=l.ehild

Thank you.:)