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

• Satvik Pandey
In summary, the conversation discusses a problem involving a chain of uniform density on a table with negligible friction. The length of the chain is 1 m and initially, one-third of the chain hangs over the edge of the table. There is a question about how long it will take for the chain to slide off the table. The conversation includes a proposed solution using conservation of energy and discusses clarifications and corrections to the approach. It also mentions the need for further clarification from the problem writer about whether the entire chain or just a part of it is moving.
ehild said:
I know that you won't. :D

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

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

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?

Satvik Pandey said:
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
haruspex said:
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?

Satvik Pandey said:
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
Orodruin said:
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.

ehild said:
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

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=eλt.
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=ekt 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)=c1ekt+c2e-kt.

Fit the constants to the initial conditions: at t=0 x(0)=x0 and v(0)=0---> c1+c2=x0.
v(t)=c1kekt-c2ke-kt
v(0)=0=k(c1-c2).

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

Satvik Pandey
ehild said:
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=eλt.
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=ekt 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)=c1ekt+c2e-kt.

Fit the constants to the initial conditions: at t=0 x(0)=x0 and v(0)=0---> c1+c2=x0.
v(t)=c1kekt-c2ke-kt
v(0)=0=k(c1-c2).

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

Thank you.:)

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