Chain Hanging Over a Table

The solution to this problem requires us to solve differential equations.

We begin with first principles (in Newtonian Mechanics). ${F}_{net} = accelerated\quad mass \times a$

Let $l$ be the length of the chain, and $x$ be the variable length hanging over the edge of the table. Since the mass is uniformly dense, we let $\rho$ be the density. Hence,

$(\rho x) g = (\rho l) \frac{dv}{dt}$

Since $\frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt} = v\frac{dv}{dx}$ , we can rewrite the above equation as

$v\frac{dv}{dx} = \frac{gx}{l}$

Let $h$ be the initial length hanging over the edge of the table. We solve the above differential equation when $v =0$ (before sliding).

${v}^{2} = \frac{g}{l} ({x}^{2} - {h}^{2})$ $v = \sqrt{\frac{g}{l} ({x}^{2} - {h}^{2})}$

Now we let $v = \frac{dx}{dt}$ and solve the new differential equation at $x = h$ at the initial time $t=0$ . (I searched up this integral.)

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

The chain slides off the table at $x=l$ so we get $t = \sqrt{\frac{l}{g}}ln\left(\frac{l+\sqrt{{l}^{2} - {h}^{2}}}{h}\right).$

Plugging in the values $h=1/3, l=1, g = 9.8$ we get $t = 0.563 seconds$ .

I will solve it with conservation of energy. Let x denote the vertical coordinate of the chain's free end. As the chains free end is at distance x so the CoM of chain will be at a distance* x/2*.

So * $M(x)=\frac { M }{ L } x$ *

and $U(x)=-\frac { M{ x }^{ 2 } g}{ 2L }$

As ${ K }_{ i }+{ U }_{ i }={ K }_{ f }+{ U }_{ f }$

So $0-\frac { mg }{ 18 } =\frac { m{ v }^{ 2 } }{ 2 } +\frac { -m{ { x }^{ 2 } }g }{ 2 }$

or $\frac { m{ v }^{ 2 } }{ 2 } =\frac { m{ x }^{ 2 }g }{ 2 } -\frac { mg }{ 18 }$

or $\frac { m{ v }^{ 2 } }{ 2 } =\frac { 9{ mx }^{ 2 }g-mg }{ 18 }$

or $9{ v }^{ 2 }=(9{ x }^{ 2 }-1)g$

or $\frac { dt }{ dx } =\sqrt { \frac { 9 }{ (9{ x }^{ 2 }-1)g } }$

or $dt=\sqrt { \frac { 9 }{ (9{ x }^{ 2 }-1)g } } dx$

or $\int _{ 0 }^{ t }{ dt } =\int _{ \frac { 1 }{ 3 } }^{ 1 }{ \sqrt { \frac { 9 }{ (9{ x }^{ 2 }-1)g } } dx }$

or $t=\frac { 3 }{ \sqrt { g } } \int _{ \frac { 1 }{ 3 } }^{ 1 }{ \sqrt { \frac { 1 }{ (9{ x }^{ 2 }-1) } } } dx$

or $t=\frac { 3 }{ \sqrt { g } } \int _{ \frac { 1 }{ 3 } }^{ 1 }{ \sqrt { \frac { 1 }{ 9\left( { x }^{ 2 }-\frac { 1 }{ 9 } \right) } } } dx$

or $t=\frac { 1 }{ \sqrt { g } } \int _{ \frac { 1 }{ 3 } }^{ 1 }{ \sqrt { \frac { 1 }{ \left( { x }^{ 2 }-\frac { 1 }{ 9 } \right) } } dx }$

or $t=\frac { 1 }{ \sqrt { g } } \log { \left| x+\sqrt { { x }^{ 2 }-\frac { 1 }{ 9 } } \right| }$

or $t=\frac { 1 }{ \sqrt { g } } \log { \left| \frac { 3x+\sqrt { 9{ x }^{ 2 }-1 } }{ 3 } \right| }$ where limits are - lower limit is 1/3 and upper limit is 1

or $frac { 1 }{ \sqrt { g } } \left\{ \log { \left| \frac { 3+2\sqrt { 2 } }{ 3 } \right| } -\log { \left| \frac { 1 }{ 3 } \right| } \right\}$

or $\frac { 1 }{ \sqrt { g } } \left\{ \log { \left| 1.94 \right| } -\log { \left| 0.33 \right| } \right\}$

or $\frac { 1 }{ \sqrt { g } } \left\{ 0.28-(-0.48) \right\}$ =0.243

By multiplying this 2.303 (to convert common logarithms into natural logarithms) we will get the answers.

2.303(0.243)=0.566 sec

This problem can be solved very easily with a little bit of knowledge of differential equations (or wolframalpha to do the job if you don't!).

If you do a bit of elementary mechanics (free-body diagram analysis) on the initial conditions given, the acceleration of the entire string at the beginning is $\frac{1}{3}g$ , with the $\frac{1}{3}$ coming from the fraction of the string hanging off the edge. With this, a nifty re-illustration of the problem can be made.

Imgur

So we have an extremely simple second order differential equation

$x''(t) = gx(t)$

with the initial conditions $x(0)=\frac{1}{3}$ and $x'(0) = 0$ . The most general solution to the differential equations is

$x(t) = c_1e^{\sqrt{g}t} + c_2e^{-\sqrt{g}t}$

and plugging in the initial conditions gives us the solution,

$x(t) = \frac{1}{6}(e^{\sqrt{g}t} + e^{-\sqrt{g}t} )$

Now we set $x=1$ , and solve for $t$ .

$t\approx.563$

Let $l$ be the length of the chain, and $x$ be the variable length hanging over the edge of the table.

initially x=1/3 m

By energy conservation

$\frac { 1 }{ 2 } { mv }^{ 2 }=\frac { m{ g }x^{ 2 } }{ l } -\frac { mg }{ 9 }$

Putting l=1 m, velocity at an instant is given by

$v=\sqrt { g{ x }^{ 2 }-\frac { g }{ 9 } }$

Since $v=\frac { dx }{ dt }$

Putting value of $v$

$\quad \frac { dx }{ dt } =\sqrt { g{ x }^{ 2 }-\frac { g }{ 9 } }$

$\frac { dx }{ \sqrt { g{ x }^{ 2 }-\frac { g }{ 9 } } } =\quad dt$

Integrating LHS from $x=\frac { 1 }{ 3 }$ to $x=1$ and RHS from $t=0$ to $t=t$

$\int _{ \frac { 1 }{ 3 } }^{ 1 }{ \frac { dx }{ \sqrt { { x }^{ 2 }-\frac { 1 }{ 9 } } } } =\quad \sqrt { g } \int _{ 0 }^{ t }{ dt }$

$\frac { [ln(x+\sqrt { { x }^{ 2 }-\frac { 1 }{ 9 } } ){ ] }_{ \frac { 1 }{ 3 } }^{ 1 } }{ \sqrt { g } } =t$

Putting g=9.8

$t=\frac { ln(3+\sqrt { 8 } ) }{ \sqrt { 9.8 } }$

Therefore $\boxed { t=0.557 }$

It's really a beautyful problem.Keep uploading such problems