Energy Question - Chain sliding off frictionless table

1. Mar 20, 2014

dsy5037

Energy Question -- Chain sliding off frictionless table

1. The problem statement, all variables and given/known data

Here is the problem that's confusing me: A frictionless chain of length 2.00m is held with 20.0% of its length hanging over the edge of a table. The chain is then released. Determine its speed the moment the entire chain comes off the table. (Answer = 4.34m/s)

2. Relevant equations

I assumed this was an energy related question and that the equations for Kinetic/Potential energy would apply.

3. The attempt at a solution

I wasn't really sure how to approach this problem but I tried

PE = KE since at first the chain is just hanging and then at the end the entire thing is off the table.

so I tried PE = KE
mgh = 1/2mv^2
(.8*mass)(g)(1.6m) = 1/2(1*mass)(v^2)

I thought you might only need to use 80% of the total mass of the string for the PE since 20% is hanging off. And I thought h would only be 1.6 since that's the remaining length of the string that still needed to fall. But I didn't get the correct answer and I'm probably totally wrong with my thinking but this question confused me :s Any help would be appreciated.

2. Mar 20, 2014

rcgldr

Assume the table is more than 2.00 meters above the ground. It may help to choose a reference height for potential energy, such as the surface of the table. (GPE below the table surface would be negative). What is the initial potential energy when 20% of the chain is hanging over the edge, and what is the potential energy of the chain when all of it is "hanging over the edge"?

Note this problem ignores chain fountain effect (you can find video examples of this by looking for "chain fountain" on youtube).

Last edited: Mar 20, 2014
3. Mar 20, 2014

tiny-tim

hi dsy5037! welcome to pf!
your method is correct, but you've not thought out the PE difference correctly …

not all the 1.6 m of string has fallen 1.6 m, has it?

4. Mar 22, 2014

Gzyousikai

Isn't it related to density?

By setting the equation:

$$Ma=m(t)g$$

solve for $m(t)$, we have
$$m(t)=\frac{1}{5}\left[e^{\sqrt{\frac{g}{L}}t}+e^{-\sqrt{\frac{g}{L}}t}\right]=\frac{1}{5}\left[e^{\sqrt{5}t}+e^{-\sqrt{5}t}\right]$$ if we solve [itex]m(t)=M[\itex] for t, the result would include density

Last edited: Mar 22, 2014
5. Mar 22, 2014

SammyS

Staff Emeritus
You don't need to use density.

As tim (That's a tiny "tim".) points out, you haven't calculated Δ(P.E.) correctly.

How far does the (center of mass of the) 80% of the chain initially on the table top fall?

How far does the other 20% of the chain fall ?

6. Mar 22, 2014

tiny-tim

™ !

7. Mar 22, 2014

haruspex

It is not a good problem because it ignores the fact that the chain will (non-uniformly) acquire horizontal speed, which means that when the trailing end leaves the table the airborne chain describes some arc, so its centre of mass has not descended as far as if it had remained vertical and straight.
However, this is not the "chain fountain" effect. As the video shows, the chain travels above the edge of the container without touching it along the way. The explanation comes from considering how the individual links are lifted from rest. If a uniform rod lying on a horizontal surface is snatched up at one end, there is an upward impulse, of half the magnitude, from the floor at the other end. This converts some the energy that would otherwise have gone into rotating the chain links (to be lost later) into upward movement. Note that this is quite different from the OP set-up, where the chain on the table is initially straight and taut.