Energy Question - Chain sliding off frictionless table

[dsy5037]

Energy Question -- Chain sliding off frictionless table

Homework Statement

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)

Homework Equations

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

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.

 

[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).

 

[tiny-tim]

hi dsy5037! welcome to pf!

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.

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?

 

[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 $m(t)=M[\itex] for t, the result would include density$

 

[SammyS]

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 $m(t)=M[\itex] for t, the result would include density$

$You don't need to use density.<br /> <br /> As <span style="font-size: 9px">_^tim (That's a tiny "tim".) points out, you haven't calculated Δ(P.E.) correctly.<br /> <br /> How far does the (center of mass of the) 80% of the chain initially on the table top fall?<br /> <br /> How far does the other 20% of the chain fall ?</span>$

 

[tiny-tim]

As _^tim (That's a tiny "tim".) …

™ !

 

[haruspex]

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

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.