# Chain on table

1. Feb 17, 2005

### No Name Required

I have this question which is confusing as it should be harder than it appears...

A uniform chain of length $$L$$ lays on a frictionless table top. Suppose one link just hangs over the table's edge so that the chain begins to fall.

Let $$x$$ be the amount of chain that has fallen. What is the chains vertical acceleration at this instant.

I started off trying to use conservation of energy
(with the zero potential line being a distance $$L$$ below the table top).

I went with

$$mgL = (L - x)mgL + xmg\left(L - \frac{1}{2}x\right) + \frac{1}{2}mv^2$$

which led to

$$v^2 = gx^2 + 2gL(1 - L)$$

Now firstly im not sure if this is correct. If it is then what is my next step?

Secondly, I started thinking is the problem maybe a pretty simple differential equation from Newtons second law?

Or can it be done both ways?

2. Feb 17, 2005

### Staff: Mentor

Realize that your units don't match--you have an extra L floating around in two of your right side terms. Redo this.

Also realize that you are assuming that the chain is laid out in a straight line so that it moves with a single speed. (Sounds like a good assumption.)
It's wrong, as noted. But once you get the correct expression for v, just take a derivative.

Sure. Just consider the forces acting on the chain.

Absolutely!

3. Feb 17, 2005

### vincentchan

At your right hand side, you may wanna replace the m by the density m/L in the first and second term, otherwise the equation seems cool

and your way is not the best way(easiest way) to attack this problem. try to find the force directly.. (the only force here is gravity) and the accelation is F/m

the force should be depend on time F=F(t), but the problem is asking you to solve F(x), therefore, you dont need do solve the differential equation

4. Feb 17, 2005

### No Name Required

Ah rite k, thankyou

It should correctly be

$$mgL = (L - x)mg + xmg\left(L - \frac{1}{2}x\right) + \frac{1}{2}mv^2$$ yes?

If this is right, you say take the derivative... Is this with respect to t?
(in order to get $$\frac{dv}{dt}$$)

If this is so i found

$$\frac{dv}{dt} = \frac{g - gL + xg}{\sqrt{2xg - 2xgl +x^2g}}\frac{dx}{dt}$$

This looks wrong except for the fact that $$\frac{dx}{dt} = v$$ i suppose....

Ill look further into the differential equation solution.

5. Feb 17, 2005

### Staff: Mentor

Getting closer--now only one term is wrong.
It should look like this:
$$mgL = (L - x)mg + xmg(L - \frac{1}{2}x)/L + \frac{1}{2}mv^2$$

You'll find that this expression simplifies nicely, giving you a simple expression for v as a function of x.

That is correct.

Wrong, for reasons stated above. Once you get the correct expression for v, taking the derivative will give you the acceleration.

Yes, you will need to use this fact.

Don't give up on the above solution. Once you get the right initial expression, the answer comes easy.

But definitely do it using Newton's 2nd law also. Don't worry about differential equations... all you need is F = ma.

Of course, you'll get the same answer.

6. Feb 17, 2005

### vincentchan

take the derivative of the WHOLE THING, the algebra will much much simpler...

$$\frac{d}{dt} [mgL] = \frac{d}{dt} [(L - x)mg + xmg/L \left(L - \frac{1}{2}x\right) + \frac{1}{2}mv^2]$$

because of the chain rule, ALL (non-zero) TERMS has a v... therefore, you can cancel the v by dividing the whole equation by v...
and you will get dv/dt as a function of x only....(which is what your original problem asking you to do)

BTW, why dont u apply F=ma directly.... obviously, F= mgx/L and a=mx/L...

7. Feb 17, 2005

### Gamma

Just a note. Problem would have been much simplified if the tabel top was used as the zero potantial level.

8. Feb 17, 2005

### Staff: Mentor

This is true! (That's how I would do it. ) But regardless of zero level, the expression simplifies immediately.

9. Feb 18, 2005

### No Name Required

Thankyou!

Ah yes, all is clear now
Thankyou all very much for your help. Its much appreciated.