Solving the Puzzling Physics Problem: Chain Falling Off Table

[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 I am 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?

 

[Doc Al]

I went with

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

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.)

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

It's wrong, as noted. But once you get the correct expression for v, just take a derivative.

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

Sure. Just consider the forces acting on the chain.

Or can it be done both ways?

Absolutely!

 

[vincentchan]

At your right hand side, you may want to 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 acceleration 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 don't need do solve the differential equation

 

[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.

 

[Doc Al]

It should correctly be

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

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.

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

That is correct.

If this is so i found

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

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

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

Yes, you will need to use this fact.

Ill look further into the differential equation solution.

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.

 

[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 don't u apply F=ma directly... obviously, F= mgx/L and a=mx/L...

 

[Gamma]

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

 

[Doc Al]

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

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

 

[No Name Required]

Thankyou!

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