Kleppner and Kolenkow 5.10 - Falling Chain

1. Apr 19, 2014

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

A chain of mass M and length $\ell$ is suspended vertically with its lowest end touching a scale. The chain is released and falls onto the scale. What is the reading of the scale when a length of chain, $x$, has fallen? (Neglect the size of individual links.)

http://gyazo.com/855deafa7a3e4bca592cb162e3c9c949 <----- Image

2. Relevant equations

$$M_{dx}=M\frac{x}{\ell}$$
$$K_i=0$$
$$U_i=Mg\frac{x}{\ell}(\ell-x)$$
$$K_f=\frac{Mxv^2}{2\ell}$$
$$U_i=0$$

3. The attempt at a solution

When I equate the initial potential energy and the final kinetic energy, it is not possible to solve for M without cancelling it; I need to find $M(x)$. Also, energy methods must be used.
Am I missing something?

2. Apr 19, 2014

Simon Bridge

I cannot see your reasoning there.

Note: "reading on the scale" is different from "mass".
If you drop something on scales, the reading is (initially) higher than the mass of the object right?

3. Apr 19, 2014

Yes. Are you saying I should use the definition of weight $W=mg$ ?

EDIT: I don't know how I can apply this to the problem but certain scales have springs in them with relatively high values of $k$ so maybe there is some elastic potential (though it seems negligible).

4. Apr 19, 2014

D H

Staff Emeritus
That's part of the answer.

Think of the chain as comprising three parts. Some hasn't yet reached the scale and is still falling, some is at rest on the scale, and a tiny third part is transitioning from falling to at rest on the scale. The scale has to support the weight of the already fallen part but it also has to provide force to change the momentum of that transitional part of the chain.

5. Apr 19, 2014

So when we're speaking in terms of energy conservation, would it be rational to examine a piece of the chain that is still falling rather than some other piece? And if that is correct, would it be right to state that the height is $(\ell-x)$ ?

6. Apr 19, 2014

$$w(\ell-x)=\frac{mv^2x}{2\ell}$$
Where $mx/\ell$ is the mass of the piece of chain.
Solving for $M(x)$ , I get:
$$M(x)=\frac{2\ell w(\ell-x)}{xv^2}$$

only I don't know what $v$ is.

7. Apr 19, 2014

Simon Bridge

As the chain falls it loses gravitational PE.
Some of it goes to kinetic energy in the remaining bit of chain in the air, you've already noticed, and some of it goes into the scales.

How do scales work?

8. Apr 19, 2014

Scales have springs within them; so the chains will have elastic potential energy.
I guess I'll have to solve for $k$ then? What about $x$ , the distance from equilibrium?

9. Apr 19, 2014

D H

Staff Emeritus
No, you don't have to solve for k. All you need to know is that scale exerts an upward force. This upward force is what a scale registers. How the scale registers that upward force is irrelevant. What you do need to know is that some part of that force goes into the part of the chain that is resting on the pan from falling, while another part goes into slowing down the chain as it hits the pan.

10. Apr 19, 2014

I know that the part of the force that goes into the chain that is resting is $mgx/\ell$ which is $wx/\ell$ , but I still don't see how that force contributes to the total energy of the system. It does no work because it is perpendicular with respect to the chain.

11. Apr 19, 2014

voko

Where does this requirement, "energy methods must be used", come from? Do you realise that energy is lost in this system?

12. Apr 19, 2014

The chapter is on conservation of energy so I thought we were meant to use energy methods to solve this problem. I know energy is lost, I just can't see what is causing that loss. Isn't the upwards force from the spring perpendicular? If it is (which seems like it) then it can't do work on the system.

13. Apr 19, 2014

voko

If the energy were conserved, then each link, upon hitting the scales, would bounce up with the same speed (recall the laws of fully elastic collisions). If, on the other hand, links come to rest immediately upon hitting the scales, their speed is lost and so is their energy.

That said, you can still use energy methods to deduce the speed at which the links hit the scales. Knowing that speed, you can deduces the additional reaction force from the scale required to counter it.

14. Apr 19, 2014

Simon Bridge

You should not need to solve for k since you can use the fact that the scales have been calibrated for weight.
Real life scales also have damping - so you need to think about the model you want to use more carefully.
i.e. can you safely neglect the distance the pan falls?

The context of the problem will be important here.

voko has an important point - to avoid mechanical energy losses you want the chain to fall completely silently, and no damping in the scales ... so after the chain has fallen, the pan will execute SHM. But you nonly care about what happens when the chain is falling.

One way of understanding the problem is to make it discrete - the chain has N links modelled as small masses joined by a classical string lengths l/N, work out the scale reading for n<N links having fallen and take the limit N gets very big.

15. Apr 19, 2014

Simon Bridge

... that is if kinetic energy is conserved... even then
1. why would the direction of the bounce affect the kinetic energy?
2. why couldn't some KE get transferred to the scales?

However: there are other forms of energy.
i.e. maybe lost gravitational potential energy ends up stored in the spring under the scale pan?

... often conservation of energy would be combined with other rules, like conservation of momentum.
But it is your textbook - which one is it BTW? - you should review the previous chapter to see how they do the examples.

16. Apr 19, 2014

D H

Staff Emeritus
It should be obvious that mechanical energy is being lost here. The chain starts at rest at some height above the pan and ends at rest on the pan. Net change in kinetic energy is zero, but there's a net loss in potential energy. (The energy stored in the spring balance does not match the initial potential energy of the suspended chain.) So where does that energy go? Heat. But that's irrelevant.

It's perpendicular to the surface. That's why it's called the normal force. The chain is also falling perpendicular to the surface. The scale does work on the chain to slow down the part that is hitting the chain.

17. Apr 19, 2014

voko

I do not think I suggested that.

The pan of the scales is stationary (not true at the beginning and the end of process, but we are not analysing those regimes here).

I think you repeated your previous argument here. Mechanical energy is very clearly lost because a chain falling from rest comes to a halt without effecting any other motion or change elsewhere.

18. Apr 19, 2014

It's Kleppner and Kolenkow

19. Apr 19, 2014

Doesn't the scale display the normal force it exerted on the object?
$$E_2-E_1=W_{nc}$$
Because the energy after the collision is zero, this reduces to:
$$-E_1=W_{nc}$$
$$-E_1=-\frac{mgx}{\ell}h=-\frac{mgx}{\ell}(\ell-x)=\int_a^bN\cdot dr$$
I'm guessing the line integral on the right reduces to $N\ell$?
So if we are looking for a function $N(x)$ then wouldn't this be correct?
Solving for $N(x)$ we get:
$$N(x)=-\frac{mgx}{\ell^2}(\ell-x)$$

I'm sure this is incorrect though.

The work kinetic energy theorem can be used to see the work done by the normal force on the chain, however it gives us velocity rather than position:

$$W=\Delta K=K_2-K_1$$
$$W_N=0-\frac{mv^2x}{2\ell}$$
$$W_N=-\frac{mv^2x}{2\ell}$$

20. Apr 19, 2014

D H

Staff Emeritus
No.

Suppose instead of holding the chain suspended and then letting it drop we hold the chain in a lump at some height above the pan and then drop the chain as a lump. Let's do this twice, once with the chain a tiny, tiny distance above pan and another time at a height $\ell/2$ above the pan. There's a net loss of mechanical energy even in the first test. The gravitational potential energy drops by $M^2g^2/k$ and the spring's potential energy increases by but half this amount. The other half? It went into heat.

The second experiment has the same initial and final conditions as the question posed in the original post. Here there's an even larger drop in gravitational potential energy while the amount stored by the spring hasn't changed a bit.