Massive springs

By Andrew Pontzen (P4476) on Wednesday, May 16, 2001 - 10:46 pm :

A uniform "slinky" spring of mass M, unstretched length L and force constant k rests in a smooth, horizontal tube. A horizontal force is applied at one end, along the spring's axis, so as to _pull_ the spring along the tube.

The force is gradually increased until the spring moves with the same horizontal acceleration a along its whole length. What is then the length of the spring?

Worryingly, I can't get a sensible solution. I must be doing something stupid. Here is what I've done:

Consider a section mass

δ m

, length

δ x

when unstretched and length

δ y

when stretched.

The difference in force acting on the front and back of this section is $δ F$ .

$\Rightarrow δ F = (δ m) a$

and $δ F = k L (δ y - δ x) / (δ x)$

When the spring is unstretched, the mass is evenly distributed

$\Rightarrow δ x = L (δ m / M)$

$\Rightarrow δ F = k M (δ y / δ m) - k L$

$\Rightarrow (δ m) a = k M (δ y / δ m) - k L$

$\Rightarrow δ y / δ m = ((δ m) a + k L)$

Let $δ m \to 0$

Then we have

$(dy / dm) = (k L) / (k M) = L / M$

Integrate both sides wrt $m$ from 0 to $M$ , to get $l = L$ , which can't be right.
This is my first post on this board, so please forgive me if I'm covering old ground - also I know this is meant to be maths rather than physics, but I think this question could just have easily appeared in the mechanics section of a maths STEP.

Thanks in advance for any help.

By Michael Doré (Md285) on Thursday, May 17, 2001 - 01:16 am :

Hi Andrew, welcome to NRICH :)

I think the error is in the statement:

$δ F = k L (δ y - δ x) / δ x$

I'm not really sure where this has come from, but if it were true then the force on the slinky would be massive. For instance suppose the slinky was locally extended to twice its natural length then $δ F$ would be of the order $k L$ . And so the total force on the slinky would be of order $k L \times L / δ x$ which tends to infinity in the limit.

Anyway, here is one approach. Suppose the slinky has ends A and B and is being dragged in the direction from A to B at a constant acceleration. We're given that all points on the spring have the same acceleration. For convenience, we'll call $P_{x}$ the point so that the natural length of the spring between A and $P_{x}$ is $x$ . Let the tension at $P_{x}$ be $T (x)$ and let the distance between A and $P_{x}$ be $D (x)$ (so that $D (L)$ = the total length of slinky).

Now consider a small strip of slinky between $P_{x}$ and $P_{x + δ x}$ . The mass of this is $M δ x / L$ so by $F = m a$ , the total force on this strip is:

$M a δ x / L$

But we also know that the total force on this strip is $T (x + δ x) - T (x)$ so we obtain:

$(T (x + δ x) - T (x)) = M a δ x / L$

Upon taking $δ \to 0$ :

$T' (x) = M a / L$

So $T (x) = M a x / L +$ constant, but the tension at A is clearly 0, so $T (x) = M a x / L$ .

Now once again, divide up the spring into small strips of length $δ x$ . What is the actual distance between $P_{x}$ and $P_{x + δ x}$ ? Well the natural length of this strip is $δ x$ , and its extension is given by $T (x) δ x / (k L)$ because the spring constant of this strip is $k L / δ x$ . Notice that I've assumed that the tension in this region is around about $T (x)$ . In fact the tension does vary slightly between $P_{x}$ and $P_{x + δ x}$ , but the error is of order $T' (x) δ x$ . This is clearly not significant in the limit. (The error term in the extension of the strip is second order.)

So:

$D (x + δ x) - D (x) = T (x) δ x / (k L) + δ x$

And:

$D' (x) = T (x) / k L + 1 = M a x / ({kL}^{2}) + 1$

Therefore:

$D (x) = M a x^{2} / (2 k L^{2}) + x +$ const

but the constant is clearly 0, so the final length is:

$D (L) = M a / (2 k) + L$

By Kerwin Hui (Kwkh2) on Thursday, May 17, 2001 - 12:55 pm :

I believe the equation

$δ F = k L (δ y - δ x) / δ x$

is perfectly OK. It just comes from $k = λ / L$ . The main problem was that you assumed mass was uniform to work out an expression for dy/dm.

Kerwin

By Andrew Pontzen (P4476) on Thursday, May 17, 2001 - 07:48 pm :

Thanks for all the help.

I believe that

δ F = k L (δ y - δ x) / δ x

is perfectly valid and, at risk of showing myself to be even more stupid, I still can't see how it is incorrect to assume that dx/dm is constant. Clearly the mass distribution won't be constant when the spring is stretched, but I don't see why it shouldn't be when it is at its natural length.
Thanks for your help

By Michael Doré (Md285) on Thursday, May 17, 2001 - 11:22 pm :

Hi,

I still don't believe the $δ F = k L (δ y - δ x) / δ x$ equation. I think you're using tension $= k L \times$ extension of strip/natural length of strip. So shouldn't the $δ F$ be an $F$ ? Why would it be the difference in tension?

Regards,

Michael

By Andrew Pontzen (P4476) on Friday, May 18, 2001 - 07:01 pm :

Ah... yes that makes sense. And I should have spotted straight away that it was wrong because if you let

δ x \to 0

then you get

0 = k L (dy / dx) - 1

which is evidently wrong.
I think this clears it all up. I will try solving the problem with my approach without that erratic line.

Thanks everyone!