Length of a spring under its own weight

1. Jul 20, 2007

gabee

1. The problem statement, all variables and given/known data
I was thinking over lunch today about this old problem from intro physics but I can't remember whether or not this is the correct solution.

What is the final length L of a spring of mass M and spring constant k, initially of length L0 after it is left to hang under its own weight?

2. Relevant equations

3. The attempt at a solution
Divide the unstretched spring into infinitesimal segments dl. Each of these segments will stretch an infinitesimal distance dx under the weight of an infinitesimal mass dm by the relation g dm = k dx. Let $$\lambda$$ be the linear density of the spring, so that $$dm = \lambda \,dl$$. Then,

$$g \lambda \,dl = k \,dx$$, and
$$dx = \frac{g}{k}\lambda dl$$.

Integrating LHS from 0 to X and RHS from 0 to L0, we find that $$X = \frac{g}{k}\lambda L_0$$, so the final length of the spring is L = L0 + X or

$$L = L_0(1 + \frac{g}{k}\lambda) = \frac{g}{k}M + L_0$$.

Is that right?

Last edited: Jul 20, 2007
2. Jul 20, 2007

PhanthomJay

I'm no student of the calculus, but your solution is identical to the case where the spring is massless and its weight is concentrated at the far end. That doesn't sound right. I believe your extension X should be 1/2 of your calculated value, i.e. X =Mg/2k. Comments welcome.

3. Jul 20, 2007

chaoseverlasting

The com is at L/2, but only half of the spring is effectively pulling that mass up, so 2kx=mg, which gives you x=mg/2k.

4. Jul 20, 2007

gabee

Yeah, now that I think about it, that should be completely wrong, since the segments at the top are under greater load than the ones at the bottom. I'll think about it again in a bit.

5. Jul 20, 2007

chaoseverlasting

I think youre forgetting that each element has more than just dm*g acting on it (the masses of the elements below it are acting on each element too).

6. Jul 20, 2007

gabee

Right, that's what I mean. Okay, so including the fact that ALL the masses below a certain point contribute to the stretching of the segment at that point, we have

$$k \,dx = \bigg( \frac{l}{L_0} \bigg) g \, dm$$

$$dx = \frac{gl}{L_0 k} \frac{M}{L_0} \, dl$$

$$dx = \frac{Mgl}{{L_0}^2 k} \, dl$$

$$x = \frac{Mg}{{L_0}^2 k} \int_0^{L_0} l \, dl$$

$$x = \frac{Mg}{{L_0}^2 k} \frac{{L_0}^2}{2}$$

$$x = \frac{Mg}{2k}$$

Ahh, thanks guys

Last edited: Jul 20, 2007