Length of a spring under its own weight

[gabee]

Homework Statement

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?

Homework Equations

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?

 

[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.

 

[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.

 

[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.

 

[chaoseverlasting]

Homework Statement

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?

Homework Equations

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?

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).

 

[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