Q:

Center of mass of heavy spring

If a heavy spring of uniform density and mass $m$ is hung vertically so that it is stretched by gravity under its own weight, where is the center of mass? 

I have had a few people tell me it is 1/2 the distance from the center of mass at equilibrium distance but this seems wrong to me. Thanks for any help.

A:

We can think of the spring as a series of masses and springs: 

Let's say we have $N$ masses, each of mass $m$. the total mass of the spring is $M = Nm$. each small spring will have a constant $k$. the springs are connected in a row, so that the total spring constant is: $$K = \frac {1}{\frac {1}{k} + \frac {1}{k} + ...} = \frac {k}{N}$$ Let's write the forces for each mass in equilibrium:
$$m_1:\ k\Delta x_1 - mg - k\Delta x_2 = 0$$
$$m_2:\ k\Delta x_2 - mg - k\Delta x_3 = 0$$
$$ \vdots$$
$$m_N:\ k\Delta x_N - mg = 0$$ 
Notice that each mass only feels it's own weight, and the springs attached to it. the last mass is only connected to one spring. 
We can add all $N$ equations to get:
$$ k\Delta x_1 - mNg = 0$$
$$ \Delta x_1 = \frac{Mg}{k}$$
To get $\Delta x_2$ we can plug this in to the second equation. in this manner we can get the general $\Delta x_i$: $$\Delta x_i = \frac {(M-(i-1)m)g}{k} = \frac {(N-i+1)mg}{k}$$ 
from here we can get anything we need, like the total elongation of the spring:
\begin{align}\Delta x &= \sum_{i=1}^{N} \Delta x_i \\&= \sum_{i=1}^{N} \frac {(N-i+1)mg}{k} \\&= \frac {(N+1)mg}{k} \sum_{i=1}^{N}  -\frac {mg}{k} \sum_{i=1}^{N} i \\&= \frac {N(N+1)mg}{k}  -\frac {mgN(N+1)}{2k} \\&= \frac {mgN(N+1)}{2k}\end{align} 
in the limit of many strings and small masses $N \gg 1$:
$$\Delta x = \frac {mgN^2}{2k} = \frac {Mg}{2K}$$ 
and this is the total elongation of the spring. if we want the center of mass it takes a bit more work but the concept is the same:
\begin{align} M_\text{cm} &=\frac{1}{M} \sum_{i=1}^N m\left(\frac {Li}{N} + \sum_{j=1}^i\Delta x_j\right)\end{align}
where $L$ is the length of the string without any force acting on it, so that $\frac {Li}{N}$ is the location of the $i$th mass when no forces act on the string. $\sum_{j=1}^i\Delta x_j$ is the total elongation of the spring up to the mass $i$.
\begin{align} M_\text{cm} &=\frac {1}{M}m  \left(\frac {L}{N}\sum_{i=1}^Ni + \sum_{i=1}^N\sum_{j=1}^i\Delta x_j\right) \\&=
 \frac {1}{M}m  \left(\frac {L}{N}\sum_{i=1}^Ni + \sum_{i=1}^N\sum_{j=1}^i\frac {(N-j+1)mg}{k}\right) \\&=
 \frac {1}{M}m  \left(\frac {L}{N}\sum_{i=1}^Ni + \frac {(N+1)mg}{k}\sum_{i=1}^N\sum_{j=1}^i - \frac {mg}{k}\sum_{i=1}^N\sum_{j=1}^ij\right) \\&= 
\frac {1}{M}m  \left(\frac {LN(N+1)}{2N} + \frac {(N+1)mg}{k}\sum_{i=1}^Ni - \frac {mg}{k}\sum_{i=1}^N \frac{i(i+1)}{2}\right) \\&= 
\frac {1}{M}m  \left(\frac {L(N+1)}{2} + \frac {N(N+1)^2mg}{2k} - \frac {mg}{2k} \left(\sum_{i=1}^Ni^2+\sum_{i=1}^Ni\right)\right) \\&= 
\frac {1}{M} m  \left(\frac {L(N+1)}{2} + \frac {N(N+1)^2mg}{2k} - \frac {mg}{2k} \left(\frac{N(N+1)(2N+1)}{6}+\frac{N(N+1)}{2}\right)\right) \\&= \frac {1}{M}m  \left(\frac {LN}{2} + \frac {N^3mg}{2k} - \frac {mg}{2k} \left(\frac{N^3}{3}+\frac{N^2}{2}\right)\right)\;\; \because N\gg 1 \\&=
   \frac {1}{M}\left(\frac {LM}{2} + \frac {N^3m^2g}{2k} -  \frac {m^2g}{2k}\frac{N^3}{3}-\frac {m^2g}{2k}\frac{N^2}{2}\right) \\&=
 \frac {1}{M}\left(\frac {LM}{2} + \frac {M^2g}{2K} -  \frac {M^2g}{6K}-\frac {Mg}{4NK}\right) \\&=
\end{align}
for $N\gg 1$ the last term is small:
$$\boxed{ M_\text{cm} =
 \frac {L}{2} + \frac {Mg}{3K}} $$
 if the total mass is small, we get that the center of mass of the hung spring is just the same as the center of mass of the spring at rest:$ \frac {L}{2}  $, that makes sense.  
I'm pretty sure my logic is correct, let me know if it's not, or if I have any calculation mistakes.  

