# 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.

• Why does it seem wrong to you? Commented Dec 2, 2015 at 17:10
• When the spring is hanging, it will stretch less at the bottom and more at the top. So when it is stretched the mass per unit length will be greater near the bottom than near the top. So it seems like the center of mass would need to be lower on the spring when it is hanging than when it is not. Commented Dec 2, 2015 at 17:15
• And that is much better insight into the problem than we usually get - congratulations! So, given that (which is correct!), can you write an equation for the stretch of the spring? Commented Dec 2, 2015 at 17:28
• Yes, I set up an integral and found that the length from equilibrium is equal to mg/(2k). Commented Dec 2, 2015 at 17:31
• A simple Chain would form a Catenary curve. Somehow I think that your "spring-chain" transfers the shape back to a parabola. Commented Dec 7, 2015 at 20:27

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.

• Nice answer, I think the last equation should be divided by $M$ (the whole mass) to get the center of mass. Commented Jan 1, 2016 at 3:36
• I'm getting a factor of $\frac{mg}{6k}$ in contrast to your $\frac{mg}{3k}$, not sure whose answer is right, may be you can look into my answer? Commented Jan 1, 2016 at 8:52
• the effective mass concept will not give you the right center of mass. it's just for oscillation calculations Commented Jan 1, 2016 at 13:28
• I don't think it's just applicable for oscillation calculations because there is no special reason for it to behave so. Commented Jan 1, 2016 at 15:26
• When you are finding the effective mass you are finding a system of a mass on a spring that has the same total kinetic energy as your system. The two systems are not equivalent in every way - they just have the same total kinetic energy! another thing - you can't say that "the center of mass of the massless spring is L/2", if it has no mass it doesn't have a center of mass.... Commented Jan 1, 2016 at 19:10

Consider the unstretched length to be $\ell$ and the total stretching to be $\delta$. The center of mass is going to be located at $c = \frac{\ell+\delta}{2}$ if the cross section remains the same through out. But I suspect you want to consider the necking that happens when you stretch something so the final deflection won't be $$\delta = \frac{\rho g \ell^2}{2 E}$$ but something different. Above $E$ is the modulus of elasticity, $\rho$ is the mass density and $g$ is gravity.

Using Hook's Law the new section area under load $T$ is

$$A' = \frac{(A E - \nu T)^2}{A E^2}$$

So the linear density of the rope is no longer $\lambda = \rho A$ but $$\lambda' = \rho \frac{(A E - \nu T)^2}{A E^2}$$

So the standard extension $u(x)$ due to gravity equation $\frac{{\rm d^2}u}{{\rm d}x^2} = \frac{g \lambda}{EA} =\frac{g \rho}{E}$ becomes

$$\frac{{\rm d^2}u}{{\rm d}x^2} = \frac{\rho g}{E} \left(1- \nu \frac{{\rm d}u}{{\rm d}x}\right)^2$$

This is solved by separation of variables (using $\epsilon =\frac{{\rm d}u}{{\rm d}x}$) as

$$\int \frac{{\rm d}\epsilon}{(1-\nu \epsilon)^2} = \int \frac{\rho g}{E} {\rm d}x + K$$ $$\frac{1}{\nu (\nu \epsilon-1)} = \frac{\rho g}{E}x + K$$

The constant of integration $K$ is found by setting the tension $T =(E A) \frac{{\rm d}}{{\rm d}x} u = (E A) \epsilon$ to zero when $x=\ell$. This makes $K=\frac{1}{\nu} - \frac{\rho g \ell}{E}$ and the above can be solved for the tension

$$T(x) = E A \theta = \frac{E A \rho g (\ell-x)}{E+ \nu \rho g (\ell-x)}$$

Now the total deflection is found by $\delta = \int_0^\ell \frac{T}{E A}\,{\rm d}x$ which ends up being

$$\delta = \frac{\nu \rho g \ell - E \ln \left(1+ \frac{\nu \rho g \ell}{E}\right)}{\rho g \nu^2}$$

To check this we take the limit of $\nu \rightarrow 0$ and it makes $\delta = \frac{\rho g \ell^2}{2 E}$ which is the standard extension with considering the necking of a stretched rope.

The expression above is a little difficult to deal with, and we can simplify it by considering situation where $E \gg \nu \rho g \ell$.

$$\boxed{ \delta \approx \frac{3 \rho g \ell^2}{2 ( 3 E + 2 \nu g \rho \ell)} }$$

The center of mass is found by

$$c \approx \frac{\ell+\delta}{2}$$

Let the spring of Young's modulus $E$, density $\rho$, with a section $S$

$E$ and $\rho$ are related by $$E=\frac{k\rho L^2}{m}$$

$$\rho=\frac{m}{SL}$$

It is convenient to use as reference position, the spring as it would be if the gravity did not act: that is, vertical and uniform. Let us consider two sections $S$ and $S’$ of the spring. Two points of the spring $P$ on $S$ and $Q$ on $S‘$, at the reference positions $x$ and $x+\Delta x$, will be displaced, under the effect of the gravity, by the quantities $y$ and $y+\Delta y$ respectively

The mass element $\Delta x$ from $P$ to $Q$ is subjected to gravity and two tensions $T$ and $T+\Delta T$

$\Delta T+\rho S g \Delta x=0$

Since the extension of the element $\Delta y$ is given by the applied force $T+\Delta T$, we have

$$T+\Delta T=SE \frac{\Delta y}{\Delta x}$$

Processing to the limit as $\Delta x→0, dT/dx = − ρSg, T = S E \frac{dy}{dx}$, we have then:

$$SE\frac{d^2 y}{dx^2}=\rho S g$$

with the boundary conditions:

$y=0$ for $x=0$, $\frac{dy}{dx}=0$ for $x=L$

That is, the tension vanishes at the lower end, this gives

$y=\frac{\rho g L}{E}x-\frac{\rho g}{2E}x^2$

so, $$y=\frac{mg}{kL}x-\frac{mg}{2kL^2}x^2$$

for $x=\frac{L}{2}$ i.e for the center

$y=\frac{3mg}{8k}$

So the center of mass will be at the point $$\frac{L}{2}+\frac{3mg}{8k}$$

• The second term of the anser dosn't units of distance Commented Jan 2, 2016 at 13:39