This article is translated from a Chinese article on my Zhihu account. The original article was posted at 2019-12-18 11:20 +0800.

Review middle school contents!

Suppose a uniform heavy elastic rope has mass mm, original length L0L_0, and stiffness kk. Find the mass distribution and length of it when hung vertically.

When the elastic rope has its original length, divide it into nn equal segments, where nn is large so that when the elastic rope is stretched, the mass distribution within every segment is uniform.

When the rope is hung vertically, the tension experienced by every segment is

Fj=mgn(n−j).F_j=\frac{mg}n\left(n-j\right).

Therefore, the length of every segment is

Δlj=Fjnk+L0n=mg(n−j)n2k+L0n.\Delta l_j=\frac{F_j}{nk}+\frac{L_0}n=\frac{mg\left(n-j\right)}{n^2k}+\frac{L_0}n.

Therefore, the position of the end of every segment is

xj=∑s=1jΔls=2nL0kj+2nmgj−mgj−mgj22n2k.x_j=\sum_{s=1}^j\Delta l_s=\frac{2nL_0kj+2nmgj-mgj-mgj^2}{2n^2k}.

The equation can be solved to get

j=2nmg+2nL0k−mg±(mg−2nL0k−2nmg)2−8n2mgkxj2mg.j=\frac{2nmg+2nL_0k-mg\pm\sqrt{\left(mg-2nL_0k-2nmg\right)^2-8n^2mgkx_j}}{2mg}.

The (linear) mass density of every segment is

ρj=mnΔlj=mn(mg(n−j)n2k+L0n).\rho_j=\frac{m}{n\Delta l_j}=\frac{m}{n\left(\frac{mg\left(n-j\right)}{n^2k}+\frac{L_0}n\right)}.

Express jj here with xjx_j, and we have

ρj=2nmkmg∓(2nL0k+mg(2n−1))2−8n2mgkxj.\rho_j=\frac{2nmk}{mg\mp\sqrt{\left(2nL_0k+mg\left(2n-1\right)\right)^2-8n^2mgkx_j}}.

The length of the rope is

L=xn=mg(n−1)2nk+L0.L=x_n=\frac{mg\left(n-1\right)}{2nk}+L_0.

Take the limit n→∞n\to\infty, and we have

L=mg2k+L0,L=\frac{mg}{2k}+L_0,

ρ ⁣(x)=±mkL02k2+m2g2+2mgk(L0−x).\rho\!\left(x\right)=\pm\frac{mk}{\sqrt{L_0^2k^2+m^2g^2+2mgk\left(L_0-x\right)}}.

The plus-minus sign here should take plus sign.

To test, calculate ∫0Lρ ⁣(x)dx=m\int_0^L\rho\!\left(x\right)\mathrm dx=m.

The conclusion can be written in a more beautiful way:

ρ ⁣(x)=m(2L−L0)2−4(L−L0)x.\rho\!\left(x\right)=\frac{m}{\sqrt{\left(2L-L_0\right)^2-4\left(L-L_0\right)x}}.