Irodov 1.184: The Hanging Length Stays Fixed – Amit & Deepali Sinha Foundation

Problem Irodov 1.184

A uniform chain of total length $l$ lies partly inside a smooth horizontal tube whose exit bends sharply downward at a fixed height $h$ above a table. A length $l-h$ of the chain lies straight inside the tube, held there by its end $A$ , while the remaining length $h$ hangs vertically below the tube’s bend, its lower end $B$ just touching the table. At a certain moment end $A$ is released. Find the speed with which end $A$ leaves the tube (that is, the speed of the chain at the instant the last of it has passed the bend).

Fig. 1.47. The chain AB of total length l: a length l-h inside the smooth horizontal tube, held at end A, and a length h hanging below the bend, with end B just touching the table. — Fig. 1.47. The chain $AB$ of total length $l$ : a length $l-h$ inside the smooth horizontal tube, held at end $A$ , and a length $h$ hanging below the bend, with end $B$ just touching the table.

Given

$l$
$h$
smooth tube and table

Find

speed of end

A

as it leaves the tube

Solution

Step 1Only a length

h

ever drives the motion

Let $x$ be the length of chain still lying in the tube, starting at $x_0=l-h$ and decreasing to $0$ as $A$ slips through. With the chain’s mass per unit length $\lambda=m/l$ , the mass still in motion at any instant, the tube’s remaining stock plus the fixed hanging length $h$ , is

\mu(x) = \lambda(x+h).

Step 2The equation of motion for the shrinking moving length

Newton’s second law for the moving length, mass $\mu(x)=\lambda(x+h)$ driven by the weight $\lambda hg$ of the hanging part alone, gives

\lambda(x+h)\,\frac{dv}{dt} = \lambda h g.

The density $\lambda$ cancels, since both the driving weight and the moving mass scale with it identically.

(x+h)\,\dfrac{dv}{dt} = hg

(1)

Step 3Trading time for the shrinking length

x

Since $x$ decreases at exactly the chain’s own speed, $dx/dt=-v$ , the chain rule turns (1) into a separable equation in $v$ and $x$ .

\begin{align*}\frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt} &= -v\frac{dv}{dx} \\ -(x+h)\,v\,\frac{dv}{dx} &= hg \\ v\,dv &= -\frac{hg}{x+h}\,dx\end{align*}

Step 4Integrating from release to the tube’s exit

The chain starts from rest at $x=x_0=l-h$ and $A$ slips out of the tube exactly when $x=0$ , the tube’s stock exhausted. Integrating between these limits, with the sign absorbed by reversing them, gives the speed $v$ at that moment.

\begin{align*}\int_0^{v} v'\,dv' &= \int_{l-h}^{0}\left(-\frac{hg}{x+h}\right)dx = hg\int_0^{l-h}\frac{dx}{x+h} \\ \frac{v^2}{2} &= hg\Bigl[\ln(x+h)\Bigr]_0^{l-h} \\ &= hg\bigl(\ln l - \ln h\bigr)\end{align*}

v = \sqrt{2gh\ln\dfrac{l}{h}}

(2)

Check

Dimensionally $[gh]=\mathrm{m^2/s^2}$ and $\ln(l/h)$ is a pure ratio, so the square root is a speed, as required. At both ends of the allowed range the result vanishes: as $h\to l$ there is no length left in the tube to pull through, and $x_0=l-h\to0$ ; as $h\to0$ , $h\ln(l/h)\to0$ (the standard limit $h\ln h\to0$ ), so a vanishing driving weight cannot build any speed either.

Remark

Because the exit speed vanishes at both $h\to0$ and $h\to l$ , it is maximized at some intermediate fraction hung. Setting $d[h\ln(l/h)]/dh=\ln(l/h)-1=0$ gives $l/h=e$ , so the release speed is greatest when the fraction $h=l/e\approx0.368\,l$ is the one left hanging at the start, a fact this problem’s numbers never suggest but its own algebra reveals.