Irodov 1.139: Falling Sleeve Stretches the Rubber Cord

Problem Irodov 1.139

An elastic (Hooke’s law) cord of natural length $l$ and elasticity coefficient $k$ hangs from a fixed point $O$ , with a small catch $B$ fixed to its lower end. A sleeve of mass $m$ is threaded onto the cord and released from rest at $O$ ; it slides freely down the cord (the cord being smooth, and the masses of the cord and catch negligible) until it reaches the catch at the cord’s natural length $l$ below $O$ . From that point on the sleeve is caught by $B$ and continues to descend, now stretching the cord elastically. Find the cord’s maximum elongation beyond its natural length $l$ .

The cord hangs from O at its natural length l, with the catch B at its lower end. The sleeve A starts from rest at O and slides freely down the cord to B, then stretches it further. — The cord hangs from $O$ at its natural length $l$ , with the catch $B$ at its lower end. The sleeve $A$ starts from rest at $O$ and slides freely down the cord to $B$ , then stretches it further.

Given

Find

the maximum elongation of the cord

Solution

Step 1Two phases, one energy balance

Let $x$ denote the elongation of the cord beyond its natural length $l$ at the lowest point of the sleeve’s motion. The sleeve falls a total distance $l+x$ from $O$ to that point, and the cord obeys Hooke’s law, so its stored elastic energy there is $kx^2/2$ .

mg(l+x) = \frac{1}{2}kx^2. \tag{1}

Step 2Rearranging into a quadratic in

x

The combination $mg/k$ recurs throughout and has the dimension of a length: it is the static extension the cord would take under the sleeve’s weight alone. Naming it shortens every line that follows.

\alpha = \dfrac{mg}{k}

(2)

Multiplying (1) through by $2/k$ and writing $mg/k$ as $\alpha$ collects every term on one side.

\begin{align*}x^2 &= 2\alpha(l+x) \\ &= 2\alpha l + 2\alpha x\end{align*}

x^2 - 2\alpha x - 2\alpha l = 0

(3)

Step 3Solving the quadratic

Applying the quadratic formula to (3),

\begin{align*}x &= \frac{1}{2}\left[2\alpha \pm \sqrt{(2\alpha)^2 + 4\cdot2\alpha l}\right] \\ &= \alpha \pm \sqrt{\alpha^2 + 2\alpha l}\end{align*}

The elongation is a length and so cannot be negative; since the square root exceeds $\alpha$ , only the $+$ root is physical.

x = \alpha + \sqrt{\alpha^2+2\alpha l}

Step 4Factoring out the static extension

Factoring $\alpha$ out of the square root puts the result in the book’s dimensionless form.

x = \alpha\left(1+\sqrt{1+\frac{2l}{\alpha}}\right)

Substituting back $\alpha = mg/k$ from (2), so that $2l/\alpha = 2kl/mg$ ,

\Delta l = \left(1+\sqrt{1+\dfrac{2kl}{mg}}\right)\dfrac{mg}{k}

(4)

Check

Dimensionally $\alpha=mg/k$ is a length (since $kx$ is a force), matching $\Delta l$ . In the stiff-cord limit $k\to\infty$ the term $2kl/mg$ dominates under the root and $\Delta l\to\sqrt{2mgl/k}\to0$ , as a very stiff cord should barely stretch at all despite the fall.

Remark

Setting $l=0$ in the answer, the sleeve starting right at the cord’s already-attached end with no free fall first, gives $\Delta l=\alpha(1+\sqrt{1})=2\alpha$ , exactly twice the static extension $\alpha=mg/k$ a mass $m$ would produce hanging at rest on the same cord. This is the familiar “sudden loading doubles the deflection” result from impact mechanics; the extra fall height $l$ here only makes the overshoot beyond $2mg/k$ larger still.