Irodov 1.293: The Loop That Pulls Itself Apart – Amit & Deepali Sinha Foundation

Problem Irodov 1.293

A circular ring of lead wire, radius $r = 25$ cm, spins about a fixed vertical axis through its own center, perpendicular to the plane of the ring, so the ring itself stays in a horizontal plane throughout. Find the rotation rate, in revolutions per second, at which the wire ruptures.

The ring of radius r seen from above; the rotation axis is vertical through its center O, perpendicular to the page. — The ring of radius $r$ seen from above; the rotation axis is vertical through its center $O$ , perpendicular to the page.

Given

$r = 25$ cm
lead wire

Find

n

, the number of revolutions per second at which the ring snaps

Solution

Step 1The tension in the wire, from a half-ring balance

The upper half of the ring as the free body, cut along the horizontal diameter. The element at angle θ from the diameter subtends dθ; each element's centripetal demand dm ω^2r points at O, and the two tensions T that replace the removed lower half pull perpendicular to the diameter. — The upper half of the ring as the free body, cut along the horizontal diameter. The element at angle $\theta$ from the diameter subtends $d\theta$ ; each element’s centripetal demand $dm\,\omega^2r$ points at $O$ , and the two tensions $T$ that replace the removed lower half pull perpendicular to the diameter.

A slice of the ring subtending angle $d\theta$ has mass $dm = \rho A r\,d\theta$ , where $A$ is the wire’s cross-sectional area and $\rho$ its density, and needs a centripetal force $dm\,\omega^2 r$ directed toward the center $O$ . Take the upper half of the ring as a free body: its total centripetal demand is met by the tension $T$ pulling along the wire, and so perpendicular to the horizontal diameter, at the two points where that diameter cuts the ring.

Only the component of each element’s demand perpendicular to the diameter is carried by those two tensions; the components along the diameter cancel between the elements at $\theta$ and at $\pi-\theta$ . With $\theta$ measured from the diameter, that surviving component is the demand times $\sin\theta$ .

dF_\perp = (dm\,\omega^2 r)\sin\theta = \rho A r^2\omega^2\sin\theta\,d\theta

Summing it over the half ring, $\theta$ running from $0$ to $\pi$ , and setting the total equal to the two tensions:

2T = \int_0^\pi \rho A r^2\omega^2\sin\theta\,d\theta = \rho A r^2\omega^2\Bigl[-\cos\theta\Bigr]_0^\pi = 2\rho A r^2\omega^2

T = \rho A r^2\omega^2

(1)

Step 2Stress, and the rupture condition

The tensile stress in the wire is $\sigma = T/A$ , which no longer depends on $A$ :

\sigma = \rho r^2\omega^2

The ring fails once $\sigma$ reaches lead’s tensile strength $\sigma_m$ . Setting $\sigma=\sigma_m$ and solving for $\omega$ , then converting to revolutions per second with $\omega=2\pi n$ :

\omega^2 = \frac{\sigma_m}{\rho r^2} \qquad\Longrightarrow\qquad n = \frac{\omega}{2\pi} = \frac{1}{2\pi r}\sqrt{\frac{\sigma_m}{\rho}}

Step 3The rupture rate for lead wire

Lead: $\rho = 1.13\times10^4$ kg/m $^3$ , $\sigma_m = 1.5\times10^7$ Pa. With $r = 0.25$ m, evaluate $\sigma_m/\rho$ , take its square root, then divide by $2\pi r$ .

\begin{aligned} \frac{\sigma_m}{\rho} &= \frac{1.5\times10^{7}}{1.13\times10^{4}} = 1.33\times10^{3}\ \text{m}^2/\text{s}^2, \qquad \sqrt{\frac{\sigma_m}{\rho}} = 36.4\ \text{m/s}\\[4pt] n &= \frac{36.4}{2\pi\times0.25} = \frac{36.4}{1.571} = 23.2\ \text{s}^{-1} \end{aligned}

n = \frac{1}{2\pi r}\sqrt{\frac{\sigma_m}{\rho}} \approx 23\ \text{rps}

(2)

Check

$\sigma_m/\rho$ has units Pa/(kg/m $^3$ ) = m $^2$ /s $^2$ , so $\sqrt{\sigma_m/\rho}/r$ carries units s $^{-1}$ , correct for $n$ . Doubling $r$ halves $n$ : a larger ring develops more stress at the same spin rate, since the element’s centripetal demand $\rho A r^2\omega^2\,d\theta$ grows as $r^2$ , so it snaps sooner.

Remark

Neither $\sigma$ nor $n$ depends on the wire’s cross-sectional area $A$ : it canceled out of $\sigma=T/A$ because both the centripetal demand carried across a cut and the area carrying it scale together, so thickening the wire buys no extra spin rate at all. Written in terms of the rim speed $v=\omega r$ , the stress is simply $\rho v^2$ , so what a spinning loop really carries is a ceiling on rim speed, $\sqrt{\sigma_m/\rho}$ , about 36 m/s for lead and a few hundred meters per second for a high strength steel. The ratio $\sigma_m/\rho$ behind it, the specific strength, is the figure of merit for every flywheel and centrifuge rotor, and it is why such parts are rated by the speed of their rim rather than by revolutions per minute.