Irodov 1.292: Where the Spinning Rod Snaps First – Amit & Deepali Sinha Foundation

Problem Irodov 1.292

A copper rod of length $l = 1.0$ m spins in a horizontal plane about a vertical axis passing through its own midpoint. The rod ruptures when the stress in it, the internal tension per unit cross-sectional area, reaches copper’s tensile strength. Find the number of complete revolutions per second at which this happens.

The rod of length l, seen from above, spinning about the vertical axis through its center O. — The rod of length $l$ , seen from above, spinning about the vertical axis through its center $O$ .

Given

$l = 1.0$ m
copper

Find

n

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

Solution

Step 1The tension a cross-section must carry

The same rod with a slice dr at distance r from O. The slice needs a centripetal pull, supplied by the tension in the rod on either side of it. — The same rod with a slice $dr$ at distance $r$ from $O$ . The slice needs a centripetal pull, supplied by the tension in the rod on either side of it.

Cut the rod at distance $r$ from the axis $O$ and look at the piece running from $r$ out to the free end of that half, at $l/2$ . Every slice $dr'$ of that outer piece has mass $dm = \rho A\,dr'$ , where $A$ is the rod’s cross-sectional area and $\rho$ its density, and needs a centripetal force $dm\,\omega^2 r'$ directed back toward $O$ . The cross-section at $r$ carries the sum of every such pull further out.

T(r) = \int_r^{l/2} \rho A \omega^2 r'\,dr' = \frac{1}{2}\rho A \omega^2\left(\frac{l^2}{4} - r^2\right)

Step 2Where the stress is greatest

The tensile stress at $r$ is $\sigma(r) = T(r)/A$ , which no longer depends on $A$ :

\sigma(r) = \frac{1}{2}\rho\omega^2\left(\frac{l^2}{4}-r^2\right)

(1)

This is largest at the center, $r = 0$ , where the pull of both halves passes through a single cross-section:

\sigma_{\max} = \sigma(0) = \frac{1}{8}\rho\omega^2 l^2

Step 3The rupture condition

The rod fails once $\sigma_{\max}$ reaches copper’s tensile strength $\sigma_m$ . Setting $\sigma_{\max} = \sigma_m$ in (1) taken at $r=0$ and solving for $\omega$ isolates the critical angular velocity.

\begin{align*}\frac{1}{8}\rho\omega^2 l^2 &= \sigma_m \\ \omega^2 &= \frac{8\sigma_m}{\rho l^2} \\ \omega &= \frac{2}{l}\sqrt{\frac{2\sigma_m}{\rho}}\end{align*}

What the problem asks for is the rate in complete revolutions per second, and one revolution is $2\pi$ radians, so the two rates are related by $\omega = 2\pi n$ .

\begin{align*}n &= \frac{\omega}{2\pi} = \frac{1}{2\pi}\cdot\frac{2}{l}\sqrt{\frac{2\sigma_m}{\rho}} \\ n &= \frac{1}{\pi l}\sqrt{\frac{2\sigma_m}{\rho}}\end{align*}

Step 4Numbers

Copper: $\rho = 8.9\times10^3$ kg/m $^3$ , $\sigma_m = 2.6\times10^8$ Pa. With $l = 1.0$ m, evaluate $2\sigma_m/\rho$ , take its square root, and divide by $\pi l$ to get $n$ in revolutions per second.

\frac{2\sigma_m}{\rho} = \frac{2\times2.6\times10^8}{8.9\times10^3} = 5.84\times10^4\ \text{m}^2/\text{s}^2, \qquad \sqrt{5.84\times10^4} = 2.42\times10^2\ \text{m/s}.

n = \frac{2.42\times10^2}{\pi\times1.0} = 76.9\ \text{rps}.

n = \frac{1}{\pi l}\sqrt{\frac{2\sigma_m}{\rho}} \approx 0.8\times10^2\ \text{rps}

(2)

Check

$\sigma_m/\rho$ has units Pa/(kg/m $^3$ ) = m $^2$ /s $^2$ , so $\sqrt{\sigma_m/\rho}/l$ carries units of s $^{-1}$ , correct for $n$ . Doubling $l$ halves $n$ : a longer rod develops more stress at the same spin rate, so it snaps sooner.

Remark

Neither $\sigma_{\max}$ nor $n$ depends on the rod’s cross-sectional area: it canceled out of $\sigma(r)=T(r)/A$ on the way to (1) because both the centripetal demand $T(r)$ and the area carrying it scale together. A thicker copper rod of the same length is therefore no more resistant to this kind of rupture than a thin one; the only levers that matter are the material’s specific strength $\sigma_m/\rho$ and the length $l$ , which is why the answer never needed the rod’s diameter at all.