Irodov 1.296: A Rod Spun from One End – Amit & Deepali Sinha Foundation

Problem Irodov 1.296

A thin, uniform copper rod of length $l$ and mass $m$ spins at a steady angular velocity $\omega$ in a horizontal plane about a vertical axis through one of its own ends. Find how the internal tension $T(r)$ varies with distance $r$ from that end, and find the rod’s total elongation $\Delta l$ .

The rod of length l, seen from above, pivoted at O, one of its own ends. — The rod of length $l$ , seen from above, pivoted at $O$ , one of its own ends.

Given

rod of length $l$ and mass $m$
rotates at angular velocity $\omega$ about a vertical axis through one end

Find

T(r)

, and the elongation

\Delta l

Solution

Step 1The tension a cross-section must carry

A slice dr at distance r from O needs a centripetal pull, supplied by the tension in the rod inward of it, toward O. — A slice $dr$ at distance $r$ from $O$ needs a centripetal pull, supplied by the tension in the rod inward of it, toward $O$ .

Let $A$ be the rod’s cross-sectional area and $\rho$ its density, so the mass per unit length is $m/l = \rho A$ . Cut the rod at distance $r$ from the pivot $O$ ( $0\le r\le l$ ) and consider the piece running from $r$ out to the free end at $l$ . A slice $dr'$ of that piece has mass $(m/l)\,dr'$ and needs a centripetal force $(m/l)\,\omega^2 r'\,dr'$ directed back toward $O$ . The cross-section at $r$ supplies the sum of all such pulls further out:

T(r) = \int_r^{l}\frac{m}{l}\omega^2 r'\,dr' = \frac{m\omega^2}{2l}\left(l^2-r^2\right).

T(r) = \frac{1}{2}m\omega^2 l\left(1-\frac{r^2}{l^2}\right)

(1)

Step 2The elongation, by integrating the local strain

Each slice stretches according to its own local stress $T(r)/A$ , so the total elongation is the integral of the local strain $T(r)/(EA)$ over the rod’s length:

\Delta l = \int_0^l \frac{T(r)}{EA}\,dr = \frac{m\omega^2 l}{2EA}\int_0^l\left(1-\frac{r^2}{l^2}\right)dr = \frac{m\omega^2 l}{2EA}\cdot\frac{2l}{3} = \frac{m\omega^2 l^2}{3EA}. \tag{2}

Step 3Eliminating the cross-section

The area does not appear in the given data, so replace it in (2) using $A = m/(\rho l)$ , from the linear density $m/l=\rho A$ :

\Delta l = \frac{m\omega^2 l^2}{3E}\cdot\frac{\rho l}{m}

\Delta l = \frac{1}{3}\frac{\rho\omega^2 l^3}{E}

(3)

Check

At $r=l$ , $T(l)=0$ by the boxed formula: the free end carries no load, which is exactly what “free end” means. And $\Delta l\to0$ as $\omega\to0$ , as it must for a rod that is not spinning.

Remark

Written in terms of the rod’s density and cross-section rather than its total mass, using $m=\rho A l$ , the peak tension is $T(0)=\rho A\omega^2 l^2/2$ , quadratic in the length $l$ , while the elongation $\Delta l=\rho\omega^2 l^3/(3E)$ is cubic in $l$ . For a rod of the same material and thickness, doubling the length quadruples the peak stress but multiplies the stretch eightfold: $T(0)$ is quadratic in $l$ because both the amount of material and the radius at which it sits grow with $l$ , and $\Delta l$ gains one further power because $T(r)/(EA)$ is then integrated over the length.