Irodov 1.308: Torque That Spins the Outer Ring – Amit & Deepali Sinha Foundation

Problem Irodov 1.308

A uniform ring of mass $m$ and outer radius $r_2$ is fitted tightly onto a shaft of radius $r_1$ , so the ring turns rigidly with the shaft. The shaft, and with it the ring, is rotated about its axis with a constant angular acceleration $\beta$ . Find the internal torque (the moment of the elastic forces) transmitted across a circular cross-section of the ring at distance $r$ from the rotation axis, with $r_1 \le r \le r_2$ , as a function of $r$ .

The ring's cross-section: shaft radius r_1, outer radius r_2, spun with angular acceleration β. The dashed circle marks the cut at radius r. — The ring’s cross-section: shaft radius $r_1$ , outer radius $r_2$ , spun with angular acceleration $\beta$ . The dashed circle marks the cut at radius $r$ .

Given

$m$
$r_1$
$r_2$
$\beta$

Find

N(r)

, the internal torque across the ring at radius

r

Solution

Step 1Mass of a thin ring element

The ring element dm, of radius r' and width dr', lying beyond the cut at radius r (dashed). — The ring element $dm$ , of radius $r'$ and width $dr'$ , lying beyond the cut at radius $r$ (dashed).

Being uniform, the ring’s areal density is its mass over its own area; a thin ring of radius $r'$ and width $dr'$ then carries a mass proportional to its own area $2\pi r'\,dr'$ .

\sigma = \frac{m}{\pi(r_2^2-r_1^2)}

dm = \dfrac{2mr'}{r_2^2-r_1^2}\,dr'

(1)

Step 2The torque that spins the material beyond

r

Cut the ring at radius $r$ . The outer piece, from $r$ to $r_2$ , has a free rim carrying no external torque, so the only torque acting on it is the internal, elastic torque $N(r)$ transmitted across the cut by the material inside $r$ . Since the whole ring turns rigidly with the shaft, that outer piece must share the shaft’s angular acceleration $\beta$ , so $N(r)$ equals $\beta$ times that piece’s own moment of inertia about the axis.

N(r) = \beta\int_r^{r_2} r'^2\,dm

Step 3Evaluating the integral

Substituting the mass element (1) and carrying out the integral.

\begin{align*}N(r) &= \beta\int_r^{r_2} r'^2\cdot\frac{2mr'}{r_2^2-r_1^2}\,dr' \\ &= \frac{2m\beta}{r_2^2-r_1^2}\int_r^{r_2} r'^3\,dr' \\ &= \frac{2m\beta}{r_2^2-r_1^2}\cdot\frac{r_2^4-r^4}{4}\end{align*}

N(r) = \dfrac{1}{2}\beta m\,\dfrac{r_2^4-r^4}{r_2^2-r_1^2}

(2)

Check

At the free outer rim, $r=r_2$ , the boxed result gives $N(r_2)=0$ directly, as it must: there is no material beyond $r_2$ for any torque to act on.

Remark

At the other extreme, $r=r_1$ , the boxed result gives $N(r_1) = \beta m(r_2^2+r_1^2)/2$ , using $r_2^4-r_1^4=(r_2^2-r_1^2)(r_2^2+r_1^2)$ . That is exactly $\beta$ times the whole ring’s own moment of inertia about its axis, $I = m(r_1^2+r_2^2)/2$ , a standard annulus result: at the shaft itself the elastic torque is the one that has to accelerate the whole ring, because everything from $r_1$ outward is ring material and the shaft is the only thing driving any of it. Note also that no instant is privileged, since $\omega$ never enters: the same $N(r)$ holds at every moment of the spin-up, not only at the start.