Irodov 1.92: Hoop Stress Splits into Two Causes – Amit & Deepali Sinha Foundation

Problem Irodov 1.92

A chain of total mass $m$ , closed into a circle of radius $R$ , rests on the smooth outer surface of a cone of half-angle $\theta$ , whose symmetry axis is vertical. The chain rotates about that vertical axis at a constant angular velocity $\omega$ , its circle staying horizontal at radius $R$ throughout. Find the tension in the chain.

The chain rests as a horizontal circle of radius R on the cone's smooth slant surface, rotating about the vertical axis at angular velocity ω. — The chain rests as a horizontal circle of radius $R$ on the cone’s smooth slant surface, rotating about the vertical axis at angular velocity $\omega$ .

Given

$m$
$R$
$\theta$
$\omega$

Find

the chain tension

T

Solution

Step 1The net inward pull of the tension on a short element

Consider a chain element subtending a small angle $d\varphi$ at the axis. Tension $T$ pulls tangentially at each end, and since the two pulls differ in direction by only $d\varphi$ , their vector sum does not cancel: it is a net force directed radially inward, toward the axis.

\left|\vect T_1+\vect T_2\right| = 2T\sin\frac{d\varphi}{2}.

For a vanishingly short element the sine may be replaced by its argument, and the factor of two cancels against the half.

\left|\vect T_1+\vect T_2\right| = T\,d\varphi

(1)

Step 2Newton’s second law for the element

Free body of one chain element, subtending angle dφ: the cone's normal reaction dN, the element's own weight dm g, and the net inward pull T dφ that Step 1 built from the tension at its two ends. — Free body of one chain element, subtending angle $d\varphi$ : the cone’s normal reaction $dN$ , the element’s own weight $dm\,\vect g$ , and the net inward pull $T\,d\varphi$ that Step 1 built from the tension at its two ends.

The element also carries the cone’s normal reaction $dN$ (perpendicular to the smooth slant, hence at angle $\theta$ above the horizontal) and its own weight $dm\,\vect g$ . There is no vertical acceleration, and the horizontal (radially inward) net force, whose tension share is (1), must supply the element’s centripetal acceleration $\omega^2 R$ .

\begin{align*} dN\sin\theta - dm\,g &= 0 \tag{2} \\ T\,d\varphi - dN\cos\theta &= dm\,\omega^2 R \tag{3}\end{align*}

Step 3Eliminating

dN

Solve the vertical equation (2) for $dN$ , then substitute that into the radial equation (3).

\begin{align*}dN &= \frac{dm\,g}{\sin\theta} \\ T\,d\varphi &= dm\,\omega^2 R + \frac{dm\,g}{\sin\theta}\cos\theta\end{align*}

Collecting the common factor $dm$ on the right leaves the element’s equation in its final form.

T\,d\varphi = dm\left(\omega^2 R+g\cot\theta\right)

(4)

Step 4Summing around the whole ring

The chain is uniform, so an element’s mass is the total mass in the same proportion as its angle is of the full turn.

dm = \frac{m\,d\varphi}{2\pi}.

Substituting that into (4),

T\,d\varphi = \frac{m\,d\varphi}{2\pi}\left(\omega^2 R+g\cot\theta\right).

Every element obeys this same equation, and $d\varphi$ appears exactly once on each side, so it cancels outright: no integration around the ring is needed, and $T$ comes out the same constant at every point of the chain.

T = \dfrac{m}{2\pi}\left(\cot\theta+\dfrac{\omega^2 R}{g}\right)g

(5)

Check

At $\theta=90^\circ$ the cone flattens into a horizontal plane, $\cot\theta\to 0$ , and the result reduces to $T=m\omega^2R/2\pi$ , exactly the tension of a free chain of that mass and radius spinning at $\omega$ with no cone beneath it at all, since $\theta=90^\circ$ means the normal reaction is purely vertical and contributes nothing to the radial balance. At $\omega=0$ the chain still needs $T=mg\cot\theta/2\pi$ just to sit on the cone, the ordinary hoop tension a ring must carry to stay curved against an outward push from the slope, even at rest.

Remark

The two terms in the answer never interact: $g\cot\theta$ is what the chain’s own weight demands of the tension through the slope’s geometry, and $\omega^2R$ is what the rotation demands for centripetal acceleration. They add linearly here for two specific reasons: the vertical equation fixes $dN$ from the weight alone, with no $\omega$ in it at all, and that same $dN$ then enters the radial equation only once, added to the single unknown $T$ . A frictionless banked turn does not share this structure: there the normal force’s vertical share (holding the car up) and its horizontal share (supplying the turn) are the same force at a fixed angle with no second unknown like $T$ to absorb them separately, so weight and speed fuse into one constraint on the banking angle, $\tan\theta = v^2/Rg$ , rather than adding as two independent terms.