Irodov 1.107: One Coordinate Carries the Whole Chain

Problem Irodov 1.107

A flexible, inextensible chain of length $l$ lies along the surface of a smooth (frictionless) sphere of radius $R$ , arranged along a single great-circle arc that starts exactly at the top of the sphere, with $l<\pi R/2$ so the whole chain stays within one quarter of the great circle. The upper end, initially held fixed at the top, is released. Find the acceleration $w$ of every element of the chain at the instant of release.

The chain (heavy arc) lying along the sphere from the top to angular position l/R, measured at the center O. Once released, every point of the chain slides along the sphere with a common speed, since the chain cannot stretch. — The chain (heavy arc) lying along the sphere from the top to angular position $l/R$ , measured at the center $O$ . Once released, every point of the chain slides along the sphere with a common speed, since the chain cannot stretch.

Given

$l$
$R$
$l<\pi R/2$

Find

w

at the instant of release

Solution

Step 1One coordinate for the whole chain

Let $\varphi$ be the angle, measured at the center $O$ , through which the chain’s trailing (originally top) end has slipped since release; at release $\varphi=0$ . The chain then occupies the arc from $\varphi$ to $\varphi+l/R$ , and every element moves at the common speed $v=R\dot\varphi$ , so the whole chain’s kinetic energy is

T = \frac{1}{2}mv^2 = \frac{1}{2}mR^2\dot\varphi^2.

Step 2Potential energy as a function of

\varphi

An element at angular position $\theta$ (measured from the top) sits at height $R\cos\theta$ above the center $O$ ; summing $g$ times this height over the chain’s mass, uniformly distributed at $m/l$ per unit arc length $R\,d\theta$ , gives the total potential energy.

\begin{aligned} U(\varphi) &= \int_{\varphi}^{\varphi+l/R} \frac{m}{l}(R\,d\theta)\,g(R\cos\theta) \\ &= \frac{mgR^2}{l}\int_{\varphi}^{\varphi+l/R}\cos\theta\,d\theta \\ &= \frac{mgR^2}{l}\Bigl[\sin\Bigl(\varphi+\frac{l}{R}\Bigr)-\sin\varphi\Bigr] \end{aligned} \tag{1}

Step 3The equation of motion for

\varphi

Differentiating $T+U=\text{const}$ with respect to time,

mR^2\dot\varphi\,\ddot\varphi + \frac{dU}{d\varphi}\dot\varphi = 0.

Factoring out the common $\dot\varphi$ ,

\dot\varphi\left[mR^2\ddot\varphi + \frac{dU}{d\varphi}\right] = 0.

Since $\dot\varphi$ is not identically zero (release is not equilibrium once $l/R\ne0$ ), the bracket it multiplies must vanish at every instant, including at release:

mR^2\ddot\varphi = -\dfrac{dU}{d\varphi}

(2)

Step 4Evaluating at the instant of release

Differentiating the potential energy (1),

\frac{dU}{d\varphi} = \frac{mgR^2}{l}\Bigl[\cos\Bigl(\varphi+\frac{l}{R}\Bigr)-\cos\varphi\Bigr].

At release, $\varphi=0$ , so substituting into (2) and solving for $\ddot\varphi$ ,

\begin{align*}mR^2\ddot\varphi\big|_{\varphi=0} &= -\frac{mgR^2}{l}\Bigl[\cos\frac{l}{R}-1\Bigr] \\ \ddot\varphi\big|_{\varphi=0} &= \frac{g}{l}\Bigl[1-\cos\frac{l}{R}\Bigr]\end{align*}

Every element shares the same tangential acceleration $w=R\ddot\varphi$ , so multiplying through by $R$ ,

w = \dfrac{Rg}{l}\Bigl[1-\cos\dfrac{l}{R}\Bigr]

(3)

Check

As $l/R\to0$ (a short chain near the top), $1-\cos(l/R)\to (l/R)^2/2$ , so $w\to gl/(2R)$ , which vanishes as $l\to0$ : an infinitesimal chain sitting exactly at the crown feels no net tangential pull, as it must by the symmetry of the top point. Dimensionally, $Rg/l$ is an acceleration and the bracket is a pure number, so $w$ has the right units.

Remark

The bracket $1-\cos(l/R)$ is the same combination that fixes, at the instant of release, how far below the top the chain’s leading (lower) end sits: with the trailing (originally top) end still exactly at the crest, $R-R\cos(l/R)$ is the leading end’s height drop below the top at that instant. The single-coordinate energy method used here works because a smooth, inextensible chain sliding along a fixed shape has only one true degree of freedom regardless of how many links or how much mass make it up; the internal tension that keeps every element moving together never appears in the final answer, precisely because it does zero net work on the chain as a whole.