Irodov 1.251: Wound or Hanging, the Inertia Holds – Amit & Deepali Sinha Foundation

Problem Irodov 1.251

A uniform cylinder of radius $R$ and mass $M$ turns freely, without friction, about a fixed horizontal axis $O$ that is its own axis of symmetry. A thin cord of total length $l$ and total mass $m$ is wound on the cylinder in a single layer, so the wound portion lies on the cylinder’s surface at radius $R$ ; the portion already unwound hangs vertically from the rim, and its hanging length is $x$ . Take the wound portion to have its center of gravity on the cylinder’s axis. Find the cylinder’s angular acceleration as a function of the hanging length $x$ .

The arrangement. The cylinder hangs from a fixed bracket and turns about the axis O. The cord is wound on its surface in a single layer, the second curve hugging the rim; the part already off hangs vertically, of length x. — The arrangement. The cylinder hangs from a fixed bracket and turns about the axis $O$ . The cord is wound on its surface in a single layer, the second curve hugging the rim; the part already off hangs vertically, of length $x$ .

Given

Find

\beta

, the cylinder’s angular acceleration as a function of

x

Solution

Step 1Divide the cord by where it is

The cord is uniform, so its mass per unit length is $m/l$ and the split between the two parts follows from their lengths alone. Both parts move at the same rate: the wound part rides on the rim, and the hanging part is drawn off the rim without stretching or slipping, so the linear acceleration $a$ of the hanging cord is the rim’s tangential acceleration.

\begin{aligned} m_{\text{hang}} &= \frac{mx}{l} \\ m_{\text{wound}} &= \frac{m(l-x)}{l} \\ a &= \beta R \end{aligned} \tag{1}

The two free bodies. Left: the cylinder with the cord still wound on it, pulled by the tension T where the cord leaves the rim, moment arm R, angular acceleration β. Right: the hanging cord, held by that tension against its own weight. — The two free bodies. Left: the cylinder with the cord still wound on it, pulled by the tension $\vect T$ where the cord leaves the rim, moment arm $R$ , angular acceleration $\vect\beta$ . Right: the hanging cord, held by that tension against its own weight.

Step 2The hanging piece of cord

The hanging piece is held up only by the tension where it leaves the rim, and is pulled down by its own weight. It does not rotate, so Newton’s second law applies to it as it stands, with the downward direction taken positive.

\dfrac{mx}{l}\,g - T = \dfrac{mx}{l}\,a

(2)

Step 3The cylinder and the cord still wound on it

Take moments about the axis $O$ for the cylinder and the wound cord together. The cylinder’s weight and the reaction at the axle act on the axis itself, and the wound cord’s center of gravity is on the axis by hypothesis, so of all the forces only the tension has a moment, with arm $R$ . The wound cord lies on the surface, so every part of it is at the distance $R$ from the axis and its moment of inertia is its mass times $R^2$ .

\begin{align*}I &= \frac{MR^2}{2} + \frac{m(l-x)}{l}R^2 \\ T R &= I\beta \\ &= \left(\frac{M}{2}+\frac{m(l-x)}{l}\right)R^2\,\frac{a}{R}\end{align*}

T = \left(\dfrac{M}{2}+\dfrac{m(l-x)}{l}\right) a

(3)

Step 4The moving inertia does not depend on

x

at all

Substituting the tension (3) into the hanging cord’s equation (2) removes $T$ and leaves one equation for $a$ . What happens next is the heart of the problem: the wound cord’s share of the inertia and the hanging cord’s share add up to the whole cord, and the length $x$ cancels out of the inertia even though it stays in the driving weight.

\begin{align*}\frac{mx}{l}g - \left(\frac{M}{2}+\frac{m(l-x)}{l}\right)a &= \frac{mx}{l}a \\ \frac{mx}{l}g &= \left(\frac{M}{2}+\frac{m(l-x)}{l}+\frac{mx}{l}\right)a\end{align*}

\left(\dfrac{M}{2}+m\right)a = \dfrac{mgx}{l}

(4)

Step 5The angular acceleration

Solving (4) for $a$ and dividing by $R$ , since $\beta = a/R$ by (1), gives the angular acceleration.

\begin{align*}a &= \frac{mgx}{l\left(\dfrac{M}{2}+m\right)} \\ \beta &= \frac{a}{R} = \frac{mgx}{Rl\left(\dfrac{M}{2}+m\right)}\end{align*}

\beta = \dfrac{2mgx}{Rl(M+2m)}

(5)

Check

The units are those of $g/(R\,l)$ times a length, that is $\mathrm{s^{-2}}$ , as an angular acceleration must be. Three special cases confirm the shape. With $x = 0$ nothing hangs, no torque acts and $\beta = 0$ . With $x = l$ the whole cord hangs free and the formula gives $\beta = 2mg/[R(M+2m)]$ , which is what a cylinder of inertia $MR^2/2$ carrying a hanging mass $m$ on its rim gives directly. And with $M = 0$ it reduces to $\beta = gx/(Rl)$ : a massless cylinder leaves the cord as the only inertia, so the hanging fraction $x/l$ of the cord accelerates the whole of it, at $gx/l$ .

Remark

The unwinding is not uniformly accelerated, it runs away. Since $\beta$ grows in proportion to $x$ , and $x$ is itself the coordinate whose second derivative is $a = \beta R$ , the boxed result says the hanging length obeys $\ddot x = \lambda^2 x$ , with $\lambda^2 = 2mg/[l(M+2m)]$ . That is the harmonic equation with the sign reversed, so instead of oscillating, $x$ grows like $\cosh(\lambda t)$ : the more cord hangs, the harder it pulls, and there is no restoring tendency anywhere in the system. The other half of the story is why the inertia stayed constant. Every element of the cord travels at the speed $\omega R$ whether it is riding on the rim or hanging in the strand, so the cord contributes $mR^2$ to the moment of inertia regardless of how it is divided, and only the driving torque, which needs the cord’s weight to act off the axis, cares where the cord actually is: the wound layer’s weight acts on the axis by hypothesis and turns nothing, while the hanging strand’s is offset from it by the whole radius $R$ .