Irodov 1.257: Which Way a Pulled Spool Rolls – Amit & Deepali Sinha Foundation

Problem Irodov 1.257

A spool of mass $m$ and outside radius $R$ rests on a rough horizontal floor, its axis horizontal. Its moment of inertia about that axis is $I = \gamma m R^2$ , with $\gamma$ a pure number. A thread is wound on the spool in a layer of radius $r$ ; the thread runs off the underside of that layer and is pulled with a constant force $F$ making an angle $\alpha$ above the horizontal, on the same side as the horizontal $x$ axis, which is taken along the floor in the direction of the horizontal component of the pull. The spool starts from rest and rolls without sliding on the floor. Find:

the projection $w_x$ of the acceleration of the spool’s axis on the $x$ axis
the work $A$ done by the force $\vect F$ during the first $t$ seconds of the motion

The spool seen along its axis. The rim of radius R meets the floor, the thread leaves the underside of the wound layer of radius r, and the pull F runs along it at α to the horizontal. Gray marks positive x. — The spool seen along its axis. The rim of radius $R$ meets the floor, the thread leaves the underside of the wound layer of radius $r$ , and the pull $\vect F$ runs along it at $\alpha$ to the horizontal. Gray marks positive $x$ .

Given

$m$
$I = \gamma m R^2$
$R$
$r$
$F$
$\alpha$
rolling without sliding, from rest

Find

(a)

w_x

, (b)

A

over the first

t

seconds

Solution

Step 1Newton’s law along the floor

Four forces act on the spool: the pull $\vect F$ along the thread, the weight $m\vect g$ at the axis, and, at the contact with the floor, the normal force $\vect N$ and the friction $\vect f$ . The vertical projection only fixes $N$ , which is not asked for; the horizontal projection carries the physics, and there only the component $F\cos\alpha$ and the friction survive. Take the friction to act backwards, against that component; the value of $f$ found in (4) below is positive and so confirms the choice.

m w_x = F\cos\alpha - f. \tag{1}

The free body. The pull acts where the thread leaves the wound layer, tangent to the inner circle; the weight acts at the axis; the normal force and the friction act at the contact point. The friction is drawn backwards.

Step 2The torque equation about the axis

Measure the angular acceleration $\beta$ in the sense that rolls the spool along $+x$ , clockwise as the figure is drawn. About the axis the weight and the normal force have no moment; the friction acts at the rim, a distance $R$ from the axis, and drives that sense, while the pull is tangent to the inner circle, a distance $r$ from the axis, and opposes it.

I\beta = Rf - Fr. \tag{2}

Step 3Rolling without sliding closes the system

Rolling on the fixed floor locks the axis’s motion to the spin: the axis advances by one rim arc per turn, so its acceleration is the angular acceleration times the outer radius. Putting that identity into (2), with the moment of inertia written out, gives the friction in terms of the acceleration, and (1) then eliminates it.

\begin{align*}w_x &= \beta R \\ \gamma m R^2\cdot\frac{w_x}{R} &= Rf - Fr \\ \gamma m R\,w_x &= Rf - Fr \\ f &= \gamma m w_x + \frac{Fr}{R}\end{align*}

That expression is not yet manifestly positive, because $w_x$ is itself unknown at this line and, as the Remark shows, can come out negative. Substituting it into (1),

\begin{align*}m w_x &= F\cos\alpha - \gamma m w_x - \frac{Fr}{R} \\ m w_x (1 + \gamma) &= F\left(\cos\alpha - \frac{r}{R}\right)\end{align*}

w_x = \dfrac{F\left(\cos\alpha - \dfrac{r}{R}\right)}{m(1 + \gamma)}

(3)

Putting that back into the friction collapses it to a form whose sign can be read off:

f = \frac{F\left(\gamma\cos\alpha + \dfrac{r}{R}\right)}{1 + \gamma} \tag{4}

which is positive for every $\alpha$ short of a right angle and every positive $\gamma$ and $r/R$ , and stays positive in the cases where $w_x$ itself is negative. So the friction does act backwards, as assumed in (1) and drawn in the figure.

Step 4Only the pull does any work

The normal force and the weight are perpendicular to the displacement of the points they act at, and the friction acts at the contact point, which is instantaneously at rest while the spool rolls, so it transports no energy either. All the work done by $\vect F$ therefore appears as kinetic energy, and starting from rest the work-energy theorem reads

A = T = \frac{1}{2}m v^2 + \frac{1}{2}I\omega^2 ,

where $v$ is the speed of the axis and $\omega$ the spin, tied to $v$ by the same rolling condition as before.

\begin{align*}\omega &= \frac{v}{R} \\ A &= \frac{1}{2}m v^2 + \frac{1}{2}\gamma m R^2\cdot\frac{v^2}{R^2}\end{align*}

A = \dfrac{1}{2}m(1 + \gamma)v^2

(5)

so the factor $1+\gamma$ that slowed the acceleration (3) is the same one that shares the energy between the motion of the axis and the spin about it.

Step 5The work after

t

seconds

The acceleration (3) is constant, so the speed after $t$ seconds is $v = w_x t$ . Putting that into (5) and then substituting $w_x$ gives the work.

\begin{align*}A &= \frac{1}{2}m(1 + \gamma)\,w_x^2\,t^2 \\ &= \frac{1}{2}m(1 + \gamma)\,t^2\left[\frac{F\left(\cos\alpha - \dfrac{r}{R}\right)}{m(1 + \gamma)}\right]^2\end{align*}

A = \dfrac{F^2 t^2\left(\cos\alpha - \dfrac{r}{R}\right)^2}{2m(1 + \gamma)}

(6)

Check

Dimensionally $F^2t^2/m$ is $(\mathrm{kg\,m\,s^{-2}})^2\,\mathrm{s^2/kg} = \mathrm{kg\,m^2s^{-2}}$ , an energy, and the bracket and $\gamma$ are pure numbers. Two special cases also come out right. With $r \to 0$ the thread leaves the axis itself and $w_x \to F\cos\alpha/[m(1+\gamma)]$ , the familiar acceleration of a body rolled along by a force applied at its center. And at $\cos\alpha = r/R$ both answers vanish together: the spool does not accelerate, so it never moves, so the pull can do no work, which is the only consistent pair of values.

Remark

The bracket $\cos\alpha - r/R$ is a piece of geometry in disguise: multiplied by $R$ it is the perpendicular distance from the floor contact to the thread’s line of action, that is, the lever arm of the pull about the contact point. So the sign of the whole answer is decided by which side of the contact point the thread’s line passes. At $\cos\alpha = r/R$ the line runs exactly through the contact and the spool stays put however hard it is pulled, so long as it does not slide, which is what the problem stipulates and not what a real floor would grant; at smaller angles the spool rolls the way it is pulled, and at steeper ones it rolls backwards, away from the puller, on a thread that is being pulled towards him. That reversal is what makes the spool a lecture demonstration rather than an exercise.