Irodov 1.255: The Rim Never Enters the Answer – Amit & Deepali Sinha Foundation

Problem Irodov 1.255

A spool rests on a smooth (frictionless) plane inclined at $\alpha = 30^\circ$ to the horizontal. Thread is wound on the spool in a layer of radius $r = 3.0$ cm; the thread leaves that wound layer running parallel to the inclined surface, up the slope, and its free end is tied to a fixed point on a wall standing at the top of the incline. The spool has mass $m = 200$ g and moment of inertia $I = 0.45$ g $\cdot$ m $^2$ about its own axis. Find the acceleration of the spool’s axis.

The arrangement. The spool rests on the smooth incline at angle α to the dashed horizontal. The thread leaves the wound layer of radius r, runs parallel to the slope, and ends on the wall; the rim of radius R touches the plane. — The arrangement. The spool rests on the smooth incline at angle $\alpha$ to the dashed horizontal. The thread leaves the wound layer of radius $r$ , runs parallel to the slope, and ends on the wall; the rim of radius $R$ touches the plane.

Given

$\alpha = 30^\circ$
$m = 200$ g
$I = 0.45$ g $\cdot$ m $^2$
$r = 3.0$ cm

Find

w

, the acceleration of the spool’s axis

Solution

Step 1What the smooth plane is allowed to exert

Only three forces act on the spool: its weight, the normal force from the plane, and the tension in the thread. Because the plane is smooth it can push but not rub, so the contact contributes no force along the surface. Resolving along the slope, with the down-slope direction taken positive, and across it, where the center has no acceleration:

\begin{aligned} m w &= mg\sin\alpha - T \\ 0 &= N - mg\cos\alpha \end{aligned} \tag{1}

The second equation only sizes $N$ ; the first carries the unknown $T$ into the answer, so a second equation for $T$ is needed.

The three forces on the spool. The weight acts at the center, the normal force at the contact on a line through the center, and the tension where the thread leaves the wound layer. No friction appears at the smooth contact.

Step 2The thread is the only thing that can turn the spool

Take moments about the spool’s own axis. The weight acts at the center and so has no moment there; the normal force acts at the contact, but the contact lies radially below the center, so the normal’s line of action passes through the axis and its moment vanishes too. The tension alone has a moment, and its arm is the radius of the wound layer.

I\beta = T\,r

T = \dfrac{I\beta}{r}

(2)

Step 3The fixed thread ties the spin to the sliding

The thread does not stretch and its far end is fixed to the wall, so every bit of thread the spool pays off the wound layer is a bit by which the axis has receded from the wall. Writing $x$ for the distance from the wall to the axis and $\varphi$ for the angle the spool has turned through, the thread paid off in a turn $d\varphi$ is $r\,d\varphi$ , and it must equal $dx$ :

dx = r\,d\varphi

Differentiating twice with respect to the time,

\frac{d^2x}{dt^2} = r\,\frac{d^2\varphi}{dt^2}

w = \beta r

(3)

Step 4Eliminate the tension and solve for the acceleration

Putting the constraint (3) into the moment result (2) expresses the tension through the acceleration alone, and substituting that into the slope equation (1) leaves one unknown.

\begin{align*}T &= \frac{I\beta}{r} = \frac{I w}{r^2} \\ m w &= mg\sin\alpha - \frac{I w}{r^2} \\ w\left(m + \frac{I}{r^2}\right) &= mg\sin\alpha\end{align*}

Dividing by $m$ turns the bracket into the dimensionless inertia ratio, whose value follows from the data:

\frac{I}{mr^2} = \frac{0.45\times 10^{-3}\ \mathrm{kg\cdot m^2}} {(0.200\ \mathrm{kg})(0.030\ \mathrm{m})^2} = 2.5

w = \dfrac{g\sin\alpha}{1 + I/mr^2} = \dfrac{(9.8)(0.5)}{1+2.5} = 1.4\ \mathrm{m/s^2}

(4)

Directed down the slope, along the incline.

Check

Put the numbers back into the two equations that produced them. The tension is $T = Iw/r^2 = (4.5\times10^{-4})(1.4)/(9.0\times10^{-4}) = 0.70$ N, while the down-slope pull of gravity is $mg\sin\alpha = (0.200)(9.8)(0.5) = 0.98$ N; their difference is $0.28$ N, and $mw = (0.200)(1.4) = 0.28$ N, so the slope equation balances. The normal force is $N = mg\cos\alpha = 1.70$ N, comfortably positive, so the spool does stay on the plane. The two limits are right as well: with $I \to 0$ the bracket becomes $1$ and $w \to g\sin\alpha$ , a body sliding freely down a smooth slope, and with $I \to \infty$ the acceleration goes to zero, since an infinitely reluctant spool cannot unwind at all.

Remark

The outer radius of the spool never appears. That is the smooth plane’s doing: the surface can only push normally, and the normal from the contact runs straight through the axis, so the ground exerts no moment and the rim is a passenger rather than a participant. The factor $1 + I/mr^2$ is the familiar effective-inertia factor of a rolling body, but here the radius inside it is the hub the thread is wound on, not the radius that touches the ground. Rough the plane up and the spool does not merely accelerate differently, it stops accelerating at all: rolling without sliding puts the thread’s departure point a perpendicular height $R + r$ above the contact line, so that point moves along the slope at $\omega(R+r)$ , and a thread of fixed length tied to the wall forces that to vanish. The spin goes with it and the axis stands still, held by $T = mg\sin\alpha/(1 + r/R)$ while the plane supplies $f = mg\sin\alpha\cdot r/(R+r)$ , which it can do as soon as the coefficient of friction reaches $\tan\alpha\cdot r/(R+r)$ . So it is the polish that lets the spool move at all, and on the smooth plane the spool would behave the same if the rim were shaved down to the wound layer, provided $I$ and $m$ were untouched.