Irodov 1.138: Winding Thread Spirals Disc Inward – Amit & Deepali Sinha Foundation

Problem Irodov 1.138

Viewed from above, a small disc $A$ slides on a frictionless horizontal plane. A thin fixed vertical cylinder of radius $R$ stands on this plane, and a thread of length $l_0$ runs from the disc to a point $B$ on the cylinder’s rim, where it lies tangent to the cylinder. The disc is given an initial speed $v_0$ directed perpendicular to the taut thread, in the sense that begins to wrap the thread around the cylinder. As the disc moves, the thread winds progressively onto the cylinder, so the straight (free) length of thread joining the disc to the cylinder steadily shortens and the disc spirals inward. Find the time that elapses until the disc reaches the cylinder’s surface.

Top view. The thread leaves the fixed cylinder of radius R tangentially at B and runs to the disc A, a free length l_0 away; the initial velocity v_0 is perpendicular to the thread. The dashed path spirals inward as the thread wraps. — Top view. The thread leaves the fixed cylinder of radius $R$ tangentially at $B$ and runs to the disc $A$ , a free length $l_0$ away; the initial velocity $\vect v_0$ is perpendicular to the thread. The dashed path spirals inward as the thread wraps.

Given

$R$
$l_0$
$v_0$

Find

the time until the disc strikes the cylinder

Solution

Step 1Parametrizing the wrap

The wrap at a general instant. The thread leaves the cylinder at P, position r_P, after turning through φ from B; the free length l runs to the disc A, whose path is dashed. T and N are drawn at A. — The wrap at a general instant. The thread leaves the cylinder at $P$ , position $\vect r_P$ , after turning through $\varphi$ from $B$ ; the free length $l$ runs to the disc $A$ , whose path is dashed. $\vect{\hat T}$ and $\vect{\hat N}$ are drawn at $A$ .

Let $P$ be the point where the thread currently leaves the cylinder, $\varphi$ the angle (in radians) through which $P$ has traveled around the cylinder since the start, and $l$ the current free (straight) length of thread from $P$ to the disc. Since the thread is inextensible, the length wound onto the cylinder, $R\varphi$ , plus the length still free must equal the original length $l_0$ .

l = l_0 - R\varphi

(1)

Step 2The disc’s velocity is perpendicular to the thread

With $\vect{\hat T}$ the unit tangent to the cylinder at $P$ , pointing from $P$ toward the disc, and $\vect{\hat N}$ the inward unit normal there, the disc sits a free length $l$ beyond $P$ along the thread, so its position vector from the center is the tangent point’s position plus that free stretch.

\vect r = \vect r_P + l\,\vect{\hat T}

The tangent point $P$ moves along the rim as the thread winds, at the rate the angle $\varphi$ grows, and its motion is along the rim’s own tangent.

\frac{d\vect r_P}{dt} = R\dot\varphi\,\vect{\hat T}

The unit tangent turns with $P$ : rotating $\vect{\hat T}$ through a small angle $d\varphi$ moves its tip by $d\varphi$ in the direction perpendicular to it, which is the inward normal.

\frac{d\vect{\hat T}}{dt} = \dot\varphi\,\vect{\hat N}

Differentiating the position with the product rule, and substituting the two rates just found,

\begin{align*}\frac{d\vect r}{dt} &= \frac{d\vect r_P}{dt} + \dot l\,\vect{\hat T} + l\,\frac{d\vect{\hat T}}{dt} \\ &= R\dot\varphi\,\vect{\hat T} + \dot l\,\vect{\hat T} + l\dot\varphi\,\vect{\hat N} \\ &= (R\dot\varphi+\dot l)\,\vect{\hat T} + l\dot\varphi\,\vect{\hat N}\end{align*}

Differentiating (1) with respect to time relates the rate at which the free length shrinks to the winding rate.

\dot l = -R\dot\varphi

(2)

Substituting (2) into the $\vect{\hat T}$ coefficient makes it vanish identically, leaving only the normal part.

\frac{d\vect r}{dt} = l\dot\varphi\,\vect{\hat N}

(3)

The velocity has no component along the thread: it is exactly perpendicular to it at every instant.

Step 3The speed stays constant at

v_0

Because the tension acts along the thread ( $\vect{\hat T}$ ) and (3) shows the velocity is always along $\vect{\hat N}$ , tension is perpendicular to velocity at every instant and does no work. With no friction and no other horizontal force, the disc’s speed never changes.

v = v_0 \text{ for all } t

(4)

Taking the magnitude of (3) and using (4),

v_0 = l\dot\varphi. \tag{5}

Step 4A differential equation for the free length

Solving (2) for $\dot\varphi=-\dot l/R$ and substituting into (5) eliminates $\varphi$ in favor of $l$ alone.

\begin{aligned} v_0 &= l\left(-\frac{\dot l}{R}\right) \\ l\,\dot l &= -v_0 R \end{aligned} \tag{6}

Step 5Integrating to the moment of impact

The disc starts with the full free length $l_0$ at $t=0$ and reaches the cylinder, $l=0$ , at the sought time $t$ . Integrating (6) over that interval,

\int_{l_0}^{0} l\,dl = -v_0 R\int_0^t dt' \;\Longrightarrow\; -\frac{l_0^2}{2} = -v_0 R\,t.

t = \dfrac{l_0^2}{2v_0 R}

(7)

Check

Dimensionally $l_0^2/(v_0R)$ is $\mathrm{m}^2/(\mathrm{m/s}\cdot\mathrm{m})=\mathrm{s}$ , a time, as required. Doubling $R$ at fixed $l_0$ and $v_0$ halves the striking time, which matches the picture: a fatter cylinder winds the same free length onto itself in fewer turns, so $\varphi$ needed to reach $l=0$ is smaller and the disc gets there sooner even though its speed never changes.

Remark

That the speed is constant, (4), even though the disc’s path is an inward spiral and its distance from the cylinder’s axis is always shrinking, is easy to mistake for a violation of energy conservation; it is not, because the tension here does no work at all, at any instant, unlike an ordinary string swung at fixed radius where tension is the sole centripetal force but still does no work only because the radius is constant. Here the radius itself changes, yet the perpendicularity in (3) holds throughout precisely because the thread is inextensible and the free segment is always tangent to the cylinder.