Irodov 1.193: Reeling a Body Toward Its Own Hole – Amit & Deepali Sinha Foundation

Problem Irodov 1.193

A small body of mass $m$ lies on a smooth (frictionless) horizontal plane, tied to a non-stretchable thread whose other end passes through a hole $O$ drilled in the plane. The thread is pulled through the hole from below at a constant speed, so the body’s distance $r$ from $O$ decreases at a constant rate. When $r=r_0$ the thread’s angular velocity is $\omega_0$ . Find the thread tension $F$ as a function of $r$ .

Top view of the plane: the thread reels in through the hole O from below at constant speed, and the body m spirals inward along it as r shrinks. The rings sketch that shrinking path. — Top view of the plane: the thread reels in through the hole $O$ from below at constant speed, and the body $m$ spirals inward along it as $r$ shrinks. The rings sketch that shrinking path.

Given

$m$
$r_0$
$\omega_0$
thread reeled in at constant speed

Find

F(r)

Solution

Step 1Angular momentum about the hole is conserved

At a general instant: the tension F, the only in-plane force, along the thread toward O, and the velocity v (drawn clear of F's line) resolved into r along the thread and rφ across it. — At a general instant: the tension $\vect F$ , the only in-plane force, along the thread toward $O$ , and the velocity $\vect v$ (drawn clear of $\vect F$ ’s line) resolved into $\dot r$ along the thread and $r\dot\varphi$ across it.

With zero torque about $O$ , the angular momentum $L=mr^2\dot\varphi$ is constant. Evaluating it at the given instant $r=r_0$ , $\dot\varphi=\omega_0$ , and equating to its value at a general $r$ ,

mr^2\dot\varphi = mr_0^2\omega_0.

\dot\varphi(r) = \dfrac{r_0^2\omega_0}{r^2}

(1)

Step 2The radial equation of motion, with

\ddot r=0

Since the thread is pulled in at a constant speed, the body’s radial velocity $\dot r$ is itself constant, so its radial acceleration vanishes, $\ddot r=0$ . In polar coordinates the radial component of acceleration is $\ddot r - r\dot\varphi^2$ , and the only in-plane force is the tension $F$ , directed inward, so Newton’s second law along the radial direction reads

-F = m\left(\ddot r - r\dot\varphi^2\right) = -mr\dot\varphi^2.

F = mr\dot\varphi^2

(2)

Step 3Combine to eliminate

\dot\varphi

Substituting (1) into (2),

\begin{align*}F &= mr\left(\frac{r_0^2\omega_0}{r^2}\right)^2 \\ &= mr\cdot\frac{r_0^4\omega_0^2}{r^4}\end{align*}

F(r) = \dfrac{m\omega_0^2r_0^4}{r^3}

(3)

Check

Dimensionally $m\omega_0^2r_0^4/r^3$ has units $\text{kg}\cdot(\text{rad/s})^2\cdot\text{m}^4/\text{m}^3=\text{kg}\,\text{m}/\text{s}^2$ , a force, as required. At $r=r_0$ the formula gives $F=m\omega_0^2r_0$ , the ordinary centripetal tension for circular motion of radius $r_0$ at angular speed $\omega_0$ , exactly matching (2) evaluated at that instant.

Remark

The constant reeling speed never entered the final answer, only the fact that it makes $\ddot r=0$ : the tension is set entirely by the instantaneous centripetal demand $mr\dot\varphi^2$ , with $\dot\varphi$ itself climbing as $1/r^2$ from angular momentum conservation. The tension therefore grows as $1/r^3$ and diverges as the body approaches the hole, a reminder that an idealized constant-speed reel-in would require unbounded tension (and so is only ever a good approximation away from $r=0$ ).