Irodov 1.76: The String That Goes Slack Mid Flight – Amit & Deepali Sinha Foundation

Problem Irodov 1.76

One end of a thread is tied to a fixed ceiling point. The thread runs down to a movable pulley, from whose axle body $1$ hangs a height $h$ above the floor, then back up and over a fixed pulley mounted on the ceiling, and finally straight down to body $2$ , which initially rests on the floor. Body $1$ ’s mass is $\eta$ times body $2$ ’s mass. All pulleys and threads are massless and frictionless. Body $2$ is released from rest on the floor and the system begins to move. Find the maximum height body $2$ rises above the floor.

The thread runs from its ceiling anchor down to the movable pulley carrying body 1, a height h above the floor, back up over the fixed ceiling pulley, then down to body 2 resting on the floor. — The thread runs from its ceiling anchor down to the movable pulley carrying body $1$ , a height $h$ above the floor, back up over the fixed ceiling pulley, then down to body $2$ resting on the floor.

Given

$\eta = 4.0$
$h = 20\text{ cm}$

Find

the maximum height

H

body

2

rises above the floor

Solution

Step 1The movable pulley fixes a 2:1 speed ratio

Let the movable pulley (and body $1$ with it) descend a distance $x_1$ below the ceiling, and let $x_2$ be body $2$ ’s own displacement measured upward from the floor. The thread’s two segments running up from the movable pulley, one to the fixed ceiling anchor and one to the fixed pulley, both lengthen by $x_1$ as the movable pulley falls, since both upper ends are fixed; that extra length is supplied by the strand running down to body $2$ shortening by twice as much. The thread’s total length cannot change, so the two gains and the one loss must cancel,

\begin{align*}x_1 + x_1 - x_2 &= 0\\ x_2 &= 2x_1\end{align*}

Differentiating twice with respect to time,

\begin{align*}\ddot x_2 &= 2\,\ddot x_1\end{align*}

so the accelerations obey the same 2:1 ratio, with body $1$ moving down while body $2$ moves up,

a_2 = 2a_1

(1)

Step 2Newton’s second law for each body

Taking $a_1$ positive downward for body $1$ and $a_2$ positive upward for body $2$ ,

\begin{align*}m_1 g - 2T &= m_1 a_1 \\ T - m_2 g &= m_2 a_2\end{align*}

Free bodies of the movable pulley system carrying body 1 (two tensions T) and of body 2 (one tension T). — Free bodies of the movable pulley system carrying body $1$ (two tensions $\vect T$ ) and of body $2$ (one tension $\vect T$ ).

Substituting $a_2 = 2a_1$ from (1), together with $m_1 = \eta m_2$ , into the pair and eliminating $T$ ,

\begin{align*}T &= \dfrac{m_1 g - m_1 a_1}{2} \\ \dfrac{m_1 g - m_1 a_1}{2} - m_2 g &= 2 m_2 a_1 \\ g(m_1 - 2m_2) &= a_1(m_1 + 4m_2)\end{align*}

a_1 = \dfrac{g(\eta - 2)}{\eta + 4}

(2)

Doubling it, as (1) requires, gives body $2$ ’s own acceleration.

a_2 = 2a_1 = \dfrac{2g(\eta-2)}{\eta+4}

Step 3Body 1 falls the height

h

Body $1$ starts from rest and falls the height $h$ before reaching the floor, at constant acceleration $a_1$ from (2).

v_1^2 = 2a_1 h

At that instant body $2$ ’s speed is $v_2 = 2v_1$ , and by the same 2:1 rule (1) body $2$ has already risen $2h$ .

v_2^2 = 4v_1^2 = 8a_1 h

Step 4Free flight once body 1 lands

The extra rise is the standard free flight height for an object launched upward at $v_2$ .

\Delta H = \frac{v_2^2}{2g} = \frac{8a_1 h}{2g} = \frac{4a_1 h}{g}

Adding the $2h$ already gained while the thread was taut and substituting $a_1$ from (2),

\begin{align*}H &= 2h + \frac{4h}{g}\cdot\frac{g(\eta-2)}{\eta+4} \\ &= 2h\left[1 + \frac{2(\eta-2)}{\eta+4}\right] \\ &= 2h\cdot\frac{(\eta+4)+2(\eta-2)}{\eta+4} \\ &= 2h\cdot\frac{3\eta}{\eta+4}\end{align*}

H = \dfrac{6h\eta}{\eta+4} = \dfrac{6(0.20\text{ m})(4.0)}{4.0+4}= 0.6\text{ m}

(3)

Check

As $\eta\to2^+$ , (2) gives $a_1\to0$ and $v_2\to0$ , so the free flight term vanishes and $H\to2h$ : body $1$ barely creeps to the floor, body $2$ barely creeps up by the $2h$ the taut thread alone provides, with nothing left to carry it further. This is exactly what (3) gives at $\eta=2$ .

Remark

The direction reversal sits at $\eta=2$ , not $\eta=1$ : because the movable pulley halves body $1$ ’s pull on body $2$ ’s side of the machine (two strands share the load), body $1$ must outweigh body $2$ by more than a factor of two, not merely be heavier, before it can lift body $2$ at all. The thread goes slack in Step 4 because body $1$ hits the floor and can fall no further, for any mass ratio; the same 2:1 ratio only sets how fast body $2$ is moving, hence how far, at the instant that happens.