Irodov 1.74: The Free Strand Pins the Friction – Amit & Deepali Sinha Foundation

Problem Irodov 1.74

A thread passes over a fixed, massless, frictionless pulley. One end is tied to the top of a vertical rod of mass $M$ and length $l$ ; the other end carries a small ball of mass $m<M$ threaded onto the thread through a hole, free to slide along it with some friction, and continuing on past the ball to a free, unattached lower end. At the moment both bodies are released from rest, the ball sits level with the rod’s lower end, and both bodies then move with constant accelerations. A time $t$ later the ball is level with the rod’s upper end. Find the friction force between the ball and the thread.

The rod M hangs from the thread's left strand; the ball m rides on the right strand, initially level with the rod's lower end, with the thread continuing free below it. — The rod $M$ hangs from the thread’s left strand; the ball $m$ rides on the right strand, initially level with the rod’s lower end, with the thread continuing free below it.

Given

$M$
$m$
$l$ (rod length)
$t$
massless pulley, frictionless axle

Find

the friction force

F_{fr}

between ball and thread

Solution

Step 1How far the rod must out-fall the ball

Let $w_M$ and $w_m$ be the (constant, downward-positive) accelerations of the rod and the ball. Starting level with the rod’s lower end, the ball ends up level with the rod’s upper end, a distance $l$ higher up the rod, so the rod must have descended exactly $l$ farther than the ball has:

x_M(t) - x_m(t) = l, \qquad x_M(t)=\dfrac{w_M t^2}{2},\ x_m(t)=\dfrac{w_m t^2}{2}.

Dividing out the common factor $t^2/2$ ,

w_M - w_m = \dfrac{2l}{t^2}

(1)

Step 2Newton’s second law for the rod

The thread pulls the rod up with tension $T$ ; its weight acts down.

Mw_M = Mg - T

Solving for the tension,

T = M(g-w_M)

(2)

Step 3The friction force equals the tension

The thread below the ball is massless with a free, unloaded lower end, so it carries no tension at all along that stretch, while above the ball the tension is the same $T$ that holds the rod, carried unchanged over the massless, frictionless pulley. The short stretch of thread inside the ball’s hole is itself massless, so the two tensions on its ends and the ball’s friction reaction on it sum to zero,

T - 0 - F_{fr} = 0

which fixes the friction at the tension’s own magnitude,

F_{fr} = T

(3)

Newton’s second law for the ball, with the friction force $F_{fr}$ acting up on it from the thread,

mw_m = mg - F_{fr}

Solving for the friction,

F_{fr} = m(g-w_m)

(4)

Step 4Solving the three relations for

F_{fr}

Inverting (2) and (4) for the two accelerations,

\begin{align*}w_M &= g - \frac{T}{M}, & w_m &= g - \frac{F_{fr}}{m}\end{align*}

Substituting both into the left side of the constraint (1), then replacing $T$ by $F_{fr}$ with (3),

\begin{align*}w_M - w_m &= \left(g-\frac{T}{M}\right) - \left(g-\frac{F_{fr}}{m}\right)\\ &= \frac{F_{fr}}{m}-\frac{T}{M} \\ &= F_{fr}\left(\frac1m-\frac1M\right)\\ &= F_{fr}\,\frac{M-m}{mM}\end{align*}

Setting this equal to the $2l/t^2$ of (1) and solving for $F_{fr}$ ,

F_{fr}\,\frac{M-m}{mM} = \frac{2l}{t^2}

F_{fr} = \dfrac{2lmM}{(M-m)t^2}

(5)

Check

Dimensionally, $l/t^2$ is an acceleration and $mM/(M-m)$ is a mass, so their product is a force, as required. As $M\to m^+$ the denominator $\to0$ and $F_{fr}\to\infty$ : from $w_M=g-T/M$ and $w_m=g-F_{fr}/m$ (Steps 2 to 3), an unboundedly growing tension drives both bodies’ accelerations far upward, away from free fall rather than toward it, which is the right runaway limit for two nearly balanced masses still required to separate by the full rod length $l$ in the same finite time $t$ : only an ever larger common tension can force even a small fixed difference $w_M-w_m$ between two nearly identical masses.

Remark

Relation (3), $F_{fr}=T$ , is the same zero-net-force-on-a-massless-element principle that fixes $T_1=2T_2$ on a massless pulley in the two-tension machines elsewhere in this chapter: a massless object carries no net force at any acceleration ( $ma=0$ automatically when $m=0$ ), so whatever acts on it, here the two tensions and the ball’s friction reaction, there the axle strand and the two rim strands, must sum to zero on its own.