Irodov 1.259: Everything Measured from the Motionless Point

Problem Irodov 1.259

A stepped pulley $B$ of mass $M$ has two coaxial cylindrical surfaces, of radii $R$ and $2R$ , and a moment of inertia $I$ about its own axis, which is horizontal. Two light threads are wound on the smaller surface of radius $R$ and their upper ends are fixed to a ceiling, so the pulley can sink while it turns and those threads unwind. A third light thread is wound on the larger surface of radius $2R$ , leaves it on the side of the pulley opposite the ceiling threads, and carries a weight $A$ of mass $m$ at its lower end. All the threads are vertical, inextensible and of negligible mass, and none of them slips on the pulley. The system is released from rest. Find the acceleration of the weight $A$ .

The arrangement seen along the pulley's axis. The ceiling threads, drawn as one strand, are wound on the inner surface of radius R; the weight A hangs from a thread on the outer surface of radius 2R, on the other side. — The arrangement seen along the pulley’s axis. The ceiling threads, drawn as one strand, are wound on the inner surface of radius $R$ ; the weight $A$ hangs from a thread on the outer surface of radius $2R$ , on the other side.

Given

$m$
$M$
$I$
radii $R$ and $2R$
threads massless and inextensible
released from rest

Find

the acceleration

w_A

of the weight

A

Solution

Step 1One angular acceleration governs the whole system

The ceiling threads are inextensible and their upper ends are fixed, so the point of the pulley from which they leave is instantaneously at rest: the pulley turns, at each instant, about that point. Every other point of the pulley then moves at the angular speed times its own distance from it, and those distances are pure geometry. The center lies $R$ away, and the point where the weight’s thread leaves the outer surface lies on the far side, $R + 2R$ away.

\begin{align*}v_B &= \omega R \\ v_A &= \omega\,(R + 2R) = 3\omega R\end{align*}

These hold at every instant, so their time derivatives hold too, with $\beta$ the angular acceleration.

w_B = \beta R, \qquad w_A = 3\beta R

(1)

The kinematics. The dot marks the point where the ceiling threads leave the pulley, momentarily at rest and so the instantaneous center of rotation. The center lies R from it and moves at v_B, the weight's thread 3R from it at v_A. — The kinematics. The dot marks the point where the ceiling threads leave the pulley, momentarily at rest and so the instantaneous center of rotation. The center lies $R$ from it and moves at $v_B$ , the weight’s thread $3R$ from it at $v_A$ .

The forces on the pulley. T_c is the combined pull of the ceiling threads at the inner surface, T_A the pull of the weight's thread at the outer surface, and M g acts at the center. The forces on A are not drawn. — The forces on the pulley. $\vect T_c$ is the combined pull of the ceiling threads at the inner surface, $\vect T_A$ the pull of the weight’s thread at the outer surface, and $M\vect g$ acts at the center. The forces on $A$ are not drawn.

Step 2Newton’s law for the weight and for the pulley’s center

The weight $A$ is an ordinary hanging body: gravity down, the thread up. The pulley’s center is pulled down by gravity and by the reaction of the weight’s thread, and held up by the ceiling threads, whose combined tension is $T_c$ . Both bodies move downwards, so both equations are written with the downward direction positive.

\begin{gathered} m w_A = mg - T_A\\ M w_B = Mg + T_A - T_c \end{gathered}

(2)

Step 3The pulley’s rotation about its own axis

About the pulley’s own axis the weight $M\vect g$ has no moment. The ceiling threads pull up on the inner surface, a distance $R$ from the axis; the weight’s thread pulls down on the outer surface, a distance $2R$ from it on the other side, so the two moments turn the pulley the same way, the way it actually turns as it sinks.

I\beta = R\,T_c + 2R\,T_A. \tag{3}

Step 4Eliminating the two tensions

The two equations (2), with the kinematics (1) put in for $w_A$ and $w_B$ , give both tensions in terms of $\beta$ alone. The first equation carries $T_A$ :

\begin{align*}T_A &= mg - m w_A \\ &= mg - 3m\beta R\end{align*}

and the second then carries $T_c$ , into which that value goes:

\begin{align*}T_c &= Mg + T_A - M w_B \\ &= Mg + (mg - 3m\beta R) - M\beta R \\ &= (M + m)g - 3m\beta R - M\beta R\end{align*}

Substituting both into the torque equation (3) and collecting the terms in $\beta$ ,

\begin{align*}I\beta &= R\bigl[(M+m)g - 3m\beta R - M\beta R\bigr] + 2R\,m\,(g - 3\beta R) \\ &= Rg\,(M + m + 2m) - \beta R^2 (3m + M + 6m) \\ &= Rg\,(M + 3m) - \beta R^2 (M + 9m) \\ I\beta + \beta R^2 (M + 9m) &= Rg\,(M + 3m)\end{align*}

\beta = \dfrac{Rg\,(M + 3m)}{I + (M + 9m)R^2}

(4)

Step 5The acceleration of the weight

The weight descends at $3R$ times the angular acceleration, by (1), so (4) carries straight over; dividing numerator and denominator by $R^2$ then clears the radius from the result.

\begin{align*}w_A &= 3R\beta \\ &= \frac{3R^2g\,(M + 3m)}{I + (M + 9m)R^2}\end{align*}

w_A = \dfrac{3g\,(M + 3m)}{M + 9m + I/R^2}

(5)

Check

Every term in the denominator is a mass, including $I/R^2$ , so the answer is $g$ times a pure number, as an acceleration must be. Take the weight away, $m = 0$ : then $w_B = \beta R = Mg/(M + I/R^2)$ , which is exactly the descent of a body unwinding on its own thread wound at radius $R$ , the standard result $g/(1 + I/MR^2)$ . And letting $I \to \infty$ sends $w_A \to 0$ : a pulley too reluctant to turn cannot let either thread unwind.

Remark

The answer is honest only while the lower thread is taut, and that is a real restriction here. Since $T_A = m(g - w_A)$ , the thread pulls rather than pushes only if $w_A \le g$ , which the boxed formula (5) turns into $I \ge 2MR^2$ . That number is precisely the moment of inertia of a uniform solid disc of radius $2R$ , so a pulley whose mass sits nearer the axis than a uniform disc of its own outer radius, an ordinary stepped pulley for instance, would carry the point $3R$ from the instantaneous center downwards faster than gravity alone could carry the weight: the thread would go slack and $A$ would simply fall freely. Notice too that the condition does not involve $m$ , so no choice of weight rescues such a pulley.