Irodov 1.246: Why the Two Tensions Cannot Match – Amit & Deepali Sinha Foundation

Problem Irodov 1.246

A uniform solid cylinder of mass $m$ and radius $R$ is free to turn about a fixed horizontal axle through its own axis. Thread is wound on the cylinder and leaves it vertically downward on each side: the section on one side, call it section 1, carries a hanging body of mass $m_1$ , and the section on the other side, section 2, carries a hanging body of mass $m_2$ . The thread does not slip on the cylinder and there is no friction in the axle. While the system is moving, find the angular acceleration of the cylinder and the ratio $T_1/T_2$ of the tensions in the two vertical sections.

A uniform solid cylinder on a fixed axle in a ceiling bracket. Thread section 1 comes down on the left carrying m_1, section 2 on the right carrying m_2. The angular acceleration β is drawn for the case m_2 > m_1. — A uniform solid cylinder on a fixed axle in a ceiling bracket. Thread section 1 comes down on the left carrying $m_1$ , section 2 on the right carrying $m_2$ . The angular acceleration $\vect\beta$ is drawn for the case $m_2 > m_1$ .

Given

$m$ (cylinder)
$R$
$m_1$ , $m_2$ (the two loads)
no slipping of the thread, no friction in the axle

Find

the angular acceleration

\beta

of the cylinder, and the ratio

T_1/T_2

Solution

Step 1One angular acceleration for both loads

No slipping means the thread moves with the rim point it touches, so each load’s speed equals the rim speed and each load’s acceleration follows the rim’s angular acceleration. Label the loads so that $m_2$ is the one that descends; nothing is lost by that, because exchanging the two labels runs exactly the same algebra and only reverses the sense of the motion. Write $a$ for the common magnitude of the two loads’ accelerations.

\begin{align*}v &= \omega R \\ a &= \frac{dv}{dt} = R\,\frac{d\omega}{dt}\end{align*}

a = \beta R

(1)

Free bodies of the two loads. Each carries its own weight and the tension in its own section of thread, T_1 and T_2. The accelerations a_1 and a_2 are equal in magnitude and opposite in direction. — Free bodies of the two loads. Each carries its own weight and the tension in its own section of thread, $\vect T_1$ and $\vect T_2$ . The accelerations $\vect a_1$ and $\vect a_2$ are equal in magnitude and opposite in direction.

Step 2Newton’s second law for each load

Each load is a point mass acted on by its own weight and its own thread tension. Resolve along the direction each one actually accelerates: upward for $m_1$ , downward for $m_2$ .

\begin{aligned} T_1 - m_1 g &= m_1 a \\ m_2 g - T_2 &= m_2 a \end{aligned} \tag{2}

The cylinder by itself. The two tensions pull down at rim points on opposite sides of the axis, each with the same moment arm R, drawn as the gray leader. The angular acceleration β is marked at the axle. — The cylinder by itself. The two tensions pull down at rim points on opposite sides of the axis, each with the same moment arm $R$ , drawn as the gray leader. The angular acceleration $\vect\beta$ is marked at the axle.

Step 3The torque equation for the cylinder

Only the two tensions have a moment about the axle, and they oppose each other. A uniform solid cylinder has moment of inertia $mR^{2}/2$ about its own axis, and the constraint (1) turns $\beta$ into $a/R$ .

\begin{align*}(T_2 - T_1)R &= I\beta \\ &= \frac{1}{2}mR^{2}\cdot\frac{a}{R}\end{align*}

T_2 - T_1 = \frac{1}{2}ma

(3)

Step 4Eliminating both tensions

Adding the two load equations (2) leaves only the combination $T_2 - T_1$ , which (3) has already expressed through $a$ .

\begin{align*}(m_2 - m_1)g - (T_2 - T_1) &= (m_1 + m_2)a \\ (m_2 - m_1)g - \frac{1}{2}ma &= (m_1 + m_2)a \\ (m_2 - m_1)g &= \left(m_1 + m_2 + \frac{m}{2}\right)a\end{align*}

a = \frac{(m_2 - m_1)g}{m_1 + m_2 + m/2}

(4)

Dividing by $R$ gives the angular acceleration. Restoring the case in which $m_1$ is the heavier body, where the whole motion runs the other way, the magnitude carries an absolute value.

\beta = \frac{|m_2 - m_1|\,g}{\left(m_1 + m_2 + m/2\right)R}

(5)

Its sense is the one in which the heavier load descends.

Step 5The two tensions and their ratio

Each load equation now gives its own tension once $a$ from (4) is put back into it. The first load is rising, so its tension exceeds its weight.

\begin{align*}T_1 &= m_1(g + a) \\ &= m_1 g\left(1 + \frac{m_2 - m_1}{m_1 + m_2 + m/2}\right) \\ &= m_1 g\,\frac{2m_2 + m/2}{m_1 + m_2 + m/2}\end{align*}

The second load is descending, so the same algebra runs with the sign of $a$ reversed and its tension falls short of its weight.

\begin{align*}T_2 &= m_2(g - a) \\ &= m_2 g\left(1 - \frac{m_2 - m_1}{m_1 + m_2 + m/2}\right) \\ &= m_2 g\,\frac{2m_1 + m/2}{m_1 + m_2 + m/2}\end{align*}

In the ratio both $g$ and the common denominator cancel, and doubling numerator and denominator clears the halves.

\frac{T_1}{T_2} = \frac{m_1\left(2m_2 + m/2\right)}{m_2\left(2m_1 + m/2\right)}

\frac{T_1}{T_2} = \frac{m_1(m + 4m_2)}{m_2(m + 4m_1)}

(6)

Check

Set $m = 0$ , the massless drum: the ratio becomes $4m_1m_2/4m_1m_2 = 1$ , so the two sections carry one and the same tension, and $\beta R$ drops to $|m_2-m_1|g/(m_1+m_2)$ , the textbook Atwood acceleration. Set $m_1 = m_2$ instead: $\beta$ vanishes and the ratio is 1, which is the static balance the arrangement must have when the two loads are equal. Dimensionally $\beta$ is (mass times acceleration) over (mass times length), that is $\mathrm{s}^{-2}$ , and the ratio is a pure number.

Remark

The answer interpolates between two familiar machines as the drum’s mass runs from nothing to everything. At $m = 0$ the ratio is 1: a massless pulley merely redirects one tension. As $m \to \infty$ the ratio tends to $m_1/m_2$ , which is what two threads tied to a wall would read, since a drum too massive to be spun up leaves each load hanging in equilibrium at $T_i = m_i g$ . In between the ratio moves monotonically from 1 toward $m_1/m_2$ : differentiating $(m+4m_2)/(m+4m_1)$ with respect to $m$ gives a numerator proportional to $m_1-m_2$ , so the ratio only ever moves in the one direction. That monotonicity is what makes the relation invertible, provided the two loads differ: with $m_1$ and $m_2$ known and unequal, measuring the two tensions weighs the drum. Equal loads carry no such information, since the derivative above vanishes with $m_1-m_2$ and the ratio is 1 whatever the drum’s mass.