Irodov 1.143: Why the Center Never Rises – Amit & Deepali Sinha Foundation

Problem Irodov 1.143

A pulley is fixed to the ceiling, and a thread passing over it carries a body of mass $m_1$ on one end and a body of mass $m_2$ on the other, with nothing else touching either body. The pulley and the thread are both massless, and there is no friction at the axle. Determine $w_C$ , the rate at which the combined center of mass of the two bodies itself accelerates.

The plain Atwood machine: a fixed, massless pulley carries two hanging masses on a massless thread. Taking m_1 > m_2, side 1 descends while side 2 rises, both with the same magnitude of acceleration. — The plain Atwood machine: a fixed, massless pulley carries two hanging masses on a massless thread. Taking $m_1 > m_2$ , side 1 descends while side 2 rises, both with the same magnitude of acceleration.

Given

$m_1$
$m_2$
massless pulley and thread
frictionless axle

Find

Find

w_C

, the acceleration of the center of inertia of the two body system

Solution

Step 1Each body’s pull against the shared tension

Take the downward direction as positive for both bodies, and let $T$ be the thread’s tension, the same on both strands because the pulley is massless and its axle frictionless. Newton’s second law along the vertical for each body gives

\begin{aligned} m_1 g - T &= m_1 a_1,\\ m_2 g - T &= m_2 a_2. \end{aligned}

Step 2The tension cancels, the shared acceleration remains

The thread is inextensible and runs over a single fixed pulley, so whatever length one side pays out the other side takes up: the two bodies move with the same speed and the same magnitude of acceleration, but along opposite senses of the downward axis just chosen, $a_2 = -a_1$ . Writing $a_1 = a$ and substituting $a_2 = -a$ into the pair above, then subtracting the second equation from the first to remove $T$ ,

(m_1 g - T) - (m_2 g - T) = m_1 a - m_2(-a) \quad\Longrightarrow\quad (m_1 - m_2) g = (m_1 + m_2) a.

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

(1)

Body 1 therefore has $a_1 = a$ and body 2 has $a_2 = -a$ , both in the downward-positive convention: for $m_1 > m_2$ , $a$ is positive and body 1 genuinely descends while body 2 rises, matching the figure.

Step 3Weighted sum gives the center of mass

The acceleration of the center of inertia of the two body system is the mass weighted average of the two accelerations, still along the same downward-positive axis:

w_C = \frac{m_1 a_1 + m_2 a_2}{m_1+m_2} = \frac{m_1 a - m_2 a}{m_1+m_2} = a\,\frac{m_1-m_2}{m_1+m_2}.

Substituting $a$ from (1),

w_C = \frac{g(m_1-m_2)}{m_1+m_2}\cdot\frac{m_1-m_2}{m_1+m_2}.

w_C = \frac{g(m_1-m_2)^2}{(m_1+m_2)^2}

(2)

Check

For $m_1 = m_2$ the numerator vanishes and $w_C = 0$ , the balanced Atwood machine, correctly at rest. As $m_2 \to 0$ , $w_C \to g$ : with nothing on the other strand, body 1 is in free fall and the thread carries no tension, which is exactly what an unloaded second end should give.

Remark

The tension found along the way is $T = m_1(g-a) = 2m_1 m_2 g/(m_1+m_2)$ , so the total upward pull on the system is $2T = 4m_1m_2 g/(m_1+m_2)$ . Since $(m_1+m_2)^2 \ge 4m_1m_2$ for any two positive masses, $2T \le (m_1+m_2)g$ always: the string can never lift the center of mass, only slow its fall, which is why the answer above is a perfect square and can never come out negative.