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Question: A pulley fixed to the ceiling carried a thread with bodies of masses $m_1$ and $m_2$ attached to its...

A pulley fixed to the ceiling carried a thread with bodies of masses m1m_1 and m2m_2 attached to its ends. The masses of the pulley and the thread are negligible and friction is absent. Find the acceleration of the centre of mass of this system.

Answer

(m1−m2)2(m1+m2)2g downwards\frac{(m_1-m_2)^2}{(m_1+m_2)^2}g \text{ downwards}

Explanation

Solution

Let m1m_1 and m2m_2 be the masses of the two bodies. Let TT be the tension in the thread. Let aa be the magnitude of the acceleration of each mass. Assuming m2>m1m_2 > m_1, mass m1m_1 accelerates upwards and mass m2m_2 accelerates downwards.

The equation of motion for mass m1m_1 (upwards is positive): T−m1g=m1aT - m_1g = m_1a (1)

The equation of motion for mass m2m_2 (downwards is positive): m2g−T=m2am_2g - T = m_2a (2)

Adding equations (1) and (2): (T−m1g)+(m2g−T)=m1a+m2a(T - m_1g) + (m_2g - T) = m_1a + m_2a (m2−m1)g=(m1+m2)a(m_2 - m_1)g = (m_1 + m_2)a a=(m2−m1)gm1+m2a = \frac{(m_2 - m_1)g}{m_1 + m_2}

If we choose the upward direction as positive, the acceleration of m1m_1 is a⃗1=aj^=(m2−m1)gm1+m2j^\vec{a}_1 = a \hat{j} = \frac{(m_2 - m_1)g}{m_1 + m_2} \hat{j}. The acceleration of m2m_2 is a⃗2=−aj^=−(m2−m1)gm1+m2j^\vec{a}_2 = -a \hat{j} = -\frac{(m_2 - m_1)g}{m_1 + m_2} \hat{j}.

The acceleration of the center of mass of the system is given by: a⃗CM=m1a⃗1+m2a⃗2m1+m2\vec{a}_{CM} = \frac{m_1 \vec{a}_1 + m_2 \vec{a}_2}{m_1 + m_2} a⃗CM=m1((m2−m1)gm1+m2j^)+m2(−(m2−m1)gm1+m2j^)m1+m2\vec{a}_{CM} = \frac{m_1 \left( \frac{(m_2 - m_1)g}{m_1 + m_2} \hat{j} \right) + m_2 \left( -\frac{(m_2 - m_1)g}{m_1 + m_2} \hat{j} \right)}{m_1 + m_2} a⃗CM=m1(m2−m1)g−m2(m2−m1)gm1+m2m1+m2j^\vec{a}_{CM} = \frac{\frac{m_1(m_2 - m_1)g - m_2(m_2 - m_1)g}{m_1 + m_2}}{m_1 + m_2} \hat{j} a⃗CM=(m2−m1)g(m1−m2)(m1+m2)2j^\vec{a}_{CM} = \frac{(m_2 - m_1)g (m_1 - m_2)}{(m_1 + m_2)^2} \hat{j} a⃗CM=−(m2−m1)2g(m1+m2)2j^\vec{a}_{CM} = -\frac{(m_2 - m_1)^2 g}{(m_1 + m_2)^2} \hat{j}

The acceleration of the center of mass is in the downward direction (since j^\hat{j} is upwards). The magnitude of the acceleration of the center of mass is (m2−m1)2g(m1+m2)2\frac{(m_2 - m_1)^2 g}{(m_1 + m_2)^2}.

If m1>m2m_1 > m_2, then m1m_1 accelerates downwards and m2m_2 accelerates upwards with magnitude a′=(m1−m2)gm1+m2a' = \frac{(m_1 - m_2)g}{m_1 + m_2}. In this case, a⃗1=−a′j^=−(m1−m2)gm1+m2j^\vec{a}_1 = -a' \hat{j} = -\frac{(m_1 - m_2)g}{m_1 + m_2} \hat{j} and a⃗2=a′j^=(m1−m2)gm1+m2j^\vec{a}_2 = a' \hat{j} = \frac{(m_1 - m_2)g}{m_1 + m_2} \hat{j}. a⃗CM=m1a⃗1+m2a⃗2m1+m2=m1(−(m1−m2)gm1+m2j^)+m2((m1−m2)gm1+m2j^)m1+m2\vec{a}_{CM} = \frac{m_1 \vec{a}_1 + m_2 \vec{a}_2}{m_1 + m_2} = \frac{m_1 \left( -\frac{(m_1 - m_2)g}{m_1 + m_2} \hat{j} \right) + m_2 \left( \frac{(m_1 - m_2)g}{m_1 + m_2} \hat{j} \right)}{m_1 + m_2} a⃗CM=−m1(m1−m2)g+m2(m1−m2)gm1+m2m1+m2j^\vec{a}_{CM} = \frac{\frac{-m_1(m_1 - m_2)g + m_2(m_1 - m_2)g}{m_1 + m_2}}{m_1 + m_2} \hat{j} a⃗CM=(m1−m2)g(−m1+m2)(m1+m2)2j^=−(m1−m2)2g(m1+m2)2j^\vec{a}_{CM} = \frac{(m_1 - m_2)g (-m_1 + m_2)}{(m_1 + m_2)^2} \hat{j} = -\frac{(m_1 - m_2)^2 g}{(m_1 + m_2)^2} \hat{j}.

In both cases, the acceleration of the center of mass is downwards with magnitude (m1−m2)2g(m1+m2)2\frac{(m_1 - m_2)^2 g}{(m_1 + m_2)^2}. We can write this as (m1−m2)2(m1+m2)2g\frac{(m_1 - m_2)^2}{(m_1 + m_2)^2} g downwards.