A pulley fixed to the ceiling carries a thread with bodies of masses $m_{1}$ and $m_{2}$ attached to its ends. The masses of the pulley and the thread are negligible, friction is absent. Find the acceleration $w_{C}$ of the centre of mass of this system :

\frac{{(m_{1} m_{2})}^{2} \to g}{{(m_{1} - m_{2})}^{2}}

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\frac{{(m_{1} + m_{2})}^{2} \to g}{{(m_{1} - m_{2})}^{2}}

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\frac{{(m_{1} m_{2})}^{2} \to g}{{(m_{1} + m_{2})}^{2}}

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\frac{{(m_{1} - m_{2})}^{2} \to g}{{(m_{1} + m_{2})}^{2}}

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Solution

The correct option is D $\frac{{(m_{1} - m_{2})}^{2} \to g}{{(m_{1} + m_{2})}^{2}}$
We know that acceleration of centre of mass of the system is given by the expression,

$_{C} = \frac{m_{1}_{1} + m_{2}_{2}}{m_{1} + m_{2}}$

Since $_{1} = -_{2}$

$_{C} = \frac{(m_{1} - m_{2})_{1}}{m_{1} + m_{2}}$ (1)

Now from Newton's second law $\to F = m \to w$ , for the bodies $m_{1}$ and $m_{2}$ respectively.

$\to T + m_{1} \to g = m_{1}_{1}$ (2)

and $\to T + m_{2} \to g = m_{2}_{2} = - m_{2}_{1}$ (3)

Solving (2) and (3)

$_{1} = \frac{(m_{1} - m_{2}) \to g}{m_{1} + m_{2}}$ (4)

Thus from (1), (2) and (4),

$_{C} = \frac{{(m_{1} - m_{2})}^{2} \to g}{{(m_{1} + m_{2})}^{2}}$

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A pulley fixed to the ceiling carries a thread with bodies of masses m1 and m2 attached to its ends. The masses of the pulley and the thread are negligible, friction is absent. Find the acceleration wC of the centre of mass of this system :

A pulley fixed to the ceiling carries a thread with bodies of masses $m_{1}$ and $m_{2}$ attached to its ends. The masses of the pulley and the thread are negligible, friction is absent. Find the acceleration $w_{C}$ of the centre of mass of this system :