A pulley fixed to the ceiling carried 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 and friction is absent. Find the acceleration of the center of mass of this system.

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

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$(m_{1} + m_{2}) g$

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

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

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Solution

The correct option is D
$(\frac{(m_{2} - m_{1})^{2} g}{(m_{1} + m_{2})^{2}})$

Let us assume that $m_{2} < m_{1}$

We can see that the masses have equal and opposite acceleration of same magnitude.

$α$ = $\frac{(m_{2} - m_{1})}{m_{2} + m_{1}}$ g and tension in string is T = $\frac{2 m_{1} m_{2} g}{m_{2} + m_{1}}$

Taking download direction as positive, $a_{1} = - a; a_{2} = + a$

$α$ = $\frac{m_{1} α_{1} + m_{2} α_{2}}{m_{1} + m_{2}}$ = $\frac{(m_{2} - m_{1}) α}{m_{1} + m_{2}}$

Substituting for the value of a, we have : $α_{c m}$ = ${(\frac{m_{2} - m_{1}}{m_{2} + m_{1}})}^{2}$ g

Alternative Method

$α_{c m}$ = $\frac{f_{e x t .}}{m_{1} + m_{2}}$ = $\frac{(m_{1} g + m_{2} g) - 2 T}{m_{1} + m_{2}}$ $\Rightarrow$ $α_{c m}$ = $g - \frac{4 m_{1} m_{2} g}{(m_{2} + m_{1})^{2}}$

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

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

A pulley fixed to the ceiling carried 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 and friction is absent. Find the acceleration of the center of mass of this system.