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Question

(A) Let the masses be $m_1 = 2m$ and $m_2 = 6m$.The acceleration of the masses in an Atwood machine is given by $a = \frac{|m_2 - m_1|}{m_1 + m_2} g$.

Vedclass · Accepted Answer

$g/4$

Vedclass · Answer

(A) Let the masses be $m_1 = 2m$ and $m_2 = 6m$.The acceleration of the masses in an Atwood machine is given by $a = \frac{|m_2 - m_1|}{m_1 + m_2} g$.Substituting the values,$a = \frac{|6m - 2m|}{6m + 2m} g = \frac{4m}{8m} g = \frac{g}{2}$.The acceleration of the centre of mass is $a_{cm} = \frac{m_1 a_1 + m_2 a_2}{m_1 + m_2}$.Here,$a_1 = a$ (upwards) and $a_2 = -a$ (downwards).So,$a_{cm} = \frac{m_1 a - m_2 a}{m_1 + m_2} = \frac{(2m - 6m) a}{2m + 6m} = \frac{-4m}{8m} a = -\frac{1}{2} a$.The magnitude of the acceleration of the centre of mass is $|a_{cm}| = \frac{1}{2} a = \frac{1}{2} 	imes \frac{g}{2} = \frac{g}{4}$.

If the system is released,then the acceleration of the centre of mass of the system is

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