Two bodies of masses m_1 and m_2 (m_1 m_2) | vedclass

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(A) For an Atwood machine,the acceleration of the masses is given by $a = \frac{m_1 - m_2}{m_1 + m_2} g$.Let the downward direction be positive for $m

Vedclass · Accepted Answer

${\left( {\frac{{{m_1} - {m_2}}}{{{m_1} + {m_2}}}} \right)^2}g$

Vedclass · Answer

(A) For an Atwood machine,the acceleration of the masses is given by $a = \frac{m_1 - m_2}{m_1 + m_2} g$.Let the downward direction be positive for $m_1$ and the upward direction be positive for $m_2$. Then the acceleration of $m_1$ is $a_1 = a$ (downward) and the acceleration of $m_2$ is $a_2 = -a$ (upward).The acceleration of the centre of mass $(a_{CM})$ is given by:$a_{CM} = \frac{m_1 a_1 + m_2 a_2}{m_1 + m_2}$Substituting the values:$a_{CM} = \frac{m_1(a) + m_2(-a)}{m_1 + m_2} = \frac{(m_1 - m_2)a}{m_1 + m_2}$Now,substitute $a = \frac{m_1 - m_2}{m_1 + m_2} g$ into the equation:$a_{CM} = \left( \frac{m_1 - m_2}{m_1 + m_2} \right) \left( \frac{m_1 - m_2}{m_1 + m_2} g \right)$$a_{CM} = \left( \frac{m_1 - m_2}{m_1 + m_2} \right)^2 g$

Two bodies of masses $m_1$ and $m_2$ $(m_1 > m_2)$ are tied to the ends of a string which passes over a light frictionless pulley. The masses are initially at rest and then released. The acceleration of the centre of mass is

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