The two bodies of mass $m_{1}$ and $m_{2}(m_{1} m_{2})$ respectively are tied to the ends of a massless string, which passes over a light and frictionless pulley. The masses are initially at rest and the released. Then acceleration of the centre of mass of the system is [image:7_image28.jpeg]

Question

The two bodies of mass $m_{1}$ and $m_{2}(m_{1} > m_{2})$ respectively are tied to the ends of a massless string, which passes over a light and frictionless pulley. The masses are initially at rest and the released. Then acceleration of the centre of mass of the system is [image:7_image28.jpeg]

Dr. Abubacker · Accepted Answer

(a) In the pulley arrangement $\left| {\overrightarrow{a}}_{1} ight| = \left| {\overrightarrow{a}}_{2} ight| = a = \left( \frac{m_{1} - m_{2}}{m_{1} + m_{2}} ight)g$
But ${\overrightarrow{a}}_{1}$ is in downward direction and in the upward direction $ie$, ${\overrightarrow{a}}_{2} = - {\overrightarrow{a}}_{1}$
$	herefore$ Acceleration of centre of mass
${\overrightarrow{a}}_{CM} = \frac{m_{1}{\overrightarrow{a}}_{1} + m_{2}{\overrightarrow{a}}_{2}}{m_{1} + m_{2}} = \frac{m_{1}\left\lbrack \frac{m_{1} - m_{2}}{m_{1} + m_{2}} ightbrack g - m_{2}\left\lbrack \frac{m_{1} - m_{2}}{m_{1} + m_{2}} ightbrack g}{(m_{1} + m_{2})}$
$= \left\lbrack \frac{m_{1} - m_{2}}{m_{1} + m_{2}} ightbrack^{2}g$

Q. The two bodies of mass $m_{1}$ and $m_{2}(m_{1} > m_{2})$ respectively are tied to the ends of a massless string, which passes over a light and frictionless pulley. The masses are initially at rest and the released. Then acceleration of the centre of mass of the system is

Explanation

Reviewed by Dr. Abubacker

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