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Centre of Gravity

two particles...

Question

two particles having masses m1 and m2 are placed at a separation r between them.locate the centre of mass of arrangement. show that the distances of centre of mass from the particles m1 amd m2 are in ratio m2:m1?

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Solution

$L e t C . M . b e a t a d i s \tan c e X f r o m m a s s m_{1} i n s u c h a w a y t h a t X = \frac{m_{1} r_{1} + m_{2} r_{2}}{m_{1} + m_{2}} r_{2} = (r - r_{1}) S o X = \frac{m_{1} r_{1} + m_{2} (r - r_{1})}{m_{1} + m_{2}} W e a r e g i v e n \frac{r_{1}}{r_{2}} = \frac{m_{1}}{m_{2}} o r r_{1} = \frac{m_{1}}{m_{2}} r_{2} = \frac{m_{1}}{m_{2}} (r - r_{1}) r_{1} (\frac{m_{1} + m_{2}}{m_{2}}) = m_{1} r o r r_{1} = \frac{m_{1} m_{2} r}{m_{1} + m_{2}} P u t t i n g t h i s v a l u e o f r_{1} i n t h e e q n o f C . M . X = \frac{m_{1} \frac{m_{1} m_{2} r}{m_{1} + m_{2}} + m_{2} (r - \frac{m_{1} m_{2} r}{m_{1} + m_{2}})}{m_{1} + m_{2}} = \frac{({m_{1}}^{2} m_{2} + m_{1} m_{2} + {m_{2}}^{2} - m_{1} {m_{2}}^{2}) r}{{(m_{1} + m_{2})}^{2}} o r X = [\frac{m_{1} m_{2} (m_{1} - m_{2}) + m_{2} (m_{1} + m_{2})}{{(m_{1} + m_{2})}^{2}}] r$

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