Question

The structure of a water molecule is shown in the figure. Find the distance of the center of mass of the molecule from the center of the oxygen atom.

Answer 1

$m_1=1gm$

$m_2=1gm$

$x_1=-(0.96\times {10}^{-10})sin{52}^o$

$x_2=-(0.96\times {10}^{-10})sin{52}^o$

$x_3=0$

$y_1=0$

$y_2=0$

$y_3=(0.96\times {10}^{-10})cos{52}^o$

So , the position of centre of mass is =

$[\frac{(m_1{x}_1+m_2{x}_2+m_3{x}_3)}{(m_1+m_2+m_3)},\frac{(m_1{y}_1+m_2{y}_2+m_3{y}_3)}{(m_1+m_2+m_3)}]$

$=[\frac{(-(0.96\times {10}^{-10})sin{52}^o+(0.96\times {10}^{-10})sin{52}^o+16\times 0)}{(1+1+16)},\frac{(0+0+16\times (0.96\times {10}^{-10}\left)cos{52}^o\right)}{\left(18\right)}]$

$=(0,\frac{8}{9}\times 0.96\times {10}^{-10}cos{52}^o)$