Question

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

Answer 1

Let OX be the x-axis, OY be the Y-axis and O be the origin.



$\mathrm{Mass}\mathrm{of}\mathrm{O}\mathrm{atom},m_1=16\mathrm{unit}$
Let the position of oxygen atom be origin.
$$\begin{gathered} \Rightarrow x_1=y_1=0 \\ \mathrm{Mass}\mathrm{of}\mathrm{H}\mathrm{atom},m_2=1\mathrm{unit} \\ {x}_2=-0.96\times {10}^{-10}\sin 52°\mathrm{m} \\ {y}_2=-0.96\times {10}^{-10}\cos 52° \\ \mathrm{Now},m_3=1\mathrm{unit} \\ {x}_3=0.96\times {10}^{-10}\sin 52°\mathrm{m} \\ {y}_3=-0.96\times {10}^{-10}\cos 52° \\ \mathrm{The}X\mathrm{coordinate}\mathrm{of}\mathrm{the}\mathrm{center}\mathrm{of}\mathrm{mass}\mathrm{is}\mathrm{given}\mathrm{by}: \\ {x}_{\mathrm{cm}}=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3}{m_1+m_2+m_3} \\ =\frac{16\times 0+1\times \left(-0.96\times {10}^{-10}\sin 52°\right)+1\times 0.96\times {{10}^{-}}^{10}\sin 52°}{1+1+16}=0 \\ \mathrm{The}Y\mathrm{coordinate}\mathrm{of}\mathrm{the}\mathrm{center}\mathrm{of}\mathrm{mass}\mathrm{is}\mathrm{given}\mathrm{by}: \\ {y}_{\mathrm{cm}}=\frac{m_1{y}_1+m_2{y}_2+m_3{y}_3}{m_1+m_2+m_3} \\ =\frac{16\times 0+2\times 0.96\times {10}^{-10}\cos 52°}{1+1+16} \\ =\frac{2\times 0.96\times {10}^{-10}\cos 52°}{18} \\ =6.4\times {10}^{-12}\mathrm{m} \end{gathered}$$

Hence, the distance of centre of mass of the molecule from the centre of the oxygen atom is $(0,6.4\times {10}^{-12}\mathrm{m})$.