Calculate the moment of inertia of an $\mathrm{H}_2 \mathrm{O}$ molecule around the axis defined by the bisector of the $\mathrm{HOH}$ angle (3). The $\mathrm{HOH}$ bond angle is $104.5^{\circ}$ and the bond length is $95.7 \mathrm{pm}$.
From eqn 12.2,
$$
I=\sum_i m_i x_i^2=m_{\mathrm{H}} x_{\mathrm{H}}^2+0+m_{\mathrm{H}} x_{\mathrm{H}}^2=2 m_{\mathrm{H}} x_{\mathrm{H}}^2
$$
If the bond angle of the molecule is denoted $2 \phi$ and the bond length is $R$, trigonometry gives $x_{\mathrm{H}}=R \sin \phi$. It follows that
$$
I=2 m_{\mathrm{H}} R^2 \sin ^2 \phi
$$
Substitution of the data gives
$$
\begin{aligned}
I & =2 \times\left(1.67 \times 10^{-27} \mathrm{~kg}\right) \times\left(9.57 \times 10^{-11} \mathrm{~m}\right)^2 \times \sin ^2\left(\frac{1}{2} \times 104.5^{\circ}\right) \\
& =1.91 \times 10^{-47} \mathrm{~kg} \mathrm{~m}^2
\end{aligned}
$$
Note that the mass of the $\mathrm{O}$ atom makes no contribution to the moment of inertia for this mode of rotation as the atom is immobile while the $\mathrm{H}$ atoms circulate around it.
