Question

Figure 8–59 illustrates an ${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$ molecule. The $\mathbf{O}\mathbf{-}\mathbf{H}$ bond length is 0.096 nm and the $\mathbf{H}\mathbf{-}\mathbf{O}\mathbf{-}\mathbf{H}$ bonds make an angle of 104°. Calculate the moment of inertia of the ${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$molecule (assume the atoms are points) about an axis passing through the center of the oxygen atom (a) perpendicular to the plane of the molecule, and (b) in the plane of the molecule, bisecting the $\mathbf{H}\mathbf{-}\mathbf{O}\mathbf{-}\mathbf{H}$ bonds.

FIGURE 8-59 Problem 82

Short answer

(a) The moment of inertia of the ${\mathrm{H}}_2\mathrm{O}$molecule when the axis of rotation passes through the center of the oxygen atom perpendicular to the plane of the molecule is $3.1\times {10}^{-47}\mathrm{k}\mathrm{g}\cdot {\mathrm{m}}^2$.

(b) The moment of inertia of the ${\mathrm{H}}_2\mathrm{O}$molecule when the axis of rotation passes through the center of the oxygen atom parallel to the plane of the molecule is $1.9\times {10}^{-47}\mathrm{k}\mathrm{g}\cdot {\mathrm{m}}^2$.

Step-by-step solution

Moment of Inertia

The rotational inertia of an object is called its moment of inertia. A rotating object consists of many particles located at different distances from the axis of rotation. The sum of the mass of each particle (m) multiplied by the square of their distances (r) from the axis of rotation gives the moment of inertia (I) of that object, i.e.,

$I=\sum m{r}^2$

In this problem,the moment of inertia of the${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$molecule will be calculated by considering the hydrogen atoms as particles.

Given information

The distance between the hydrogen atom and the oxygen atom is $r=0.096\mathrm{n}\mathrm{m}=0.096\times {10}^{-9}\mathrm{m}$.

The angle formed between the$\mathrm{H}-\mathrm{O}-\mathrm{H}$bonds is $\theta ={104}^{\circ }$.

The mass of the hydrogen atom is:

$\begin{array}{r}cm=1.01\mathrm{a}.\mathrm{m}.\mathrm{u}.\\ =1.01\times \left(1.66\times {10}^{-27}\right)\mathrm{k}\mathrm{g}\\ =1.68\times {10}^{-27}\mathrm{k}\mathrm{g}\end{array}$

(a) Calculation of the moment of inertia when the axis of rotation is perpendicular to the plane of the molecule

Since the axis of rotation passes through the oxygen atom, theoxygen atom will not contribute to the total moment of inertia.

When the axis of rotation passes through the center of the oxygen atom perpendicular to the plane, themoment of inertia of the ${\mathrm{H}}_2\mathrm{O}$moleculeis:

$\begin{array}{r}cI=m{r}^2+m{r}^2\\ =2m{r}^2\\ =2\left(1.68\times {10}^{-27}\mathrm{k}\mathrm{g}\right){\left(0.096\times {10}^{-9}\mathrm{m}\right)}^2\\ =3.1\times {10}^{-47}\mathrm{k}\mathrm{g}\cdot {\mathrm{m}}^2\end{array}$

(b) Calculation of the moment of inertia when the axis of rotation is parallel to the plane of the molecule

When the axis of rotation passes through the center of the oxygen atom parallel to the plane of the molecule, the axis bisects the $\mathrm{H}-\mathrm{O}-\mathrm{H}$angle into two parts of angle ${\theta }^{\prime }={52}^{\circ }$ each. This is shown in the figure below:

From$\mathrm{\Delta }\mathrm{O}\mathrm{H}\mathrm{P}$, the distance between the hydrogen atom and the axis of rotation is:

$\begin{array}{r}c\sin {\theta }^{\prime }=\frac{\mathrm{H}\mathrm{P}}{\mathrm{O}\mathrm{H}}\\ =\frac{x}{r}\\ x=r\sin {52}^{\circ }\\ =\left(0.096\times {10}^{-9}\mathrm{m}\right)\left(0.788\right)\\ =7.56\times {10}^{-11}\mathrm{m}\end{array}$

Thus,the moment of inertiaof the${\mathrm{H}}_2\mathrm{O}$molecule when the axis of rotation is parallel to the plane of the molecule is:

$\begin{array}{r}c{I}^{\prime }=m{x}^2+m{x}^2\\ =2m{x}^2\\ =2\left(1.68\times {10}^{-27}\mathrm{k}\mathrm{g}\right){\left(7.56\times {10}^{-11}\mathrm{m}\right)}^2\\ =1.9\times {10}^{-47}\mathrm{k}\mathrm{g}\cdot {\mathrm{m}}^2\end{array}$