CENTER OF MASS OF A WATER MOLECULE
Figure 8.28 shows a simple model of a water molecule. The oxygen–hydrogen separation is d = 9.57 × 10^{-11} m. Each hydrogen atom has mass 1.0 u, and the oxygen atom has mass 16.0 u. Find the position of the center of mass.
IDENTIFY and SET UP:
Nearly all the mass of each atom is concentrated in its nucleus, whose radius is only about 10^{-5} times the overall radius of the atom. Hence we can safely represent each atom as a point particle. Figure 8.28 shows our coordinate system, with the x-axis chosen to lie along the molecule’s symmetry axis. We’ll use Eqs. (8.28):
\begin{aligned}&x_{\mathrm{cm}}=\frac{m_{1} x_{1}+m_{2} x_{2}+m_{3} x_{3}+\cdots}{m_{1}+m_{2}+m_{3}+\cdots}=\frac{\sum_{i} m_{i} x_{i}}{\sum_{i} m_{i}} \\&y_{\mathrm{cm}}=\frac{m_{1} y_{1}+m_{2} y_{2}+m_{3} y_{3}+\cdots}{m_{1}+m_{2}+m_{3}+\cdots}=\frac{\sum_{i} m_{i} y_{i}}{\sum_{i}m_{i}}\end{aligned}to find x_{\mathrm{cm}} \text { and } y_{\mathrm{cm}}.
EXECUTE:
The oxygen atom is at x = 0, y = 0. The x-coordinate of each hydrogen atom is d cos (105°/2); the y-coordinates are ±d sin (105°/2). From Eqs. (8.28),
\begin{aligned}&x_{\mathrm{cm}}=\frac{\left[\begin{array}{c}(1.0 \mathrm{u})\left(d \cos 52.5^{\circ}\right)+(1.0 \mathrm{u})\left(d \cos 52.5^{\circ}\right) \\+(16.0 \mathrm{u})(0)\end{array}\right]}{1.0 \mathrm{u}+1.0 \mathrm{u}+16.0 \mathrm{u}}=0.068 d \\&y_{\mathrm{cm}}=\frac{\left[\begin{array}{c}(1.0\mathrm{u})\left(d \sin 52.5^{\circ}\right)+(1.0 \mathrm{u})\left(-d \sin 52.5^{\circ}\right) \\+(16.0\mathrm{u})(0)\end{array}\right]}{1.0 \mathrm{u}+1.0 \mathrm{u}+16.0 \mathrm{u}}=0\end{aligned}Substituting d=9.57 \times 10^{-11} \mathrm{~m}, we find
x_{\mathrm{cm}}=(0.068)\left(9.57 \times 10^{-11} \mathrm{~m}\right)=6.5 \times 10^{-12} \mathrm{~m}
EVALUATE: The center of mass is much closer to the oxygen atom (located at the origin) than to either hydrogen atom because the oxygen atom is much more massive. The center of mass lies along the molecule’s axis of symmetry. If the molecule is rotated 180° around this axis, it looks exactly the same as before. The position of the center of mass can’t be affected by this rotation, so it must lie on the axis of symmetry.