#
(p.403)
A8 Reduced Mass

- Source:
- Bonding, Structure and Solid-State Chemistry
- Author(s):
### Mark Ladd

- Publisher:
- Oxford University Press

The *reduced mass* is an effective inertial mass of a system, considered here as a two-particle problem in classical mechanics. Let the particles of masses ${m}_{1}$ and ${m}_{2}$ be at position vectors ${\mathbf{\text{r}}}_{1}$ and ${\mathbf{\text{r}}}_{2}$ with respect to an origin at the centre of mass *C*; *r* is the linear distance between ${m}_{1}$ and ${m}_{2}$ (Fig. A8.1).

Then:

(A8.1)$$r={r}_{1}+{r}_{2}$$

and

(A8.2)$${m}_{1}{r}_{1}={m}_{2}{r}_{2}$$

Now:

(A8.3)$${r}_{2}=r-{r}_{1}$$

Hence:

(A8.4)$${m}_{1}{r}_{1}={m}_{2}(r-{r}_{1})$$

so that

(A8.5)$${r}_{1}={m}_{2}r/({m}_{1}+{m}_{2})$$

and

(A8.6)$${r}_{2}={m}_{1}r/({m}_{1}+{m}_{2})$$

The moment of inertia *I* of the system is given by

(A8.7)$$\begin{array}{rl}I={m}_{1}{r}_{1}^{2}+{m}_{2}{r}_{2}^{2}& ={m}_{1}{\left(\frac{{m}_{2}r}{{m}_{1}+{m}_{2}}\right)}^{2}+{m}_{2}{\left(\frac{{m}_{1}r}{{m}_{1}+{m}_{2}}\right)}^{2}\\ & =\frac{{m}_{1}{m}_{2}^{2}{r}^{2}}{({m}_{1}+{m}_{2}{)}^{2}}+\frac{{m}_{1}^{2}{m}_{2}{r}^{2}}{({m}_{1}+{m}_{2}{)}^{2}}=\frac{{m}_{1}{m}_{2}({m}_{1}+{m}_{2}){r}^{2}}{({m}_{1}+{m}_{2}{)}^{2}}\end{array}$$

(A8.8)$$I=\frac{{m}_{1}{m}_{2}{r}^{2}}{({m}_{1}+{m}_{2})}=\mathrm{\mu}{r}^{2}$$

(p.404)
where *μ* is the *reduced mass* of the system:

(A8.9)$$\mathrm{\mu}=\frac{{m}_{1}{m}_{2}}{({m}_{1}+{m}_{2})}$$

A similar expression arises for the moment of inertia of a diatomic molecule, in which case *m*_{1} and *m*_{2} are the masses of the two atoms and *r* is the length of the bond between them.