# Derivation of reduced mass

1. Jun 25, 2006

### Jex

Does anyone know how to actually derive the reduced mass formula? I have to prove the formula: inertia = (reduced mass)r^2 and am having some difficulties.

To be more specific I'm working with the inertia between to atoms where r is the bond length.

Any help with this is much appreciated, really.

2. Jun 26, 2006

### Hootenanny

Staff Emeritus
3. Jun 26, 2006

### tim_lou

are you talking about the 2 body problem or the moment of inertial? if you are talking about the moment, then around which axis is it calculated? the axis orthogonal to the plane of rotation and at the center of mass of the system?

4. Jun 26, 2006

### tim_lou

if it is what i think it is, then:
let a=r1, b=r2, a+b=r, m be the mass of the first body, M be the mass of the second body, and r be the bond length:
$I=ma^2+Mb^2$
relative to the center of mass:
$ma=Mb$
$I=(ma)a+(Mb)b$
$I=ma(a+b)=mar$
$a=r-b=r-ma/M$
$a(1+m/M)=r$
$I=mar={m\over{(1+m/M)}}r^2$
so
I=reduced mass*r^2

Last edited: Jun 26, 2006