# Moment of Inertia of a molecule of collinear atoms

1. Nov 15, 2013

### raopeng

1. The problem statement, all variables and given/known data
Landau&Lifshitz Vol. Mechanics, p101 Q1

Find the moment of inertia of a molecule of collinear atoms

2. Relevant equations

3. The attempt at a solution
I defined the origin alone the orientation of the molecule. $I_3=0$ obviously. For $I_2$ I wrote $I_2=Ʃm_b[x_b-\frac{Ʃm_a x_a}{μ}]^2$ where μ is the total mass. But it cannot give the desired answer of $\frac{1}{μ}Ʃm_a m_b l^2_{ab}$. Thanks guys!

Last edited: Nov 15, 2013
2. Nov 15, 2013

### TSny

Hello, raopeng.

You can show that $I_2=Ʃm_b[x_b-\frac{Ʃm_a x_a}{μ}]^2$ will reduce to $\frac{1}{μ}Ʃm_a m_b l^2_{ab}$. But it's somewhat tedious.

It's a little easier if you introduce coordinates $\overline{x}_b$ relative to the center of mass: $\overline{x}_b = x_b-\frac{Ʃm_a x_a}{μ}$.

So, $I_2=\sum_bm_b\overline{x}_b^2$. To get started, note that

$I_2=\sum_bm_b\overline{x}_b^2 = \frac{\mu}{\mu}\sum_bm_b\overline{x}_b^2 = \frac{1}{\mu}\sum_a\sum_b m_a m_b\overline{x}_b^2 = \frac{1}{\mu}\sum_a\sum_b m_a m_b\overline{x}_a^2$

Consider $\frac{1}{\mu}\sum_a\sum_b m_a m_b(\overline{x}_b-\overline{x}_a)^2$ .

3. Nov 15, 2013

### raopeng

Thanks I got there in the end using that expansion though.