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Moment of Inertia of a molecule of collinear atoms

  1. Nov 15, 2013 #1
    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. [itex]I_3=0[/itex] obviously. For [itex]I_2[/itex] I wrote [itex]I_2=Ʃm_b[x_b-\frac{Ʃm_a x_a}{μ}]^2[/itex] where μ is the total mass. But it cannot give the desired answer of [itex]\frac{1}{μ}Ʃm_a m_b l^2_{ab}[/itex]. Thanks guys!
    Last edited: Nov 15, 2013
  2. jcsd
  3. Nov 15, 2013 #2


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    Hello, raopeng.

    You can show that [itex]I_2=Ʃm_b[x_b-\frac{Ʃm_a x_a}{μ}]^2[/itex] will reduce to [itex]\frac{1}{μ}Ʃm_a m_b l^2_{ab}[/itex]. 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## .
  4. Nov 15, 2013 #3
    Thanks I got there in the end using that expansion though.
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