Moment of Inertia of a molecule of collinear atoms

AI Thread Summary
The discussion focuses on calculating the moment of inertia of a molecule composed of collinear atoms. The original poster attempts to derive the moment of inertia using the equation I_2=Ʃm_b[x_b-Ʃm_a x_a/μ]^2 but struggles to reach the expected result of (1/μ)Ʃm_a m_b l^2_{ab}. Other participants suggest simplifying the problem by introducing coordinates relative to the center of mass, which leads to a more manageable form of the equation. Clarifications are made regarding the orientation of the molecule along the z-axis versus the x-y plane, emphasizing the importance of correctly defining the coordinate system. Ultimately, the conversation revolves around refining the approach to achieve the correct calculation of moment of inertia.
raopeng
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Homework Statement


Landau&Lifshitz Vol. Mechanics, p101 Q1Find the moment of inertia of a molecule of collinear atoms

Homework Equations


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!
 
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Hello, raopeng.

raopeng said:
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!

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

I was working my way through this problem but couldn't even begin. Could someone explain more explicitly?

My attempt at a solution:
Assuming the molecule lies along the z axis (z=0), the principle moment of inertia should be:
1. I1 = summation(m*ysquare)
2. I2 = summation (m*xsquare)
3. I3 = summation (m*(xsquare + ysquare))

But this is nowhere close to the answer! Please help
 
It would help if you posted the question itself too (not all of us have the book).

Your assumption is confusing. If it lies along the z-axis, that means all x and y are 0 .
If it lies in the x-y plane, that means all z are 0.
From the context I guess you mean the latter.

Start with stating the relevant equation for ##I##.
 
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