Moment of Inertia of a molecule of collinear atoms

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Homework Help Overview

The problem involves calculating the moment of inertia of a molecule composed of collinear atoms, as presented in a question from Landau & Lifshitz. The discussion centers around the mathematical formulation and assumptions related to the orientation of the molecule.

Discussion Character

  • Mathematical reasoning, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore different formulations for the moment of inertia, with one attempting to define the origin along the molecule's orientation and expressing the moment of inertia in terms of mass distributions. Others question the assumptions regarding the molecule's orientation and the implications for the calculations.

Discussion Status

The discussion includes various attempts to derive the moment of inertia, with some participants providing insights into simplifying the expressions. There is a recognition of the complexity involved in the calculations, and some participants express confusion about the assumptions made regarding the molecule's orientation.

Contextual Notes

There is a mention of the need for clarity regarding the orientation of the molecule, specifically whether it lies along the z-axis or in the x-y plane, which affects the calculations. Additionally, one participant notes the importance of stating the relevant equations for the 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. [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!
 
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Hello, raopeng.

raopeng said:
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!

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## .
 
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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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