Moment of inertia of a collinear system of particles

[Physgeek64]

Homework Statement

Landau&Lifshitz Vol. Mechanics, p101 Q1

Find the moment of inertia of a collinear set of molecules

Homework Equations

I=sum m_i*(r_i^2)

The Attempt at a Solution

r_i= r_i'-r_cm where r_cm=sum m_i*r_i/sum m_i
I=sum m_i r_i^2
I= 1/M sum m_i m_j (r_i' - r_cm)^2
I=1/M sum m_i m_j (r_i'^2 -2r_i' r_cm + r_cm^2)

The central term will reduce to zero by definition of the COM leaving

I=1/M sum m_i m_j (r_i'^2 + r_cm^2)

The answer of 1/M sum m_i m_j r_ij is given, where r_ij is the distance between the ith and jth particle, and the summation is over i not equal to j, but I'm not sure how to simplify this into that expression- I have attempted it a few different ways but none have yet to work. Any help would be much appreciated

Thanks in advance :)

 

[Fred Wright]

Using L&L's notation I write
I₁ = I₂ = I = Σ_am_a(l_cm-l_a)²
l_cm = (1/∑_bm_b)∑_bm_bl_b
I = Σ_am_a((1/∑_bm_b)∑_bm_bl_b - l_a)² = Σ_am_a[(1/∑_bm_b)²(∑_bm_bl_b-∑_bm_bl_a)²]
= Σ_am_a[(1/∑_bm_b)²(∑_bm_bl_ab)²]
where l_ab = l_a-l_b
This double sum looks pretty intractable so L&L suggest that the summation includes one term for every pair of atoms in the molecule and l_ab is the distance between the pair. I assert that the sum in the denominator is only over each pair. That is,
μ_ab = ∑_I=a^I=bm_i = m_a + m_b
I = Σ_am_a[(1/μ_ab)²](∑_bm_bl_ab)²]
For example, expanding this double sum for 4 atoms,
m₁(m₂l₁₂)²/(μ₁₂)² + m₂(m₁l₂₁)²/(μ₂₁)² + m₃(m₄l₃₄)²/(μ₃₄)² + m₄(m₃l₄₃)²/(μ₄₃)²
= m₁m₂(l₁₂)²/μ₁₂ + m₃m₄(l₃₄)²/μ₃₄

 

[Sturk200]

I was just stuck on this problem for a while, but I think I got it now and hope that I can help someone else see how it goes.

The thing about the formula that is given in the textbook is that it is valid for any system of coordinates -- i.e. one need not assume that the coordinates are defined relative to the center of mass of the system. Of course, if one does assume that the origin is at the center of mass, the solution is trivially $\Sigma m x^2$. But we will not make this assumption, and therefore will not be allowed to eliminate terms like $\Sigma m x$. I suppose the advantage to having a formula that is valid for any coordinate system is obvious; one need not calculate the center of mass and the positions of atoms with respect to it.

Start with the expression $$I=\Sigma_a m_a (x_a - \frac{1}{\mu}\Sigma_b m_b x_b)^2 $$ Here $\mu$ is the total mass of the system, $m_b$ is the mass of the $b^{th}$ particle and $x_b$ is its position. We can expand the square to give $$\Sigma_a m_a (x_a^2 + \frac{1}{\mu^2}\Sigma_b\Sigma_j m_b m_j x_b x_j - \frac{2}{\mu}x_a\Sigma m_b x_b) =$$ $$\Sigma_a m_a x_a^2 + \frac{1}{\mu^2}\Sigma_a m_a \Sigma_b \Sigma_j m_b m_j x_b x_j - \frac{2}{\mu}\Sigma_a m_a x_a \Sigma_b m_b x_b =$$ $$\Sigma_a m_a x_a^2 + \frac{1}{\mu}\Sigma_a \Sigma_b m_b m_a x_b x_a - \frac{2}{\mu}\Sigma_a\Sigma_b m_a x_a m_b x_b=$$ $$ \Sigma_a m_a x_a^2 - \frac{1}{\mu}\Sigma_a\Sigma_b m_a x_a m_b x_b =$$ $$\frac{1}{\mu}\Sigma_a \Sigma_b m_a m_b (x_a^2 - x_a x_b).$$

Getting from the second to the third line I used the fact that $\mu = \Sigma_a m_a$ and then changed the index in one of the remaining sums from $j$ to $a$ to make clear that the second and third terms are proportional. And in getting from the fourth to the fifth line I multiplied the first term by $\mu$ in the form of $\Sigma_a m_a$ and divided it by $\mu$ in the form of $\mu$.

Good, now comes the tricky part where you have to know the answer to get the answer. $\Sigma_a \Sigma_b m_a m_b x_a^2 = \Sigma_a \Sigma_b m_a m_b x_b^2$, so we can rewrite our last expression as $$\frac{1}{2\mu}\Sigma_a \Sigma_b m_a m_b (x_a^2 + x_b^2 - 2x_a x_b) = \frac{1}{2\mu}\Sigma_a \Sigma_b m_a m_b (x_a - x_b)^2.$$ Finally, one must specify verbally that the sum is meant to be taken only over pairs of particles, which allows one to eliminate the factor of 1/2 (the sum as it is written counts each pair twice, of course). This gives us the final result ($l_{ab}$ is the distance between atoms $a$ and $b$) $$I=\frac{1}{\mu}\Sigma_{pairs \; a\neq b}m_a m_b l_{ab}^2.$$

And that's how it goes.