Moment of Inertia tensor - displaced axes theorem:

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SUMMARY

The discussion focuses on calculating the moment of inertia tensor for a system of two spheres (m1 and m2) connected by a massless rod, utilizing the displaced axes theorem. The moment of inertia for each sphere is given by I3(sphere a) = 2/5Mi/r^2, leading to I3 = 2/5(M1a^2 + M2b^2). The application of the displaced axes theorem is clarified, emphasizing that it is applied to axes through the center of each sphere, not to I3 directly. The justification for summing the moments of inertia of both spheres is based on their symmetry and the integral definition of moment of inertia.

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  • Understanding of moment of inertia tensor
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  • Basic calculus for integrating mass distributions
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binbagsss
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Ok, so the system consists of two massive spheres, m1 and m2, of radii a and b respectively, connected by a massless rod of length R, as seen in the diagram attached.

The question is to calculate the moment of inertia tensor.

Sol:
Set the origin at the centre of mass . So that we are in the principal axis frame we chose I3 to be along the line connecting the masses.
=> M1=M2R/(M1+M2)]k and M2=-M1R/(M1+M2)]k

Now I have some questions on the next part of my book's solution...
It says, we are to find the moment of inetia tensor for a sphere and apply the displaced axes theorem. So we do this and it comes out as I3(sphere a) = 2/5Mi/r^2, where r is the radius of the sphere.

=> I3=2/5(M1a^2+M2b^2)*

And by symmetry we can see that I1=I2, and we apply the displaced axes theorem.

(Theorem: If we know the moment of inertia tensor using the center of mass as the origin (Icm), and we wish to know the moment of inertia of a tensor about the origin displaced by a, a constant vector, it is given by: Ia=Icm+M(a^2δαβ -aαaβ)

It then deduces that: I1=I2=2/5(M1a^2)+M1(M2R/(M1+M2))^2+2/5(M2a^2)+M2(M1R/(M1+M2))^2

My questions:

1) I3=I3(sphere a) + I3(sphere b) from * ; just wondering what the justification is, as 2/5Mi/r^2 is I for a sphere through it's centre of mass, which is any possible axes passing through the centre of axe, I believe, due to the symmetry of the body. So justifying why we can simply add these two, is because I3 passes through the centre of BOTH bodies.

2) I'm not really understanding the application of the displaced axes theorem. I believe it is being applied to I3. And I don't really understand the part of the theorem ' If we know the moment of inertia tensor using the center of mass as the origin and wish to know the moment of inertia of a tensor about the origin displaced by a, a constant vector, it is given by..'. As applied to I1/I2 from I3 because , as daft as it sounds, don't I1,I2,I3 share the same origin - the centre of mass, as defined in the beginning.
So perhaps I am going wrong in the way that we need to apply this theorem to the spheres in turn, starting with an arbitrary axis through the centre of the sphere (anything due to the symmetry) and ending with I2/I3?

Many many thanks to anyone who can shed some light on this, greatly appreciated :) !
 

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binbagsss said:
It then deduces that: I1=I2=2/5(M1a^2)+M1(M2R/(M1+M2))^2+2/5(M2a^2)+M2(M1R/(M1+M2))^2
2/5(M1a2)+M1(M2R/(M1+M2))2+2/5(M2b2)+M2(M1R/(M1+M2))2 I hope
So justifying why we can simply add these two, is because I3 passes through the centre of BOTH bodies.
Yes. Remember that the M of I is the integral ∫r2.dm where r is the distance of the mass element from the axis. Clearly the integral for the whole structure is just the sum of the integrals for the two spheres taken separately.
2) I'm not really understanding the application of the displaced axes theorem. I believe it is being applied to I3.
No, it's being applied to axes through the centre of the sphere and parallel to I1 and I2.
I3 is not relevant here.
 

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