Question on Moment of Inertia Tensor of a Rotating Rigid Body

[warhammer]

Homework Statement

A rigid body consists of three thin rigid (assume massless) rods of length 21, welded at their midpoints, in three mutually perpendicular directions, with point masses at their ends. If this object is placed such that the rod along the x axis has masses 3M at the ends, the rods along y-axis have masses M at their ends and the rod along z axis has masses 2M at the ends, calculate the principal moments of inertia.

(See picture attached below)

Relevant Equations

Picture attached below showcasing my attempt at the solution.

Hi.

So I was asked the following question whose picture is attached below along with my attempt at the solution.

Now my doubt is, since the question refers to the whole system comprising of these thin rigid body 'mini systems', should the Principle moments of Inertia about the respective axes be added together to give the solution in context of the 'total' rigid body systems? Or should we leave it by simply calculating the Principle moments of inertia for each body aligned at some axis about all the 3 axes?

 

[BvU]

what does this try to say ?

21 should be $2L$ (and $a$ should be $0$) !

I don't see where the $2\over 5$ comes from ?

$\$

 

[warhammer]

what does this try to say ?

For each pair of masses, the products of inertia turn out to be zero since the alignment is with the coordinate axes.

I don't see where the 2/5 comes from ?

That's the form for moment of inertia of a solid sphere about it's diameter which comes out as 2/5 M*R^2

 

[BvU]

For each pair of masses, the products of inertia turn out to be zero since the alignment is with the coordinate axes.

Ah, 'pair' was hard to read. The off-diagonal elements of the $I$ tensor are zero, yes.

That's the form for moment of inertia of a solid sphere about it's diameter which comes out as 2/5 M*R^2

There are no spheres, only point masses. There is no R.

So that is zero.

should the Principle moments of Inertia about the respective axes be added together

Yes. For the complete rigid object you have to add all three $I_{zz}$. See the definition here .

this last one is $I_zz$

so you get $I_{zz} = 8ML^2$. etc. All in all, not a very hard exercise, but good to go through at least once !$\$

 

[warhammer]

Ah, 'pair' was hard to read. The off-diagonal elements of the $I$ tensor are zero, yes.
View attachment 281476There are no spheres, only point masses. There is no R.So that is zero.

Yes. For the complete rigid object you have to add all three $I_{zz}$. See the definition here .this last one is $I_zz$

so you get $I_{zz} = 8ML^2$. etc. All in all, not a very hard exercise, but good to go through at least once !$\$

Thank you so much for your response. I'm summarising all the mistakes I made just to ensure all holes have been plugged:

1) Rather than spheres those are point masses. So moment of inertia of each is mr^2 where r is the perpendicular distance to rotation axis. Since 3M mass (for instance) is aligned at x-axis so it's I(xx) would be zero. Same would be true for 2M whose I(zz) & M whose I(yy) would also be zero respectively.

2) For the whole body, I need to add all the I(xx) , I(yy) & the I(zz) components together of the 3 rigid body pairs that are specified in figure.
Therefore for the whole the body,
I(xx)=6M*l^2
I(yy)=10M*l^2
I(zz)=8*M*l^2

Hope I've finally got this one right