Find the Mass Moment of Inertia about the x,y,z axes

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SUMMARY

The discussion focuses on calculating the mass moments of inertia for an assembly of two small spheres connected by a light rigid rod in the x-z plane. The key formulae used include the parallel axis theorem and the mass moment of inertia for a sphere, specifically I = 2/3 * m * r² for each sphere. The final calculations yield Ixx = 3mL², Iyy = 2mL², and Izz = 2mL², with the slender rod's mass being negligible. Participants clarified the distances of the masses from the axes and confirmed that the slender rods do not contribute to the mass moment of inertia.

PREREQUISITES
  • Understanding of mass moment of inertia concepts
  • Familiarity with the parallel axis theorem
  • Knowledge of the mass moment of inertia for spheres
  • Basic coordinate geometry in three dimensions
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  • Learn about the mass moment of inertia for different geometric shapes
  • Explore applications of mass moments of inertia in mechanical systems
  • Practice problems involving mass moments of inertia in three-dimensional space
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Northbysouth
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Homework Statement


The two small spheres of mass m each are connected by the light rigid rod which lies in the x-z plane. Determine the mass moments of inertia of the assembly about the x, y, z axes.

I have attached an image of the question


Homework Equations





The Attempt at a Solution



The answer is given but I don't understand how the answer is reached.

For Ixx The slender bar of length 2L has no mass moment of inertia about x. Then the other two rods each have a mass moment of inertia of 1/3*ml2 or do these cancel out and it's only the spheres which give the system a mass moment of inertia?

I think that, for the spheres, the mass moment of inertia should be:

2*(2/3)mr2 + 2ml2

Is the first portion, 2*(2/3)mr2, 0 because the radius is negligible small?

This would leave me with 2mL2

Any guidance would be appreciated. Thank you.
 

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Northbysouth said:

Homework Statement


The two small spheres of mass m each are connected by the light rigid rod which lies in the x-z plane. Determine the mass moments of inertia of the assembly about the x, y, z axes.

I have attached an image of the question

Homework Equations



The Attempt at a Solution



The answer is given but I don't understand how the answer is reached.

For Ixx The slender bar of length 2L has no mass moment of inertia about x. Then the other two rods each have a mass moment of inertia of 1/3*ml2 or do these cancel out and it's only the spheres which give the system a mass moment of inertia?

I think that, for the spheres, the mass moment of inertia should be:

2*(2/3)mr2 + 2ml2

Is the first portion, 2*(2/3)mr2, 0 because the radius is negligible small?

This would leave me with 2mL2

Any guidance would be appreciated. Thank you.
Anyone helping you will want a nice size image of that.

attachment.php?attachmentid=58355&d=1367294931.png



How far is each mass from each of the three possible axes of rotation?

.
 
So, I'm only really using the parallel axis theorem?

For the mass moment of inertia about the x axis, the centers of the slender rods perpendicular to the x-axis are a distance L/2 away and the spheres are each a distance L away.

But summing these would give me 3mL^2
 
Northbysouth said:
So, I'm only really using the parallel axis theorem?

For the mass moment of inertia about the x axis, the centers of the slender rods perpendicular to the x-axis are a distance L/2 away and the spheres are each a distance L away.

But summing these would give me 3mL^2
Light rods usually implies that you can ignore the mass of the rods. After all, that mass is not given.
 
Ahh, that makes sense now. I think I misread the question, because I was thinking that the slender rods also had mass m.

But, then isn't it the same case for Iyy? The spheres are still a distance L from the y axis.
 
Northbysouth said:
Ahh, that makes sense now. I think I misread the question, because I was thinking that the slender rods also had mass m.

But, then isn't it the same case for Iyy? The spheres are still a distance L from the y axis.

Not for the y-axis, look again.

What are the coordinates of the two masses?
 
On the x-z plane, the spheres have coordinates (L, L) and (-L, -L). Should I use Pythagoras here? Or am I using mL^2 twice for each sphere to account for the x and z components?
 
Northbysouth said:
On the x-z plane, the spheres have coordinates (L, L) and (-L, -L). Should I use Pythagoras here? Or am I using mL^2 twice for each sphere to account for the x and z components?
Use Pythagoras.

I would write the coordinates as ordered triples, i.e. (L, 0, L) and (-L, 0, -L) .
 

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