# Moment of Inertia of question

1. Nov 27, 2003

### iceburn182

I am trying to find the moment of inertia for an assymetrical barbell shape. The mass as one end of the 2 meter boom is 9kg and at the other is 5.59kg. The boom or rod is a balloon-like tube and I am going to assume it is massless. I am using this information to calculate thrust impulses for a satellite. Can you help? Thanks.

Last edited: Nov 27, 2003
2. Nov 27, 2003

### Ambitwistor

What shape are the masses? Are they spherical? Solid?

If they're symmetric about axes that are perpendicular to the rod, then you can find their individual moments of inertia, and then use the parallel-axis theorem to find the moment of inertia of the combined system about a single axis that is parallel to those two (also perpendicular to the rod), but passes through the center of mass of the system.

3. Dec 2, 2003

### iceburn182

The two halves are octogonal prisms with height 9" and a distance of 18" vertex to vertex. We are assuming for now that the mass is evenly distributed about the volume, because hardware hasn't been placed within it. However, the Icm will remain blanced for all angles about the axis of lowest inertia (that running lengthwise through the boom).

I could use the shape of a point masses or spherical masses separated by a massless rod, but I was curious about doing it as the actual shape. Any comments?

4. Dec 2, 2003

### Ambitwistor

Regular right octagonal prisms? (i.e., a volume swept out by vertically translating a regular octagon?)

Do you mean, the boom runs through the symmetry axes of the prisms, so the entire structure has a (eightfold) rotational symmetry about the boom axis?

It's certainly feasible to just calculate the moment of inertia integral for that shape, if I understand you correctly. You could probably just look up what it is for a triangular prism of the appropriate size (maybe applying the parallel axis theorem if it gives the moment about the central axis of the prism rather than one of the three edges), and multiply that by eight.

5. Dec 3, 2003

### enigma

Staff Emeritus

If you cant calculate it, you should be able to empirically determine it (assuming it isn't too delicate).

6. Dec 3, 2003

### iceburn182

I will look into your suggestions as soon as I have time to get back to this. (School is busy right now)

enigma,

The satellite isn't built yet - still on the design board. In the future, we will find its inertial properties with a rigid boom and placed components.

thanks, for the ideas. If you have any more I'll check back.

7. Dec 3, 2003

### HallsofIvy

The simplest thing to do- and it will probably be pretty accurate- is to assume that the two end masses are "point masses". That is, that their mass is concentrated at a point at the end of the barbell.

Let "x" be the distance from the 9 kg mass end to the center of gravity. Then there is a "torque" (twisting force) around the center of gravity of 9gx (g is the acceleration of gravity so that 9g is the weight). Since the boom has length two meters, the distance from the other, 5.59 kg mass, at the other end, to the center of gravity is 2- x meters. The torque around the center of gravity due to that weight is 5.59g(2-x). In order that the two "balance" (which is the whole point of "center of gravity"), we must have 5.59g(2-x)= 9gx. Of course, the "g"s cancel and we have 11.18- 5.59x= 9x so the equation is 14.59x= 11.18 or x= .77. The center of gravity is approximately 0.77 meters from the heavier 9 kg end.

8. Dec 3, 2003

### Ambitwistor

Nearly as easy and even more accurate is to approximate them as solid cylinders ... but an exact solution probably isn't much harder.

9. Dec 3, 2003

### enigma

Staff Emeritus
You know, I think you can actually build the thing in Pro-E and have it compute the moment of inertia for you.