Calculating Moment of Inertia for an Asymmetrical Barbell Shape

In summary, the conversation discusses finding the moment of inertia for an asymmetrical barbell shape with masses at each end. The shape of the masses is clarified to be regular right octagonal prisms with a rotational symmetry around the boom axis. Various methods for calculating the moment of inertia are suggested, including using the shape of the actual masses or approximating them as point masses or solid cylinders. The option of building the structure in a design program to compute the moment of inertia is also mentioned.
  • #1
iceburn182
3
0
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.
 
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  • #2
Originally posted by 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.

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
Thanks for the reply,

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
Originally posted by iceburn182
The two halves are octogonal prisms with height 9" and a distance of 18" vertex to vertex.

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


However, the Icm will remain blanced for all angles about the axis of lowest inertia (that running lengthwise through the boom).

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?

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.

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
Is the thing built already?

If you can't calculate it, you should be able to empirically determine it (assuming it isn't too delicate).
 
  • #6
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?


correct on both of your comments, Ambitwistor

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
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
Nearly as easy and even more accurate is to approximate them as solid cylinders ... but an exact solution probably isn't much harder.
 
  • #9
You know, I think you can actually build the thing in Pro-E and have it compute the moment of inertia for you.
 

1. What is moment of inertia and why is it important in science?

Moment of inertia is a measure of an object's resistance to rotational motion. It is important in science because it helps us understand how objects behave when they are in motion and how they respond to external forces.

2. How is moment of inertia calculated?

The moment of inertia of an object can be calculated by multiplying the mass of the object by the square of its distance from the axis of rotation. This can be represented by the equation I = mr^2, where I is the moment of inertia, m is the mass, and r is the distance from the axis of rotation.

3. What factors affect the moment of inertia of an object?

The moment of inertia of an object depends on its mass, shape, and distribution of mass. Objects with a larger mass or a greater distance from the axis of rotation will have a higher moment of inertia. Objects with a more compact or symmetrical shape will have a lower moment of inertia.

4. How does moment of inertia relate to rotational energy?

Moment of inertia is directly related to an object's rotational energy. As the moment of inertia increases, the rotational energy also increases. This means that objects with a higher moment of inertia will require more energy to rotate at the same speed as objects with a lower moment of inertia.

5. How is moment of inertia used in real-life applications?

Moment of inertia has various applications in real life, such as in calculating the stability of bridges and buildings, designing vehicles and machinery, and understanding the motion of objects in space. It is also used in sports, such as determining the moment of inertia of a tennis racket to improve its performance.

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