Moment of inertia formula for a propeller

AI Thread Summary
A moment of inertia formula for a two-bladed propeller can be approximated using the formula for a rod rotating about its center, which is 1/12 ml², where m is mass and l is length. The suitability of this estimate depends on the specific geometry and mass distribution of the propeller in question. For basic physics problems, treating the propeller as a simple rod is often acceptable. However, for more accurate calculations, detailed models from manufacturers that account for the actual mass distribution are preferred. In some discussions, an alternative formula of 2/3 mr² has been mentioned, but its derivation remains unclear.
localrob
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Is there a moment of inertia formula for a 2 bladed propeller?
If you only have the mass and the length of the propeller, I would think a good estimate would be using a rod rotating about the middle. 1/12ml2
 
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localrob said:
Is there a moment of inertia formula for a 2 bladed propeller?
If you only have the mass and the length of the propeller, I would think a good estimate would be using a rod rotating about the middle. 1/12ml2

"Good" depends upon what you want to use the estimate for. Does a simple rod adequately describe the geometry and mass distribution of the propeller you have in mind?
 
I think that for a high school physics problem, it can be assumed to act like a rod. So when finding the rotational kinetic energy of a spinning propeller, I could use the rod formula.
But I was curious if there were another generally accepted formula for a propeller that I could use. I'm not sure what the distribution of mass is for an actual propeller, and nothing other than the mass and length are provided.
 
No doubt manufacturers of propellers will have detailed models of their propellers, including all the geometry and mass distribution. But for a problem where the only information you have is mass and length you're rather constrained to choose the rod as a model.
 
That's how I feel too. I did see a post on these forums where someone found the formula to be 2/3mr^2, but I don't know where that came from.
 
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