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In Reply to: Re: Effective Mass and Moment of Inertia (again) posted by John Elison on November 18, 2005 at 11:42:48:
So that we don't have to continually solve integrals we'll use an algebraic conversion. For a body with uniform weight distribution per unit length and negligible width in proportion to its length, we can write dM = M/l dxSo ∫x 2 dM becomes ∫x 2 M/l dx. This solves to M.l 2 /12. This is the moment of inertia around the centre of mass, I cm .
The moment of inertia for any body around any point is I cm + M.x 2 where x is the displacement between the centre of mass of the body and the pivot in question.
Thus for the first example given
Headshell 7g x (40 2 /12 + 220 2 ) = 339733
Armtube 15g x (180 2 /12 + 110 2 ) = 222000
Bearing 60g x 40 2 /12 = 8000
Counterweight 150g x 21.27 2 = 67862Total = 637594, divided by 225 2 gives 637594/50625 = 12.5.
The approximation is probably not really valid for the pivot structure which does have significant width in porportion to length but it is such a small part of the sum it doesn't really matter. Similarly considering the counterweight as a point mass is an approximation but it also has little effect unless the counterweight is very close to the pivot.
Mark Kelly
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Follow Ups
- OK , here goes. - Mark Kelly 11/18/0514:35:57 11/18/05 (2)
- Now I understand.... - John Elison 18:13:34 11/18/05 (0)
- On that day of Engineering school . . . - jlep 17:50:13 11/18/05 (0)