How Do You Calculate Moment of Inertia for Complex Shapes?

AI Thread Summary
The discussion revolves around calculating the moment of inertia (MI) for a complex shape, with participants addressing a specific problem. The correct formula for the MI of a rod about its end is highlighted as (ML^2)/3, but there were initial misunderstandings regarding unit conversions and mass distribution. The final answer provided is 3.97×10-3 kg*m^2, derived from the correct application of the formulas and proper unit handling. Participants emphasize the importance of calculating in CGS before converting to SI units for accuracy. The thread concludes with an acknowledgment of the initial confusion and a commitment to careful calculations in the future.
nikkou
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[SOLVED] a tricky Moment of Inertia question, help needed...

The question is in the JPG attachment.
This is what i tried:
Obviously the side which is right on the axis has no moment of inertia so we're left with the 2 horizontal ones and the other vertical. for the two horizontal sides we've got an equation: (M*L^2)/3 and we consider it twice.
the last side is probably where i went wrong - i thought it's simply M*L^2 because it can be regarded as a point mass at distance L but that didn't work. then i tried using the parallel axis theorem but that didn't help too.
Any thoughts? I'm sure its really easy for most of you guys...
Thanx.
 

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The m given in the picture is that of the whole mass -- I hope you have taken that into account. The MI of the vertical side is equal to (m/4)a^2, as you had correctly deduced.

The MI of a rod about its end is ML^2/3, but you have written a different formula.
 
Well, I did use 1/4 of the given mass for each rod. sorry for not mentioning it. and the formula i wrote is the same one you did: (ML^2)/3 . still it doesn't work...
The real answer, which i forgot to give, is: 3.97×10-3 kg*m^2
I never get it...
 
(Sorry, you had written the correct formula. I misread it.)

MI = (m/4)(2a²/3 + a²) = (m/4)(5/3)a² = 39703.125 g*cm², which is your answer.

In this sort of problems, where the values are in CGS, it's better to calculate in CGS and then convert at one stroke to SI.
 
Last edited:
nikkou said:
Well, I did use 1/4 of the given mass for each rod. sorry for not mentioning it. and the formula i wrote is the same one you did: (ML^2)/3 . still it doesn't work...
The real answer, which i forgot to give, is: 3.97×10-3 kg*m^2
I never get it...

What value are you using for the mass of each rod, in kg? And for the length of each rod in m? Perhaps you are just messing the units up. Post a solution showing how you plugged the numbers in.
 
man, that's embarrassing... It's such a simple question, I just messed up the units.
please excuse my inattention, I promise not to be so hasty next time and make sure i really need help with something substantial before posting :-)
Thank you very very much!
 
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