How Do You Calculate Moment of Inertia for Complex Shapes?

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Homework Help Overview

The discussion revolves around calculating the moment of inertia for complex shapes, specifically focusing on the contributions from different segments of the shape. Participants are examining the correct application of formulas and the implications of mass distribution.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the moment of inertia for various segments of a shape, questioning the application of the parallel axis theorem and the correct formulas for rods. There is an exploration of mass distribution and unit conversion issues.

Discussion Status

The conversation is ongoing, with participants providing insights into the calculations and formulas used. Some guidance has been offered regarding unit conversions and the correct interpretation of mass for the components involved.

Contextual Notes

There are indications of confusion regarding the mass values and units used in calculations, which may affect the results. Participants are also reflecting on their assumptions and the need for careful attention to detail in their calculations.

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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